CHEM 121 - GENERAL CHEMISTRY I - Use of
Amount of Substance on Volumes of Gases
Question Bank
Question 1
A sample of gas occupies a volume of 2.50 L at a pressure of 1.50 atm
and a temperature of 25
°
C. If the pressure is increased to 2.00 atm and the
temperature is raised to 35
°
C, what will be the new volume of the gas? Assume
that the amount of gas remains constant.
Solution:
Given: Initial volume, V1= 2.50 L
Initial pressure, P1= 1.50 atm
Initial temperature, T1= 25◦C = 25 + 273 = 298 K
Final pressure, P2= 2.00 atm
Final temperature, T2= 35◦C = 35 + 273 = 308 K
We can use the combined gas law to find the new volume:
P1·V1
T1
=P2·V2
T2
Solving for V2:
V2=P1·V1·T2
P2·T1
Substitute the given values:
V2=1.50 atm ×2.50 L ×308 K
2.00 atm ×298 K
V2=3.525 atm ·L·K
2.00 atm = 5.325 L
Therefore, the new volume of the gas will be 5.325 L.Question 1:
A sample of gas occupies a volume of 2.50 L at a pressure of 1.50
atm and a temperature of 25
°
C. If the pressure is increased to 2.00
atm and the temperature is raised to 35
°
C, what will be the new
volume of the gas? Assume that the amount of gas remains constant.
Solution:
1
Given: Initial volume, V1= 2.50 L
Initial pressure, P1= 1.50 atm
Initial temperature, T1= 25◦C= 25 + 273 = 298 K
Final pressure, P2= 2.00 atm
Final temperature, T2= 35◦C= 35 + 273 = 308 K
We can use the combined gas law to find the new volume:
P1·V1
T1
=P2·V2
T2
Solving for V2:
V2=P1·V1·T2
P2·T1
Substitute the given values:
V2=1.50 atm ×2.50 L×308 K
2.00 atm ×298 K
V2=3.525 atm ·L·K
2.00 atm = 5.325 L
Therefore, the new volume of the gas will be 5.325 L.
Question 2
A gas sample has a volume of 2.50 L at a temperature of 25
°
C and
a pressure of 1.20 atm. If the pressure is increased to 1.50 atm while
keeping the temperature constant, what will be the new volume of
the gas?
Step-by-step Solution:
Given: Initial volume, Vinitial = 2.50 L Initial pressure, Pinitial =
1.20 atm Final pressure, Pfinal = 1.50 atm
Since the temperature is constant, we can use the combined gas
law equation: Pinitial ×Vinitial
Tinitial
=Pfinal ×Vfinal
Tfinal
Since the temperature is constant, we can cancel out the temper-
atures from both sides of the equation:
Pinitial ×Vinitial =Pfinal ×Vfinal
We can now plug in the values to find the new volume:
1.20 atm ×2.50 L= 1.50 atm ×Vfinal
3.00 = 1.50 ×Vfinal
2
Vfinal =3.00
1.50
Vfinal = 2.00 L
Therefore, the new volume of the gas will be 2.00 L when the
pressure is increased to 1.50 atm while keeping the temperature con-
stant.Question 2:
A gas sample has a volume of 2.50 L at a temperature of 25
°
C and
a pressure of 1.20 atm. If the pressure is increased to 1.50 atm while
keeping the temperature constant, what will be the new volume of
the gas?
Step-by-step Solution:
Given: Initial volume, Vinitial = 2.50 L Initial pressure, Pinitial =
1.20 atm Final pressure, Pfinal = 1.50 atm
Since the temperature is constant, we can use the combined gas
law equation: Pinitial ×Vinitial
Tinitial
=Pfinal ×Vfinal
Tfinal
Since the temperature is constant, we can cancel out the temper-
atures from both sides of the equation:
Pinitial ×Vinitial =Pfinal ×Vfinal
We can now plug in the values to find the new volume:
1.20 atm ×2.50 L= 1.50 atm ×Vfinal
3.00 = 1.50 ×Vfinal
Vfinal =3.00
1.50
Vfinal = 2.00 L
Therefore, the new volume of the gas will be 2.00 L when the pres-
sure is increased to 1.50 atm while keeping the temperature constant.
Question 3
A gas occupies a volume of 4.0 L at a temperature of 300 K and a
pressure of 1.2 atm. Calculate the amount of the gas in moles.
Step-by-step solution:
Given: Volume of gas, V= 4.0L, Temperature, T= 300 K, Pres-
sure, P= 1.2atm.
Using the ideal gas law equation:
P V =nRT
3
Where: n= amount of gas in moles, R= 0.0821 L·atm/mol ·K
(universal gas constant).
Plug in the given values:
(1.2atm)(4.0L) = n(0.0821 L·atm/mol ·K)(300 K)
4.8 = 24.63n
n=4.8
24.63
n≈0.195 mol
Therefore, the amount of the gas in moles is approximately 0.195
mol.Question 3:
A gas occupies a volume of 4.0 L at a temperature of 300 K and a
pressure of 1.2 atm. Calculate the amount of the gas in moles.
Step-by-step solution:
Given: Volume of gas, V= 4.0L, Temperature, T= 300 K, Pres-
sure, P= 1.2atm.
Using the ideal gas law equation:
P V =nRT
Where: n= amount of gas in moles, R= 0.0821 L·atm/mol ·K
(universal gas constant).
Plug in the given values:
(1.2atm)(4.0L) = n(0.0821 L·atm/mol ·K)(300 K)
4.8 = 24.63n
n=4.8
24.63
n≈0.195 mol
Therefore, the amount of the gas in moles is approximately 0.195
mol.
4
Question 4
A certain amount of gas occupies a volume of 500 mL at a pressure
of 1 atm and a temperature of 300 K. If the pressure is increased to 2
atm and the temperature is raised to 400 K, what is the new volume
of the gas?
Step-by-step solution: 1. Determine the initial amount of gas
using the ideal gas law: P V =nRT , where Pis the pressure, Vis the
volume, nis the amount of gas in moles, Ris the gas constant, and T
is the temperature in Kelvin. 2. Rearrange the ideal gas law to solve
for n:n=P V
RT . 3. Calculate the initial amount of gas when pressure is
1 atm and temperature is 300 K. 4. Use the given conditions of the
pressure being increased to 2 atm and the temperature raised to 400
K to calculate the new volume of the gas using the ideal gas law. 5.
Substitute the new pressure, temperature, and the initial amount of
gas into the ideal gas law and solve for the new volume. 6. Present
the final answer with appropriate units.Question 4:
A certain amount of gas occupies a volume of 500 mL at a pressure
of 1 atm and a temperature of 300 K. If the pressure is increased to 2
atm and the temperature is raised to 400 K, what is the new volume
of the gas?
Step-by-step solution: 1. Determine the initial amount of gas
using the ideal gas law: P V =nRT , where Pis the pressure, Vis the
volume, nis the amount of gas in moles, Ris the gas constant, and T
is the temperature in Kelvin. 2. Rearrange the ideal gas law to solve
for n:n=P V
RT . 3. Calculate the initial amount of gas when pressure is
1 atm and temperature is 300 K. 4. Use the given conditions of the
pressure being increased to 2 atm and the temperature raised to 400
K to calculate the new volume of the gas using the ideal gas law. 5.
Substitute the new pressure, temperature, and the initial amount of
gas into the ideal gas law and solve for the new volume. 6. Present
the final answer with appropriate units.
Question 5
Given that a certain amount of gas occupies a volume of 2.5 L at a
temperature of 25
°
C and a pressure of 1 atm, determine the volume
this gas will occupy if the temperature is increased to 75
°
C and the
pressure is increased to 2 atm.
Step-by-step solution:
1. Write down the given values: Initial volume, V1= 2.5L Initial
temperature, T1= 25C= 25 + 273.15 K Initial pressure, P1= 1 atm
2. Use the ideal gas law equation: P V =nRT , where Pis pressure,
Vis volume, nis the amount of substance, Ris the ideal gas constant,
and Tis temperature in Kelvin.
5
3. Calculate the initial amount of substance: n=P1V1
RT1
4. Determine the final volume using the ideal gas law and the
calculated amount of substance: V2=nRT2
P2, where Final temperature,
T2= 75C= 75 + 273.15 K Final pressure, P2= 2 atm
5. Substitute the values into the equation to find the final volume
occupied by the gas.
6. Calculate the final volume to find the answer.Question 5:
Given that a certain amount of gas occupies a volume of 2.5 L at a
temperature of 25
°
C and a pressure of 1 atm, determine the volume
this gas will occupy if the temperature is increased to 75
°
C and the
pressure is increased to 2 atm.
Step-by-step solution:
1. Write down the given values: Initial volume, V1= 2.5L Initial
temperature, T1= 25C= 25 + 273.15 K Initial pressure, P1= 1 atm
2. Use the ideal gas law equation: P V =nRT , where Pis pressure,
Vis volume, nis the amount of substance, Ris the ideal gas constant,
and Tis temperature in Kelvin.
3. Calculate the initial amount of substance: n=P1V1
RT1
4. Determine the final volume using the ideal gas law and the
calculated amount of substance: V2=nRT2
P2, where Final temperature,
T2= 75C= 75 + 273.15 K Final pressure, P2= 2 atm
5. Substitute the values into the equation to find the final volume
occupied by the gas.
6. Calculate the final volume to find the answer.
Question 6
Step-by-step Solution: Given: Volume of gas, V= 2.50LTemper-
ature, T= 25C= 298K(Convert Celsius to Kelvin by adding 273)
Pressure, P= 1.00atm
We can use the ideal gas law equation to find the number of moles
of gas:
P V =nRT
where: P= pressure V= volume n= number of moles R= ideal
gas constant (0.0821L·atm/(mol ·K))T= temperature in Kelvin
First, convert the temperature from Celsius to Kelvin: T= 25C+
273 = 298K
Now, plug in the values into the ideal gas law equation: 1.00atm ×
2.50L=n×0.0821L·atm/(mol ·K)×298K
Solving for n:2.50atm·L= 0.0821×298×n2.50 = 24.4678×n n =2.50
24.4678
n= 0.1021 moles
Therefore, the number of moles of gas present in the sample is
0.1021 moles.Question 6: A gas sample has a volume of 2.50 L at
a temperature of 25
°
C and a pressure of 1.00 atm. Calculate the
number of moles of gas present in the sample.
6
Step-by-step Solution: Given: Volume of gas, V= 2.50LTemper-
ature, T= 25C= 298K(Convert Celsius to Kelvin by adding 273)
Pressure, P= 1.00atm
We can use the ideal gas law equation to find the number of moles
of gas:
P V =nRT
where: P= pressure V= volume n= number of moles R= ideal
gas constant (0.0821L·atm/(mol ·K))T= temperature in Kelvin
First, convert the temperature from Celsius to Kelvin: T= 25C+
273 = 298K
Now, plug in the values into the ideal gas law equation: 1.00atm ×
2.50L=n×0.0821L·atm/(mol ·K)×298K
Solving for n:2.50atm·L= 0.0821×298×n2.50 = 24.4678×n n =2.50
24.4678
n= 0.1021 moles
Therefore, the number of moles of gas present in the sample is
0.1021 moles.
Question 7
A gas sample has a volume of 2.50 L at a temperature of 300 K and
a pressure of 1.50 atm. If the number of moles of the gas is 0.075 mol,
calculate the molar volume of the gas at STP (standard temperature
and pressure).
Step-by-step solution:
Given: - Volume of gas (V) = 2.50 L - Temperature (T) = 300 K
- Pressure (P) = 1.50 atm - Number of moles of gas (n) = 0.075 mol
To calculate the molar volume of the gas at STP, we first need to
convert the given conditions to STP conditions.
STP conditions: - Temperature (TSTP) = 273 K - Pressure (PSTP )
= 1 atm - Number of moles at STP (nSTP ) = 1 mol
We can use the combined gas law to find the molar volume at STP:
P·V
n·T=PSTP ·VSTP
nSTP ·TSTP
Given:
P= 1.50 atm, V = 2.50 L, n = 0.075 mol, PSTP = 1 atm, T = 300 K, TSTP = 273 K, nSTP = 1 mol
Substitute the values into the equation and solve for VSTP:
VSTP =P·V·nSTP ·TSTP
n·PSTP
VSTP =1.50 atm ×2.50 L×1mol ×273 K
0.075 mol ×1atm
7
VSTP =1.50 ×2.50 ×273
0.075
VSTP = 135.0L
Therefore, the molar volume of the gas at STP is 135.0 L.Question
7:
A gas sample has a volume of 2.50 L at a temperature of 300 K and
a pressure of 1.50 atm. If the number of moles of the gas is 0.075 mol,
calculate the molar volume of the gas at STP (standard temperature
and pressure).
Step-by-step solution:
Given: - Volume of gas (V) = 2.50 L - Temperature (T) = 300 K
- Pressure (P) = 1.50 atm - Number of moles of gas (n) = 0.075 mol
To calculate the molar volume of the gas at STP, we first need to
convert the given conditions to STP conditions.
STP conditions: - Temperature (TSTP) = 273 K - Pressure (PSTP )
= 1 atm - Number of moles at STP (nSTP ) = 1 mol
We can use the combined gas law to find the molar volume at STP:
P·V
n·T=PSTP ·VSTP
nSTP ·TSTP
Given:
P= 1.50 atm, V = 2.50 L, n = 0.075 mol, PSTP = 1 atm, T = 300 K, TSTP = 273 K, nSTP = 1 mol
Substitute the values into the equation and solve for VSTP:
VSTP =P·V·nSTP ·TSTP
n·PSTP
VSTP =1.50 atm ×2.50 L×1mol ×273 K
0.075 mol ×1atm
VSTP =1.50 ×2.50 ×273
0.075
VSTP = 135.0L
Therefore, the molar volume of the gas at STP is 135.0 L.
8
Question 8
A sample of helium gas occupies a volume of 2.00 L at a temper-
ature of 300 K and a pressure of 2.50 atm. Calculate the amount of
substance (in moles) of the helium gas present in the sample.
Step-by-step solution:
Given data: - Volume (V) = 2.00 L - Temperature (T)=300K-
Pressure (P) = 2.50 atm
We can use the ideal gas law equation to find the amount of sub-
stance (moles) of the gas present:
P V =nRT
where: - Pis the pressure - Vis the volume - nis the amount of
substance (in moles) - Ris the gas constant (0.0821 L atm/mol K) -
Tis the temperature
Substitute the given values into the equation:
2.50 atm ×2.00 L=n×0.0821 L atm/mol K ×300 K
5.00 = 24.63n
n=5.00
24.63
n≈0.203 moles
Therefore, the amount of substance of the helium gas present in
the sample is approximately 0.203 moles.Question 8:
A sample of helium gas occupies a volume of 2.00 L at a temper-
ature of 300 K and a pressure of 2.50 atm. Calculate the amount of
substance (in moles) of the helium gas present in the sample.
Step-by-step solution:
Given data: - Volume (V) = 2.00 L - Temperature (T)=300K-
Pressure (P) = 2.50 atm
We can use the ideal gas law equation to find the amount of sub-
stance (moles) of the gas present:
P V =nRT
where: - Pis the pressure - Vis the volume - nis the amount of
substance (in moles) - Ris the gas constant (0.0821 L atm/mol K) -
Tis the temperature
Substitute the given values into the equation:
2.50 atm ×2.00 L=n×0.0821 L atm/mol K ×300 K
9
5.00 = 24.63n
n=5.00
24.63
n≈0.203 moles
Therefore, the amount of substance of the helium gas present in
the sample is approximately 0.203 moles.
Question 9
A gas sample has a volume of 2.50 L at a temperature of 25
°
C and
a pressure of 1.20 atm. If the pressure is increased to 2.00 atm while
keeping the temperature constant, what will be the new volume of
the gas sample?
Step-by-step Solution:
Given: Initial volume, Vi= 2.50 L Initial pressure, Pi= 1.20 atm
Final pressure, Pf= 2.00 atm Temperature is constant
According to the Boyle’s Law, which states that for a given amount
of gas at constant temperature, the pressure and volume are inversely
proportional:
Pi×Vi=Pf×Vf
Substitute the given values into the equation:
1.20 ×2.50 = 2.00 ×Vf
Solve for Vf:
3.00 = 2.00 ×Vf
Vf=3.00
2.00
Vf= 1.50 L
Therefore, the new volume of the gas sample will be 1.50 L when
the pressure is increased to 2.00 atm while keeping the temperature
constant.Question 9:
A gas sample has a volume of 2.50 L at a temperature of 25
°
C and
a pressure of 1.20 atm. If the pressure is increased to 2.00 atm while
keeping the temperature constant, what will be the new volume of
the gas sample?
Step-by-step Solution:
10
Given: Initial volume, Vi= 2.50 L Initial pressure, Pi= 1.20 atm
Final pressure, Pf= 2.00 atm Temperature is constant
According to the Boyle’s Law, which states that for a given amount
of gas at constant temperature, the pressure and volume are inversely
proportional:
Pi×Vi=Pf×Vf
Substitute the given values into the equation:
1.20 ×2.50 = 2.00 ×Vf
Solve for Vf:
3.00 = 2.00 ×Vf
Vf=3.00
2.00
Vf= 1.50 L
Therefore, the new volume of the gas sample will be 1.50 L when
the pressure is increased to 2.00 atm while keeping the temperature
constant.
Question 10
A gas occupies a volume of 4.50 L at a temperature of 27
°
C and
a pressure of 0.900 atm. If the pressure is increased to 1.20 atm and
the temperature is decreased to 20
°
C, what will be the new volume
of the gas? Assume the amount of substance remains constant.
Step-by-step solution:
Given: Initial volume, V1= 4.50 L Initial temperature, T1= 27C=
27 + 273 = 300 K Initial pressure, P1= 0.900 atm Final pressure, P2=
1.20 atm Final temperature, T2= 20C= 20 + 273 = 293 K
According to the combined gas law equation:
P1·V1
T1
=P2·V2
T2
Substitute the given values into the equation and solve for V2:
0.900 ·4.50
300 =1.20 ·V2
293
V2=0.900 ·4.50 ·293
1.20 ·300
V2= 4.60 L
11
Therefore, the new volume of the gas will be 4.60 L.Question 10:
A gas occupies a volume of 4.50 L at a temperature of 27
°
C and
a pressure of 0.900 atm. If the pressure is increased to 1.20 atm and
the temperature is decreased to 20
°
C, what will be the new volume
of the gas? Assume the amount of substance remains constant.
Step-by-step solution:
Given: Initial volume, V1= 4.50 L Initial temperature, T1= 27C=
27 + 273 = 300 K Initial pressure, P1= 0.900 atm Final pressure, P2=
1.20 atm Final temperature, T2= 20C= 20 + 273 = 293 K
According to the combined gas law equation:
P1·V1
T1
=P2·V2
T2
Substitute the given values into the equation and solve for V2:
0.900 ·4.50
300 =1.20 ·V2
293
V2=0.900 ·4.50 ·293
1.20 ·300
V2= 4.60 L
Therefore, the new volume of the gas will be 4.60 L.
Question 11
A 2.50 L container has 0.500 mol of helium gas at a certain tem-
perature and pressure. If the pressure is increased while the temper-
ature remains constant, what will be the new volume of the gas if the
amount of substance is kept constant?
Step-by-step solution: 1. The ideal gas law is given by the equation
P V =nRT , where: - Pis the pressure of the gas, - Vis the volume of
the gas, - nis the amount of substance in moles, - Ris the ideal gas
constant (0.0821 L.atm/mol.K), - Tis the temperature in Kelvin.
2. Initially, the helium gas is in a 2.50 L container with 0.500 mol
of helium. The pressure and temperature are constant.
3. When the pressure is increased, the amount of substance (moles
of gas) and temperature are kept constant.
4. Since P V =nRT and nis constant, we can rearrange the equa-
tion to solve for the new volume Vnew:
PinitialVinitial =Pf inalVnew
5. Substitute the initial values:
Pinitial =Pfinal ×Vnew
Vinitial
12
6. The new pressure is higher than the initial pressure, so we have
Pfinal > Pinitial, meaning the volume will decrease.
7. To find the new volume, use the ratio of the pressures:
Pfinal
Pinitial
=Vinitial
Vnew
8. Substitute the values:
Pfinal
Pinitial
=Vinitial
Vnew
Pfinal
Pinitial
=2.50 L
Vnew
9. If the pressure is doubled, then Pf inal
Pinitial = 2. Substitute this value
into the equation and solve for Vnew:
2 = 2.50 L
Vnew
Vnew =2.50 L
2
Vnew = 1.25 L
The new volume of the helium gas when the pressure is doubled
while keeping the temperature and amount of substance constant will
be 1.25 L.Question 11:
A 2.50 L container has 0.500 mol of helium gas at a certain tem-
perature and pressure. If the pressure is increased while the temper-
ature remains constant, what will be the new volume of the gas if the
amount of substance is kept constant?
Step-by-step solution: 1. The ideal gas law is given by the equation
P V =nRT , where: - Pis the pressure of the gas, - Vis the volume of
the gas, - nis the amount of substance in moles, - Ris the ideal gas
constant (0.0821 L.atm/mol.K), - Tis the temperature in Kelvin.
2. Initially, the helium gas is in a 2.50 L container with 0.500 mol
of helium. The pressure and temperature are constant.
3. When the pressure is increased, the amount of substance (moles
of gas) and temperature are kept constant.
4. Since P V =nRT and nis constant, we can rearrange the equa-
tion to solve for the new volume Vnew:
PinitialVinitial =Pf inalVnew
5. Substitute the initial values:
Pinitial =Pfinal ×Vnew
Vinitial
13
6. The new pressure is higher than the initial pressure, so we have
Pfinal > Pinitial, meaning the volume will decrease.
7. To find the new volume, use the ratio of the pressures:
Pfinal
Pinitial
=Vinitial
Vnew
8. Substitute the values:
Pfinal
Pinitial
=Vinitial
Vnew
Pfinal
Pinitial
=2.50 L
Vnew
9. If the pressure is doubled, then Pf inal
Pinitial = 2. Substitute this value
into the equation and solve for Vnew:
2 = 2.50 L
Vnew
Vnew =2.50 L
2
Vnew = 1.25 L
The new volume of the helium gas when the pressure is doubled
while keeping the temperature and amount of substance constant will
be 1.25 L.
Question 12
Step-by-step solution: Given: Initial volume, V1= 2.50 L Initial
pressure, P1= 1.50 atm Final pressure, P2= 3.00 atm Initial tempera-
ture, T1= 25C
Since the temperature remains constant, the change in pressure
and volume can be related by Boyle’s Law: P1V1=P2V2
Solving for the final volume, V2:
P1V1=P2V2
2.50 ×1.50 = 3.00 ×V2
3.75 = 3.00V2
V2=3.75
3.00
V2= 1.25 L
Therefore, the new volume of the gas will be 1.25 L when the pres-
sure is increased to 3.00 atm.Question 12: A gas occupies a volume
of 2.50 L at a pressure of 1.50 atm and a temperature of 25
°
C. If
14
the pressure is increased to 3.00 atm while the temperature remains
constant, what will be the new volume of the gas?
Step-by-step solution: Given: Initial volume, V1= 2.50 L Initial
pressure, P1= 1.50 atm Final pressure, P2= 3.00 atm Initial tempera-
ture, T1= 25C
Since the temperature remains constant, the change in pressure
and volume can be related by Boyle’s Law: P1V1=P2V2
Solving for the final volume, V2:
P1V1=P2V2
2.50 ×1.50 = 3.00 ×V2
3.75 = 3.00V2
V2=3.75
3.00
V2= 1.25 L
Therefore, the new volume of the gas will be 1.25 L when the
pressure is increased to 3.00 atm.
Question 13
A gas sample occupying 2.50 L at 1.00 atm and 25.0
°
C is heated
to 125
°
C at constant pressure. What is the new volume of the gas
sample?
Step-by-step Solution:
Given: Initial volume (V1)=2.50 L Initial pressure (P1)=1.00 atm
Initial temperature (T1) = 25.0C= 298.15 K
Final temperature (T2) = 125C= 398.15 K
Using the ideal gas law P V =nRT , where nis the amount of
substance and Ris the gas constant, we can rearrange the formula to
solve for the final volume V2:
P1V1
T1
=P2V2
T2
Substitute the given values into the equation:
1.00atm ×2.50L
298.15K =1.00atm ×V2
398.15K
Solve for V2:
V2=1.00 atm ×2.50 L×398.15 K
298.15 K
V2≈3.34 L
15
Therefore, the new volume of the gas sample is approximately 3.34
L.Question 13:
A gas sample occupying 2.50 L at 1.00 atm and 25.0
°
C is heated
to 125
°
C at constant pressure. What is the new volume of the gas
sample?
Step-by-step Solution:
Given: Initial volume (V1)=2.50 L Initial pressure (P1)=1.00 atm
Initial temperature (T1) = 25.0C= 298.15 K
Final temperature (T2) = 125C= 398.15 K
Using the ideal gas law P V =nRT , where nis the amount of
substance and Ris the gas constant, we can rearrange the formula to
solve for the final volume V2:
P1V1
T1
=P2V2
T2
Substitute the given values into the equation:
1.00atm ×2.50L
298.15K =1.00atm ×V2
398.15K
Solve for V2:
V2=1.00 atm ×2.50 L×398.15 K
298.15 K
V2≈3.34 L
Therefore, the new volume of the gas sample is approximately 3.34
L.
Question 14
A gas sample has a volume of 2.5 L at a pressure of 1.2 atm and a
temperature of 25
°
C. If the pressure is increased to 1.5 atm and the
temperature is kept constant, what will be the new volume of the gas
sample?
Step-by-step Solution:
Given: Initial volume, V1= 2.5L Initial pressure, P1= 1.2atm Final
pressure, P2= 1.5atm
We can use the combined gas law equation to find the final volume:
P1·V1
T1
=P2·V2
T2
Since the temperature is kept constant, T1=T2, and we can rear-
range the equation to solve for the final volume, V2:
V2=P1·V1
P2
16
Substitute the given values into the equation:
V2=(1.2atm ·2.5L)
1.5atm
V2=3atm ·L
1.5
V2= 2 L
Therefore, the new volume of the gas sample will be 2.0 L when
the pressure is increased to 1.5 atm while keeping the temperature
constant.Question 14:
A gas sample has a volume of 2.5 L at a pressure of 1.2 atm and a
temperature of 25
°
C. If the pressure is increased to 1.5 atm and the
temperature is kept constant, what will be the new volume of the gas
sample?
Step-by-step Solution:
Given: Initial volume, V1= 2.5L Initial pressure, P1= 1.2atm Final
pressure, P2= 1.5atm
We can use the combined gas law equation to find the final volume:
P1·V1
T1
=P2·V2
T2
Since the temperature is kept constant, T1=T2, and we can rear-
range the equation to solve for the final volume, V2:
V2=P1·V1
P2
Substitute the given values into the equation:
V2=(1.2atm ·2.5L)
1.5atm
V2=3atm ·L
1.5
V2= 2 L
Therefore, the new volume of the gas sample will be 2.0 L when
the pressure is increased to 1.5 atm while keeping the temperature
constant.
Question 15
A gaseous compound composed of nitrogen and oxygen has a vol-
ume of 3.50 L at a pressure of 1.20 atm and a temperature of 25
°
C.
If the compound is found to contain 30
Step-by-step solution:
17
1. Use the ideal gas law equation: P V =nRT where - P= pressure
in atm - V= volume in liters - n= amount of substance in moles -
R= ideal gas constant = 0.0821 L
·
atm/mol
·
K - T= temperature in
Kelvin
2. Convert the given temperature from Celsius to Kelvin:
T= 25C+ 273.15 = 298.15K
3. Calculate the total amount of substance using the ideal gas law:
ntotal =P V
RT =1.20 atm ×3.50 L
0.0821 L
·
atm/mol
·
K×298.15 K
4. Since the compound contains 30
nN2= 0.30 ×ntotal
5. Use the molar volume of gases at STP (Standard Temperature
and Pressure) to find the volume of nitrogen: - Molar volume of gases
at STP = 22.4 L/mol
Volume of nitrogen =nN2×Molar volume of gases at STP
Substitute the calculated values and solve to find the volume of
nitrogen in the compound.Question 15:
A gaseous compound composed of nitrogen and oxygen has a vol-
ume of 3.50 L at a pressure of 1.20 atm and a temperature of 25
°
C.
If the compound is found to contain 30
Step-by-step solution:
1. Use the ideal gas law equation: P V =nRT where - P= pressure
in atm - V= volume in liters - n= amount of substance in moles -
R= ideal gas constant = 0.0821 L
·
atm/mol
·
K - T= temperature in
Kelvin
2. Convert the given temperature from Celsius to Kelvin:
T= 25C+ 273.15 = 298.15K
3. Calculate the total amount of substance using the ideal gas law:
ntotal =P V
RT =1.20 atm ×3.50 L
0.0821 L
·
atm/mol
·
K×298.15 K
4. Since the compound contains 30
nN2= 0.30 ×ntotal
5. Use the molar volume of gases at STP (Standard Temperature
and Pressure) to find the volume of nitrogen: - Molar volume of gases
at STP = 22.4 L/mol
Volume of nitrogen =nN2×Molar volume of gases at STP
Substitute the calculated values and solve to find the volume of
nitrogen in the compound.
18
Question 16
A gas sample at 0.89 atm and 24
°
C occupies a volume of 2.50 L. If
the pressure is increased to 1.20 atm while the temperature remains
constant, what will be the new volume of the gas sample?
Step-by-step Solution:
Given: Initial pressure, Pi= 0.89 atm Initial volume, Vi= 2.50 L
Final pressure, Pf= 1.20 atm Initial volume, Vf=?
Step 1: Use the combined gas law formula to find the final volume.
Pi·Vi
Ti
=Pf·Vf
Tf
Since the temperature remains constant, we can simplify the for-
mula to:
Pi·Vi=Pf·Vf
Step 2: Solve for the final volume (Vf).
Vf=Pi·Vi
Pf
Step 3: Substitute the given values to find the final volume.
Vf=0.89 atm ×2.50 L
1.20 atm
Vf=2.225
1.20 L
Vf= 1.854 L
Therefore, the new volume of the gas sample will be 1.854 L when
the pressure is increased to 1.20 atm.Question 16:
A gas sample at 0.89 atm and 24
°
C occupies a volume of 2.50 L. If
the pressure is increased to 1.20 atm while the temperature remains
constant, what will be the new volume of the gas sample?
Step-by-step Solution:
Given: Initial pressure, Pi= 0.89 atm Initial volume, Vi= 2.50 L
Final pressure, Pf= 1.20 atm Initial volume, Vf=?
Step 1: Use the combined gas law formula to find the final volume.
Pi·Vi
Ti
=Pf·Vf
Tf
Since the temperature remains constant, we can simplify the for-
mula to:
Pi·Vi=Pf·Vf
19
Step 2: Solve for the final volume (Vf).
Vf=Pi·Vi
Pf
Step 3: Substitute the given values to find the final volume.
Vf=0.89 atm ×2.50 L
1.20 atm
Vf=2.225
1.20 L
Vf= 1.854 L
Therefore, the new volume of the gas sample will be 1.854 L when
the pressure is increased to 1.20 atm.
Question 17
A gas sample occupies a volume of 2.00 L at a pressure of 2.00
atm and a temperature of 300 K. If the pressure is increased to 3.00
atm while keeping the temperature constant, what will be the new
volume of the gas sample?
Step-by-step solution:
Given: Initial volume (V1) = 2.00 L Initial pressure (P1) = 2.00
atm Initial temperature (T1) = 300 K Final pressure (P2) = 3.00 atm
Since the temperature remains constant, we can use Boyle’s Law to
solve for the final volume:
P1×V1=P2×V2
2.00 atm ×2.00 L= 3.00 atm ×V2
4.00 = 3.00 ×V2
V2=4.00
3.00
V2= 1.33 L
Therefore, the new volume of the gas sample will be 1.33 L when
the pressure is increased to 3.00 atm while keeping the temperature
constant.Question 17:
A gas sample occupies a volume of 2.00 L at a pressure of 2.00
atm and a temperature of 300 K. If the pressure is increased to 3.00
20
atm while keeping the temperature constant, what will be the new
volume of the gas sample?
Step-by-step solution:
Given: Initial volume (V1) = 2.00 L Initial pressure (P1) = 2.00
atm Initial temperature (T1) = 300 K Final pressure (P2) = 3.00 atm
Since the temperature remains constant, we can use Boyle’s Law to
solve for the final volume:
P1×V1=P2×V2
2.00 atm ×2.00 L= 3.00 atm ×V2
4.00 = 3.00 ×V2
V2=4.00
3.00
V2= 1.33 L
Therefore, the new volume of the gas sample will be 1.33 L when
the pressure is increased to 3.00 atm while keeping the temperature
constant.
Question 18
A gas sample occupies a volume of 3.50 L at a temperature of
30.0
°
C and a pressure of 0.800 atm. Calculate the volume the gas will
occupy if the pressure is increased to 1.20 atm and the temperature
is raised to 50.0
°
C.
Step-by-step solution:
Given data: Initial volume (V1) = 3.50 L Initial temperature (T1)
= 30.0
°
C = 303.15 K Initial pressure (P1) = 0.800 atm
Final pressure (P2) = 1.20 atm Final temperature (T2) = 50.0
°
C
= 323.15 K
Using the combined gas law, P1V1/T1=P2V2/T2
Substitute the known values: (0.800 atm)(3.50 L)/(303.15 K) = (1.20 atm)(V2)/(323.15 K)
Solve for V2:V2= (0.800 ×3.50 ×323.15)/(1.20 ×303.15) V2= 2.809 L
Therefore, the volume the gas will occupy at a pressure of 1.20
atm and a temperature of 50.0
°
C is 2.809 L.Question 18:
A gas sample occupies a volume of 3.50 L at a temperature of
30.0
°
C and a pressure of 0.800 atm. Calculate the volume the gas will
occupy if the pressure is increased to 1.20 atm and the temperature
is raised to 50.0
°
C.
Step-by-step solution:
21
Given data: Initial volume (V1) = 3.50 L Initial temperature (T1)
= 30.0
°
C = 303.15 K Initial pressure (P1) = 0.800 atm
Final pressure (P2) = 1.20 atm Final temperature (T2) = 50.0
°
C
= 323.15 K
Using the combined gas law, P1V1/T1=P2V2/T2
Substitute the known values: (0.800 atm)(3.50 L)/(303.15 K) = (1.20 atm)(V2)/(323.15 K)
Solve for V2:V2= (0.800 ×3.50 ×323.15)/(1.20 ×303.15) V2= 2.809 L
Therefore, the volume the gas will occupy at a pressure of 1.20
atm and a temperature of 50.0
°
C is 2.809 L.
Question 19
A gas sample has a volume of 3.5 L at a pressure of 2.0 atm and a
temperature of 25
°
C. If the pressure is increased to 3.0 atm and the
temperature is raised to 50
°
C, what will be the new volume of the
gas sample?
Step-by-step Solution:
Given: Initial volume, V1= 3.5L Initial pressure, P1= 2.0atm
Initial temperature, T1= 25C= 25 + 273.15 K= 298.15 K
Final pressure, P2= 3.0atm Final temperature, T2= 50C= 50 +
273.15 K= 323.15 K
Using the ideal gas law equation:
P V =nRT
Where: P= pressure V= volume n= amount of substance R=
ideal gas constant T= temperature
The amount of substance remains constant, so we can write:
P1V1=P2V2
Now, substitute the given values into the equation:
2.0×3.5=3.0×V2
7.0=3.0V2
V2=7.0
3.0= 2.33 L
Therefore, the new volume of the gas sample will be 2.33 L after
the pressure is increased to 3.0 atm and the temperature is raised to
50
°
C.Question 19:
A gas sample has a volume of 3.5 L at a pressure of 2.0 atm and a
temperature of 25
°
C. If the pressure is increased to 3.0 atm and the
22
temperature is raised to 50
°
C, what will be the new volume of the
gas sample?
Step-by-step Solution:
Given: Initial volume, V1= 3.5L Initial pressure, P1= 2.0atm
Initial temperature, T1= 25C= 25 + 273.15 K= 298.15 K
Final pressure, P2= 3.0atm Final temperature, T2= 50C= 50 +
273.15 K= 323.15 K
Using the ideal gas law equation:
P V =nRT
Where: P= pressure V= volume n= amount of substance R=
ideal gas constant T= temperature
The amount of substance remains constant, so we can write:
P1V1=P2V2
Now, substitute the given values into the equation:
2.0×3.5=3.0×V2
7.0=3.0V2
V2=7.0
3.0= 2.33 L
Therefore, the new volume of the gas sample will be 2.33 L after
the pressure is increased to 3.0 atm and the temperature is raised to
50
°
C.
Question 20
A gas sample has a volume of 2.50 L and contains 0.0500 moles of
a certain gas at a temperature of 300 K and a pressure of 1.20 atm.
Calculate the molar volume of this gas sample at STP.
Step-by-step Solution: 1. Convert the given conditions to STP
(Standard Temperature and Pressure): - Temperature at STP is 273
K - Pressure at STP is 1 atm
2. Use the combined gas law equation:
P1V1
n1T1
=P2V2
n2T2
3. Substitute the given values into the equation to find the volume
of the gas sample at STP:
1.20 ×2.50
0.0500 ×300 =1×VST P
0.0500 ×273
23
Given: Initial volume, V1= 2.50 L
Initial pressure, P1= 1.50 atm
Initial temperature, T1= 25◦C= 25 + 273 = 298 K
Final pressure, P2= 2.00 atm
Final temperature, T2= 35◦C= 35 + 273 = 308 K
We can use the combined gas law to find the new volume:
P1·V1
T1
=P2·V2
T2
Solving for V2:
V2=P1·V1·T2
P2·T1
Substitute the given values:
V2=1.50 atm ×2.50 L×308 K
2.00 atm ×298 K
V2=3.525 atm ·L·K
2.00 atm = 5.325 L
Therefore, the new volume of the gas will be 5.325 L.
Question 2
A gas sample has a volume of 2.50 L at a temperature of 25
°
C and
a pressure of 1.20 atm. If the pressure is increased to 1.50 atm while
keeping the temperature constant, what will be the new volume of
the gas?
Step-by-step Solution:
Given: Initial volume, Vinitial = 2.50 L Initial pressure, Pinitial =
1.20 atm Final pressure, Pfinal = 1.50 atm
Since the temperature is constant, we can use the combined gas
law equation: Pinitial ×Vinitial
Tinitial
=Pfinal ×Vfinal
Tfinal
Since the temperature is constant, we can cancel out the temper-
atures from both sides of the equation:
Pinitial ×Vinitial =Pfinal ×Vfinal
We can now plug in the values to find the new volume:
1.20 atm ×2.50 L= 1.50 atm ×Vfinal
3.00 = 1.50 ×Vfinal
2
Vfinal =3.00
1.50
Vfinal = 2.00 L
Therefore, the new volume of the gas will be 2.00 L when the
pressure is increased to 1.50 atm while keeping the temperature con-
stant.Question 2:
A gas sample has a volume of 2.50 L at a temperature of 25
°
C and
a pressure of 1.20 atm. If the pressure is increased to 1.50 atm while
keeping the temperature constant, what will be the new volume of
the gas?
Step-by-step Solution:
Given: Initial volume, Vinitial = 2.50 L Initial pressure, Pinitial =
1.20 atm Final pressure, Pfinal = 1.50 atm
Since the temperature is constant, we can use the combined gas
law equation: Pinitial ×Vinitial
Tinitial
=Pfinal ×Vfinal
Tfinal
Since the temperature is constant, we can cancel out the temper-
atures from both sides of the equation:
Pinitial ×Vinitial =Pfinal ×Vfinal
We can now plug in the values to find the new volume:
1.20 atm ×2.50 L= 1.50 atm ×Vfinal
3.00 = 1.50 ×Vfinal
Vfinal =3.00
1.50
Vfinal = 2.00 L
Therefore, the new volume of the gas will be 2.00 L when the pres-
sure is increased to 1.50 atm while keeping the temperature constant.
Question 3
A gas occupies a volume of 4.0 L at a temperature of 300 K and a
pressure of 1.2 atm. Calculate the amount of the gas in moles.
Step-by-step solution:
Given: Volume of gas, V= 4.0L, Temperature, T= 300 K, Pres-
sure, P= 1.2atm.
Using the ideal gas law equation:
P V =nRT
3
Where: n= amount of gas in moles, R= 0.0821 L·atm/mol ·K
(universal gas constant).
Plug in the given values:
(1.2atm)(4.0L) = n(0.0821 L·atm/mol ·K)(300 K)
4.8 = 24.63n
n=4.8
24.63
n≈0.195 mol
Therefore, the amount of the gas in moles is approximately 0.195
mol.Question 3:
A gas occupies a volume of 4.0 L at a temperature of 300 K and a
pressure of 1.2 atm. Calculate the amount of the gas in moles.
Step-by-step solution:
Given: Volume of gas, V= 4.0L, Temperature, T= 300 K, Pres-
sure, P= 1.2atm.
Using the ideal gas law equation:
P V =nRT
Where: n= amount of gas in moles, R= 0.0821 L·atm/mol ·K
(universal gas constant).
Plug in the given values:
(1.2atm)(4.0L) = n(0.0821 L·atm/mol ·K)(300 K)
4.8 = 24.63n
n=4.8
24.63
n≈0.195 mol
Therefore, the amount of the gas in moles is approximately 0.195
mol.
4
Question 4
A certain amount of gas occupies a volume of 500 mL at a pressure
of 1 atm and a temperature of 300 K. If the pressure is increased to 2
atm and the temperature is raised to 400 K, what is the new volume
of the gas?
Step-by-step solution: 1. Determine the initial amount of gas
using the ideal gas law: P V =nRT , where Pis the pressure, Vis the
volume, nis the amount of gas in moles, Ris the gas constant, and T
is the temperature in Kelvin. 2. Rearrange the ideal gas law to solve
for n:n=P V
RT . 3. Calculate the initial amount of gas when pressure is
1 atm and temperature is 300 K. 4. Use the given conditions of the
pressure being increased to 2 atm and the temperature raised to 400
K to calculate the new volume of the gas using the ideal gas law. 5.
Substitute the new pressure, temperature, and the initial amount of
gas into the ideal gas law and solve for the new volume. 6. Present
the final answer with appropriate units.Question 4:
A certain amount of gas occupies a volume of 500 mL at a pressure
of 1 atm and a temperature of 300 K. If the pressure is increased to 2
atm and the temperature is raised to 400 K, what is the new volume
of the gas?
Step-by-step solution: 1. Determine the initial amount of gas
using the ideal gas law: P V =nRT , where Pis the pressure, Vis the
volume, nis the amount of gas in moles, Ris the gas constant, and T
is the temperature in Kelvin. 2. Rearrange the ideal gas law to solve
for n:n=P V
RT . 3. Calculate the initial amount of gas when pressure is
1 atm and temperature is 300 K. 4. Use the given conditions of the
pressure being increased to 2 atm and the temperature raised to 400
K to calculate the new volume of the gas using the ideal gas law. 5.
Substitute the new pressure, temperature, and the initial amount of
gas into the ideal gas law and solve for the new volume. 6. Present
the final answer with appropriate units.
Question 5
Given that a certain amount of gas occupies a volume of 2.5 L at a
temperature of 25
°
C and a pressure of 1 atm, determine the volume
this gas will occupy if the temperature is increased to 75
°
C and the
pressure is increased to 2 atm.
Step-by-step solution:
1. Write down the given values: Initial volume, V1= 2.5L Initial
temperature, T1= 25C= 25 + 273.15 K Initial pressure, P1= 1 atm
2. Use the ideal gas law equation: P V =nRT , where Pis pressure,
Vis volume, nis the amount of substance, Ris the ideal gas constant,
and Tis temperature in Kelvin.
5
3. Calculate the initial amount of substance: n=P1V1
RT1
4. Determine the final volume using the ideal gas law and the
calculated amount of substance: V2=nRT2
P2, where Final temperature,
T2= 75C= 75 + 273.15 K Final pressure, P2= 2 atm
5. Substitute the values into the equation to find the final volume
occupied by the gas.
6. Calculate the final volume to find the answer.Question 5:
Given that a certain amount of gas occupies a volume of 2.5 L at a
temperature of 25
°
C and a pressure of 1 atm, determine the volume
this gas will occupy if the temperature is increased to 75
°
C and the
pressure is increased to 2 atm.
Step-by-step solution:
1. Write down the given values: Initial volume, V1= 2.5L Initial
temperature, T1= 25C= 25 + 273.15 K Initial pressure, P1= 1 atm
2. Use the ideal gas law equation: P V =nRT , where Pis pressure,
Vis volume, nis the amount of substance, Ris the ideal gas constant,
and Tis temperature in Kelvin.
3. Calculate the initial amount of substance: n=P1V1
RT1
4. Determine the final volume using the ideal gas law and the
calculated amount of substance: V2=nRT2
P2, where Final temperature,
T2= 75C= 75 + 273.15 K Final pressure, P2= 2 atm
5. Substitute the values into the equation to find the final volume
occupied by the gas.
6. Calculate the final volume to find the answer.
Question 6
Step-by-step Solution: Given: Volume of gas, V= 2.50LTemper-
ature, T= 25C= 298K(Convert Celsius to Kelvin by adding 273)
Pressure, P= 1.00atm
We can use the ideal gas law equation to find the number of moles
of gas:
P V =nRT
where: P= pressure V= volume n= number of moles R= ideal
gas constant (0.0821L·atm/(mol ·K))T= temperature in Kelvin
First, convert the temperature from Celsius to Kelvin: T= 25C+
273 = 298K
Now, plug in the values into the ideal gas law equation: 1.00atm ×
2.50L=n×0.0821L·atm/(mol ·K)×298K
Solving for n:2.50atm·L= 0.0821×298×n2.50 = 24.4678×n n =2.50
24.4678
n= 0.1021 moles
Therefore, the number of moles of gas present in the sample is
0.1021 moles.Question 6: A gas sample has a volume of 2.50 L at
a temperature of 25
°
C and a pressure of 1.00 atm. Calculate the
number of moles of gas present in the sample.
6
Step-by-step Solution: Given: Volume of gas, V= 2.50LTemper-
ature, T= 25C= 298K(Convert Celsius to Kelvin by adding 273)
Pressure, P= 1.00atm
We can use the ideal gas law equation to find the number of moles
of gas:
P V =nRT
where: P= pressure V= volume n= number of moles R= ideal
gas constant (0.0821L·atm/(mol ·K))T= temperature in Kelvin
First, convert the temperature from Celsius to Kelvin: T= 25C+
273 = 298K
Now, plug in the values into the ideal gas law equation: 1.00atm ×
2.50L=n×0.0821L·atm/(mol ·K)×298K
Solving for n:2.50atm·L= 0.0821×298×n2.50 = 24.4678×n n =2.50
24.4678
n= 0.1021 moles
Therefore, the number of moles of gas present in the sample is
0.1021 moles.
Question 7
A gas sample has a volume of 2.50 L at a temperature of 300 K and
a pressure of 1.50 atm. If the number of moles of the gas is 0.075 mol,
calculate the molar volume of the gas at STP (standard temperature
and pressure).
Step-by-step solution:
Given: - Volume of gas (V) = 2.50 L - Temperature (T) = 300 K
- Pressure (P) = 1.50 atm - Number of moles of gas (n) = 0.075 mol
To calculate the molar volume of the gas at STP, we first need to
convert the given conditions to STP conditions.
STP conditions: - Temperature (TSTP) = 273 K - Pressure (PSTP )
= 1 atm - Number of moles at STP (nSTP ) = 1 mol
We can use the combined gas law to find the molar volume at STP:
P·V
n·T=PSTP ·VSTP
nSTP ·TSTP
Given:
P= 1.50 atm, V = 2.50 L, n = 0.075 mol, PSTP = 1 atm, T = 300 K, TSTP = 273 K, nSTP = 1 mol
Substitute the values into the equation and solve for VSTP:
VSTP =P·V·nSTP ·TSTP
n·PSTP
VSTP =1.50 atm ×2.50 L×1mol ×273 K
0.075 mol ×1atm
7
VSTP =1.50 ×2.50 ×273
0.075
VSTP = 135.0L
Therefore, the molar volume of the gas at STP is 135.0 L.Question
7:
A gas sample has a volume of 2.50 L at a temperature of 300 K and
a pressure of 1.50 atm. If the number of moles of the gas is 0.075 mol,
calculate the molar volume of the gas at STP (standard temperature
and pressure).
Step-by-step solution:
Given: - Volume of gas (V) = 2.50 L - Temperature (T) = 300 K
- Pressure (P) = 1.50 atm - Number of moles of gas (n) = 0.075 mol
To calculate the molar volume of the gas at STP, we first need to
convert the given conditions to STP conditions.
STP conditions: - Temperature (TSTP) = 273 K - Pressure (PSTP )
= 1 atm - Number of moles at STP (nSTP ) = 1 mol
We can use the combined gas law to find the molar volume at STP:
P·V
n·T=PSTP ·VSTP
nSTP ·TSTP
Given:
P= 1.50 atm, V = 2.50 L, n = 0.075 mol, PSTP = 1 atm, T = 300 K, TSTP = 273 K, nSTP = 1 mol
Substitute the values into the equation and solve for VSTP:
VSTP =P·V·nSTP ·TSTP
n·PSTP
VSTP =1.50 atm ×2.50 L×1mol ×273 K
0.075 mol ×1atm
VSTP =1.50 ×2.50 ×273
0.075
VSTP = 135.0L
Therefore, the molar volume of the gas at STP is 135.0 L.
8
Question 8
A sample of helium gas occupies a volume of 2.00 L at a temper-
ature of 300 K and a pressure of 2.50 atm. Calculate the amount of
substance (in moles) of the helium gas present in the sample.
Step-by-step solution:
Given data: - Volume (V) = 2.00 L - Temperature (T)=300K-
Pressure (P) = 2.50 atm
We can use the ideal gas law equation to find the amount of sub-
stance (moles) of the gas present:
P V =nRT
where: - Pis the pressure - Vis the volume - nis the amount of
substance (in moles) - Ris the gas constant (0.0821 L atm/mol K) -
Tis the temperature
Substitute the given values into the equation:
2.50 atm ×2.00 L=n×0.0821 L atm/mol K ×300 K
5.00 = 24.63n
n=5.00
24.63
n≈0.203 moles
Therefore, the amount of substance of the helium gas present in
the sample is approximately 0.203 moles.Question 8:
A sample of helium gas occupies a volume of 2.00 L at a temper-
ature of 300 K and a pressure of 2.50 atm. Calculate the amount of
substance (in moles) of the helium gas present in the sample.
Step-by-step solution:
Given data: - Volume (V) = 2.00 L - Temperature (T)=300K-
Pressure (P) = 2.50 atm
We can use the ideal gas law equation to find the amount of sub-
stance (moles) of the gas present:
P V =nRT
where: - Pis the pressure - Vis the volume - nis the amount of
substance (in moles) - Ris the gas constant (0.0821 L atm/mol K) -
Tis the temperature
Substitute the given values into the equation:
2.50 atm ×2.00 L=n×0.0821 L atm/mol K ×300 K
9
5.00 = 24.63n
n=5.00
24.63
n≈0.203 moles
Therefore, the amount of substance of the helium gas present in
the sample is approximately 0.203 moles.
Question 9
A gas sample has a volume of 2.50 L at a temperature of 25
°
C and
a pressure of 1.20 atm. If the pressure is increased to 2.00 atm while
keeping the temperature constant, what will be the new volume of
the gas sample?
Step-by-step Solution:
Given: Initial volume, Vi= 2.50 L Initial pressure, Pi= 1.20 atm
Final pressure, Pf= 2.00 atm Temperature is constant
According to the Boyle’s Law, which states that for a given amount
of gas at constant temperature, the pressure and volume are inversely
proportional:
Pi×Vi=Pf×Vf
Substitute the given values into the equation:
1.20 ×2.50 = 2.00 ×Vf
Solve for Vf:
3.00 = 2.00 ×Vf
Vf=3.00
2.00
Vf= 1.50 L
Therefore, the new volume of the gas sample will be 1.50 L when
the pressure is increased to 2.00 atm while keeping the temperature
constant.Question 9:
A gas sample has a volume of 2.50 L at a temperature of 25
°
C and
a pressure of 1.20 atm. If the pressure is increased to 2.00 atm while
keeping the temperature constant, what will be the new volume of
the gas sample?
Step-by-step Solution:
10
Given: Initial volume, Vi= 2.50 L Initial pressure, Pi= 1.20 atm
Final pressure, Pf= 2.00 atm Temperature is constant
According to the Boyle’s Law, which states that for a given amount
of gas at constant temperature, the pressure and volume are inversely
proportional:
Pi×Vi=Pf×Vf
Substitute the given values into the equation:
1.20 ×2.50 = 2.00 ×Vf
Solve for Vf:
3.00 = 2.00 ×Vf
Vf=3.00
2.00
Vf= 1.50 L
Therefore, the new volume of the gas sample will be 1.50 L when
the pressure is increased to 2.00 atm while keeping the temperature
constant.
Question 10
A gas occupies a volume of 4.50 L at a temperature of 27
°
C and
a pressure of 0.900 atm. If the pressure is increased to 1.20 atm and
the temperature is decreased to 20
°
C, what will be the new volume
of the gas? Assume the amount of substance remains constant.
Step-by-step solution:
Given: Initial volume, V1= 4.50 L Initial temperature, T1= 27C=
27 + 273 = 300 K Initial pressure, P1= 0.900 atm Final pressure, P2=
1.20 atm Final temperature, T2= 20C= 20 + 273 = 293 K
According to the combined gas law equation:
P1·V1
T1
=P2·V2
T2
Substitute the given values into the equation and solve for V2:
0.900 ·4.50
300 =1.20 ·V2
293
V2=0.900 ·4.50 ·293
1.20 ·300
V2= 4.60 L
11
Therefore, the new volume of the gas will be 4.60 L.Question 10:
A gas occupies a volume of 4.50 L at a temperature of 27
°
C and
a pressure of 0.900 atm. If the pressure is increased to 1.20 atm and
the temperature is decreased to 20
°
C, what will be the new volume
of the gas? Assume the amount of substance remains constant.
Step-by-step solution:
Given: Initial volume, V1= 4.50 L Initial temperature, T1= 27C=
27 + 273 = 300 K Initial pressure, P1= 0.900 atm Final pressure, P2=
1.20 atm Final temperature, T2= 20C= 20 + 273 = 293 K
According to the combined gas law equation:
P1·V1
T1
=P2·V2
T2
Substitute the given values into the equation and solve for V2:
0.900 ·4.50
300 =1.20 ·V2
293
V2=0.900 ·4.50 ·293
1.20 ·300
V2= 4.60 L
Therefore, the new volume of the gas will be 4.60 L.
Question 11
A 2.50 L container has 0.500 mol of helium gas at a certain tem-
perature and pressure. If the pressure is increased while the temper-
ature remains constant, what will be the new volume of the gas if the
amount of substance is kept constant?
Step-by-step solution: 1. The ideal gas law is given by the equation
P V =nRT , where: - Pis the pressure of the gas, - Vis the volume of
the gas, - nis the amount of substance in moles, - Ris the ideal gas
constant (0.0821 L.atm/mol.K), - Tis the temperature in Kelvin.
2. Initially, the helium gas is in a 2.50 L container with 0.500 mol
of helium. The pressure and temperature are constant.
3. When the pressure is increased, the amount of substance (moles
of gas) and temperature are kept constant.
4. Since P V =nRT and nis constant, we can rearrange the equa-
tion to solve for the new volume Vnew:
PinitialVinitial =Pf inalVnew
5. Substitute the initial values:
Pinitial =Pfinal ×Vnew
Vinitial
12
6. The new pressure is higher than the initial pressure, so we have
Pfinal > Pinitial, meaning the volume will decrease.
7. To find the new volume, use the ratio of the pressures:
Pfinal
Pinitial
=Vinitial
Vnew
8. Substitute the values:
Pfinal
Pinitial
=Vinitial
Vnew
Pfinal
Pinitial
=2.50 L
Vnew
9. If the pressure is doubled, then Pf inal
Pinitial = 2. Substitute this value
into the equation and solve for Vnew:
2 = 2.50 L
Vnew
Vnew =2.50 L
2
Vnew = 1.25 L
The new volume of the helium gas when the pressure is doubled
while keeping the temperature and amount of substance constant will
be 1.25 L.Question 11:
A 2.50 L container has 0.500 mol of helium gas at a certain tem-
perature and pressure. If the pressure is increased while the temper-
ature remains constant, what will be the new volume of the gas if the
amount of substance is kept constant?
Step-by-step solution: 1. The ideal gas law is given by the equation
P V =nRT , where: - Pis the pressure of the gas, - Vis the volume of
the gas, - nis the amount of substance in moles, - Ris the ideal gas
constant (0.0821 L.atm/mol.K), - Tis the temperature in Kelvin.
2. Initially, the helium gas is in a 2.50 L container with 0.500 mol
of helium. The pressure and temperature are constant.
3. When the pressure is increased, the amount of substance (moles
of gas) and temperature are kept constant.
4. Since P V =nRT and nis constant, we can rearrange the equa-
tion to solve for the new volume Vnew:
PinitialVinitial =Pf inalVnew
5. Substitute the initial values:
Pinitial =Pfinal ×Vnew
Vinitial
13
6. The new pressure is higher than the initial pressure, so we have
Pfinal > Pinitial, meaning the volume will decrease.
7. To find the new volume, use the ratio of the pressures:
Pfinal
Pinitial
=Vinitial
Vnew
8. Substitute the values:
Pfinal
Pinitial
=Vinitial
Vnew
Pfinal
Pinitial
=2.50 L
Vnew
9. If the pressure is doubled, then Pf inal
Pinitial = 2. Substitute this value
into the equation and solve for Vnew:
2 = 2.50 L
Vnew
Vnew =2.50 L
2
Vnew = 1.25 L
The new volume of the helium gas when the pressure is doubled
while keeping the temperature and amount of substance constant will
be 1.25 L.
Question 12
Step-by-step solution: Given: Initial volume, V1= 2.50 L Initial
pressure, P1= 1.50 atm Final pressure, P2= 3.00 atm Initial tempera-
ture, T1= 25C
Since the temperature remains constant, the change in pressure
and volume can be related by Boyle’s Law: P1V1=P2V2
Solving for the final volume, V2:
P1V1=P2V2
2.50 ×1.50 = 3.00 ×V2
3.75 = 3.00V2
V2=3.75
3.00
V2= 1.25 L
Therefore, the new volume of the gas will be 1.25 L when the pres-
sure is increased to 3.00 atm.Question 12: A gas occupies a volume
of 2.50 L at a pressure of 1.50 atm and a temperature of 25
°
C. If
14
the pressure is increased to 3.00 atm while the temperature remains
constant, what will be the new volume of the gas?
Step-by-step solution: Given: Initial volume, V1= 2.50 L Initial
pressure, P1= 1.50 atm Final pressure, P2= 3.00 atm Initial tempera-
ture, T1= 25C
Since the temperature remains constant, the change in pressure
and volume can be related by Boyle’s Law: P1V1=P2V2
Solving for the final volume, V2:
P1V1=P2V2
2.50 ×1.50 = 3.00 ×V2
3.75 = 3.00V2
V2=3.75
3.00
V2= 1.25 L
Therefore, the new volume of the gas will be 1.25 L when the
pressure is increased to 3.00 atm.
Question 13
A gas sample occupying 2.50 L at 1.00 atm and 25.0
°
C is heated
to 125
°
C at constant pressure. What is the new volume of the gas
sample?
Step-by-step Solution:
Given: Initial volume (V1)=2.50 L Initial pressure (P1)=1.00 atm
Initial temperature (T1) = 25.0C= 298.15 K
Final temperature (T2) = 125C= 398.15 K
Using the ideal gas law P V =nRT , where nis the amount of
substance and Ris the gas constant, we can rearrange the formula to
solve for the final volume V2:
P1V1
T1
=P2V2
T2
Substitute the given values into the equation:
1.00atm ×2.50L
298.15K =1.00atm ×V2
398.15K
Solve for V2:
V2=1.00 atm ×2.50 L×398.15 K
298.15 K
V2≈3.34 L
15
Therefore, the new volume of the gas sample is approximately 3.34
L.Question 13:
A gas sample occupying 2.50 L at 1.00 atm and 25.0
°
C is heated
to 125
°
C at constant pressure. What is the new volume of the gas
sample?
Step-by-step Solution:
Given: Initial volume (V1)=2.50 L Initial pressure (P1)=1.00 atm
Initial temperature (T1) = 25.0C= 298.15 K
Final temperature (T2) = 125C= 398.15 K
Using the ideal gas law P V =nRT , where nis the amount of
substance and Ris the gas constant, we can rearrange the formula to
solve for the final volume V2:
P1V1
T1
=P2V2
T2
Substitute the given values into the equation:
1.00atm ×2.50L
298.15K =1.00atm ×V2
398.15K
Solve for V2:
V2=1.00 atm ×2.50 L×398.15 K
298.15 K
V2≈3.34 L
Therefore, the new volume of the gas sample is approximately 3.34
L.
Question 14
A gas sample has a volume of 2.5 L at a pressure of 1.2 atm and a
temperature of 25
°
C. If the pressure is increased to 1.5 atm and the
temperature is kept constant, what will be the new volume of the gas
sample?
Step-by-step Solution:
Given: Initial volume, V1= 2.5L Initial pressure, P1= 1.2atm Final
pressure, P2= 1.5atm
We can use the combined gas law equation to find the final volume:
P1·V1
T1
=P2·V2
T2
Since the temperature is kept constant, T1=T2, and we can rear-
range the equation to solve for the final volume, V2:
V2=P1·V1
P2
16
Substitute the given values into the equation:
V2=(1.2atm ·2.5L)
1.5atm
V2=3atm ·L
1.5
V2= 2 L
Therefore, the new volume of the gas sample will be 2.0 L when
the pressure is increased to 1.5 atm while keeping the temperature
constant.Question 14:
A gas sample has a volume of 2.5 L at a pressure of 1.2 atm and a
temperature of 25
°
C. If the pressure is increased to 1.5 atm and the
temperature is kept constant, what will be the new volume of the gas
sample?
Step-by-step Solution:
Given: Initial volume, V1= 2.5L Initial pressure, P1= 1.2atm Final
pressure, P2= 1.5atm
We can use the combined gas law equation to find the final volume:
P1·V1
T1
=P2·V2
T2
Since the temperature is kept constant, T1=T2, and we can rear-
range the equation to solve for the final volume, V2:
V2=P1·V1
P2
Substitute the given values into the equation:
V2=(1.2atm ·2.5L)
1.5atm
V2=3atm ·L
1.5
V2= 2 L
Therefore, the new volume of the gas sample will be 2.0 L when
the pressure is increased to 1.5 atm while keeping the temperature
constant.
Question 15
A gaseous compound composed of nitrogen and oxygen has a vol-
ume of 3.50 L at a pressure of 1.20 atm and a temperature of 25
°
C.
If the compound is found to contain 30
Step-by-step solution:
17
1. Use the ideal gas law equation: P V =nRT where - P= pressure
in atm - V= volume in liters - n= amount of substance in moles -
R= ideal gas constant = 0.0821 L
·
atm/mol
·
K - T= temperature in
Kelvin
2. Convert the given temperature from Celsius to Kelvin:
T= 25C+ 273.15 = 298.15K
3. Calculate the total amount of substance using the ideal gas law:
ntotal =P V
RT =1.20 atm ×3.50 L
0.0821 L
·
atm/mol
·
K×298.15 K
4. Since the compound contains 30
nN2= 0.30 ×ntotal
5. Use the molar volume of gases at STP (Standard Temperature
and Pressure) to find the volume of nitrogen: - Molar volume of gases
at STP = 22.4 L/mol
Volume of nitrogen =nN2×Molar volume of gases at STP
Substitute the calculated values and solve to find the volume of
nitrogen in the compound.Question 15:
A gaseous compound composed of nitrogen and oxygen has a vol-
ume of 3.50 L at a pressure of 1.20 atm and a temperature of 25
°
C.
If the compound is found to contain 30
Step-by-step solution:
1. Use the ideal gas law equation: P V =nRT where - P= pressure
in atm - V= volume in liters - n= amount of substance in moles -
R= ideal gas constant = 0.0821 L
·
atm/mol
·
K - T= temperature in
Kelvin
2. Convert the given temperature from Celsius to Kelvin:
T= 25C+ 273.15 = 298.15K
3. Calculate the total amount of substance using the ideal gas law:
ntotal =P V
RT =1.20 atm ×3.50 L
0.0821 L
·
atm/mol
·
K×298.15 K
4. Since the compound contains 30
nN2= 0.30 ×ntotal
5. Use the molar volume of gases at STP (Standard Temperature
and Pressure) to find the volume of nitrogen: - Molar volume of gases
at STP = 22.4 L/mol
Volume of nitrogen =nN2×Molar volume of gases at STP
Substitute the calculated values and solve to find the volume of
nitrogen in the compound.
18
Question 16
A gas sample at 0.89 atm and 24
°
C occupies a volume of 2.50 L. If
the pressure is increased to 1.20 atm while the temperature remains
constant, what will be the new volume of the gas sample?
Step-by-step Solution:
Given: Initial pressure, Pi= 0.89 atm Initial volume, Vi= 2.50 L
Final pressure, Pf= 1.20 atm Initial volume, Vf=?
Step 1: Use the combined gas law formula to find the final volume.
Pi·Vi
Ti
=Pf·Vf
Tf
Since the temperature remains constant, we can simplify the for-
mula to:
Pi·Vi=Pf·Vf
Step 2: Solve for the final volume (Vf).
Vf=Pi·Vi
Pf
Step 3: Substitute the given values to find the final volume.
Vf=0.89 atm ×2.50 L
1.20 atm
Vf=2.225
1.20 L
Vf= 1.854 L
Therefore, the new volume of the gas sample will be 1.854 L when
the pressure is increased to 1.20 atm.Question 16:
A gas sample at 0.89 atm and 24
°
C occupies a volume of 2.50 L. If
the pressure is increased to 1.20 atm while the temperature remains
constant, what will be the new volume of the gas sample?
Step-by-step Solution:
Given: Initial pressure, Pi= 0.89 atm Initial volume, Vi= 2.50 L
Final pressure, Pf= 1.20 atm Initial volume, Vf=?
Step 1: Use the combined gas law formula to find the final volume.
Pi·Vi
Ti
=Pf·Vf
Tf
Since the temperature remains constant, we can simplify the for-
mula to:
Pi·Vi=Pf·Vf
19
Step 2: Solve for the final volume (Vf).
Vf=Pi·Vi
Pf
Step 3: Substitute the given values to find the final volume.
Vf=0.89 atm ×2.50 L
1.20 atm
Vf=2.225
1.20 L
Vf= 1.854 L
Therefore, the new volume of the gas sample will be 1.854 L when
the pressure is increased to 1.20 atm.
Question 17
A gas sample occupies a volume of 2.00 L at a pressure of 2.00
atm and a temperature of 300 K. If the pressure is increased to 3.00
atm while keeping the temperature constant, what will be the new
volume of the gas sample?
Step-by-step solution:
Given: Initial volume (V1) = 2.00 L Initial pressure (P1) = 2.00
atm Initial temperature (T1) = 300 K Final pressure (P2) = 3.00 atm
Since the temperature remains constant, we can use Boyle’s Law to
solve for the final volume:
P1×V1=P2×V2
2.00 atm ×2.00 L= 3.00 atm ×V2
4.00 = 3.00 ×V2
V2=4.00
3.00
V2= 1.33 L
Therefore, the new volume of the gas sample will be 1.33 L when
the pressure is increased to 3.00 atm while keeping the temperature
constant.Question 17:
A gas sample occupies a volume of 2.00 L at a pressure of 2.00
atm and a temperature of 300 K. If the pressure is increased to 3.00
20
atm while keeping the temperature constant, what will be the new
volume of the gas sample?
Step-by-step solution:
Given: Initial volume (V1) = 2.00 L Initial pressure (P1) = 2.00
atm Initial temperature (T1) = 300 K Final pressure (P2) = 3.00 atm
Since the temperature remains constant, we can use Boyle’s Law to
solve for the final volume:
P1×V1=P2×V2
2.00 atm ×2.00 L= 3.00 atm ×V2
4.00 = 3.00 ×V2
V2=4.00
3.00
V2= 1.33 L
Therefore, the new volume of the gas sample will be 1.33 L when
the pressure is increased to 3.00 atm while keeping the temperature
constant.
Question 18
A gas sample occupies a volume of 3.50 L at a temperature of
30.0
°
C and a pressure of 0.800 atm. Calculate the volume the gas will
occupy if the pressure is increased to 1.20 atm and the temperature
is raised to 50.0
°
C.
Step-by-step solution:
Given data: Initial volume (V1) = 3.50 L Initial temperature (T1)
= 30.0
°
C = 303.15 K Initial pressure (P1) = 0.800 atm
Final pressure (P2) = 1.20 atm Final temperature (T2) = 50.0
°
C
= 323.15 K
Using the combined gas law, P1V1/T1=P2V2/T2
Substitute the known values: (0.800 atm)(3.50 L)/(303.15 K) = (1.20 atm)(V2)/(323.15 K)
Solve for V2:V2= (0.800 ×3.50 ×323.15)/(1.20 ×303.15) V2= 2.809 L
Therefore, the volume the gas will occupy at a pressure of 1.20
atm and a temperature of 50.0
°
C is 2.809 L.Question 18:
A gas sample occupies a volume of 3.50 L at a temperature of
30.0
°
C and a pressure of 0.800 atm. Calculate the volume the gas will
occupy if the pressure is increased to 1.20 atm and the temperature
is raised to 50.0
°
C.
Step-by-step solution:
21
Given data: Initial volume (V1) = 3.50 L Initial temperature (T1)
= 30.0
°
C = 303.15 K Initial pressure (P1) = 0.800 atm
Final pressure (P2) = 1.20 atm Final temperature (T2) = 50.0
°
C
= 323.15 K
Using the combined gas law, P1V1/T1=P2V2/T2
Substitute the known values: (0.800 atm)(3.50 L)/(303.15 K) = (1.20 atm)(V2)/(323.15 K)
Solve for V2:V2= (0.800 ×3.50 ×323.15)/(1.20 ×303.15) V2= 2.809 L
Therefore, the volume the gas will occupy at a pressure of 1.20
atm and a temperature of 50.0
°
C is 2.809 L.
Question 19
A gas sample has a volume of 3.5 L at a pressure of 2.0 atm and a
temperature of 25
°
C. If the pressure is increased to 3.0 atm and the
temperature is raised to 50
°
C, what will be the new volume of the
gas sample?
Step-by-step Solution:
Given: Initial volume, V1= 3.5L Initial pressure, P1= 2.0atm
Initial temperature, T1= 25C= 25 + 273.15 K= 298.15 K
Final pressure, P2= 3.0atm Final temperature, T2= 50C= 50 +
273.15 K= 323.15 K
Using the ideal gas law equation:
P V =nRT
Where: P= pressure V= volume n= amount of substance R=
ideal gas constant T= temperature
The amount of substance remains constant, so we can write:
P1V1=P2V2
Now, substitute the given values into the equation:
2.0×3.5=3.0×V2
7.0=3.0V2
V2=7.0
3.0= 2.33 L
Therefore, the new volume of the gas sample will be 2.33 L after
the pressure is increased to 3.0 atm and the temperature is raised to
50
°
C.Question 19:
A gas sample has a volume of 3.5 L at a pressure of 2.0 atm and a
temperature of 25
°
C. If the pressure is increased to 3.0 atm and the
22
temperature is raised to 50
°
C, what will be the new volume of the
gas sample?
Step-by-step Solution:
Given: Initial volume, V1= 3.5L Initial pressure, P1= 2.0atm
Initial temperature, T1= 25C= 25 + 273.15 K= 298.15 K
Final pressure, P2= 3.0atm Final temperature, T2= 50C= 50 +
273.15 K= 323.15 K
Using the ideal gas law equation:
P V =nRT
Where: P= pressure V= volume n= amount of substance R=
ideal gas constant T= temperature
The amount of substance remains constant, so we can write:
P1V1=P2V2
Now, substitute the given values into the equation:
2.0×3.5=3.0×V2
7.0=3.0V2
V2=7.0
3.0= 2.33 L
Therefore, the new volume of the gas sample will be 2.33 L after
the pressure is increased to 3.0 atm and the temperature is raised to
50
°
C.
Question 20
A gas sample has a volume of 2.50 L and contains 0.0500 moles of
a certain gas at a temperature of 300 K and a pressure of 1.20 atm.
Calculate the molar volume of this gas sample at STP.
Step-by-step Solution: 1. Convert the given conditions to STP
(Standard Temperature and Pressure): - Temperature at STP is 273
K - Pressure at STP is 1 atm
2. Use the combined gas law equation:
P1V1
n1T1
=P2V2
n2T2
3. Substitute the given values into the equation to find the volume
of the gas sample at STP:
1.20 ×2.50
0.0500 ×300 =1×VST P
0.0500 ×273
23
VST P =1.20 ×2.50 ×0.0500 ×273
0.0500 ×300
VST P = 1.22 L/mol
Therefore, the molar volume of this gas sample at STP is 1.22
L/mol.Question 20:
A gas sample has a volume of 2.50 L and contains 0.0500 moles of
a certain gas at a temperature of 300 K and a pressure of 1.20 atm.
Calculate the molar volume of this gas sample at STP.
Step-by-step Solution: 1. Convert the given conditions to STP
(Standard Temperature and Pressure): - Temperature at STP is 273
K - Pressure at STP is 1 atm
2. Use the combined gas law equation:
P1V1
n1T1
=P2V2
n2T2
3. Substitute the given values into the equation to find the volume
of the gas sample at STP:
1.20 ×2.50
0.0500 ×300 =1×VST P
0.0500 ×273
VST P =1.20 ×2.50 ×0.0500 ×273
0.0500 ×300
VST P = 1.22 L/mol
Therefore, the molar volume of this gas sample at STP is 1.22
L/mol.
24
