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AP Physics 2 : Thermodynamics

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Example Questions

Example Question #2 : Ideal Gas Law

How many moles of nitrogen gas are in a 10L tank with a pressure of 5atm at 29^oC ?

$R=0.0821\frac{L\cdot atm}{mol\cdot K}$

Possible Answers:

4.04mol

2.02mol

21.0mol

2.10mol

Correct answer:

2.02mol

Explanation:

We will use the ideal gas equation:

PV=nRT

Where is the pressure in atm, is the volume in liters, is the number of moles of gas, $R=0.0821\frac{L\cdot atm}{mol\cdot K}$ , and is the temperature in Kelvin.

Rearrange the equation to solve for moles:

$\frac{PV}{RT}=n$

Plug in known values and solve.

$n=\frac{5*10}{0.0821*302}$

n=2.02mol

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Example Question #11 : Ideal Gas Law

What is the volume of 1kg of nitrogen gas (diatomic) at $24^{\circ}C$ and .98 atm ?

$R=.0821 \frac{(atm*L)}{(K*mol)}$

Possible Answers:

888 L

80.4 L

40.4 L

951 L

Correct answer:

888 L

Explanation:

We will use the equation:

PV=nRT

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atm*L)}{(K*mol)}$

is the temperature in Kelvin.

We need to convert the temperature to Kelvin.

$24^{\circ}C+273=298^{\circ}K$

Rearrange our original equation for volume:

$V=\frac{nRT}{P}$

We need to find the moles of nitrogen gas. We divide the mass by the molar mass of nitrogen, which is $28 \frac{g}{mol}$ .

$\frac{(1000 g)}{(28 g/mol)}=35.7 moles$

Plug in our values to our original equation and solve.

$V=\frac{35.7*.0821*297}{.98}$

888 L=V

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Example Question #51 : Thermodynamics

What is the mass of 16.5 L of helium gas at $21^\circ C$ and 1.03 atm ?

$R=.0821 \frac{(atm*L)}{(K*mol)}$

Possible Answers:

2.81g

.704 g

28 g

22.4 g

Correct answer:

2.81g

Explanation:

We will use the equation:

PV=nRT

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atm*L)}{(K*mol)}$

is the temperature in Kelvin

We need to convert the temperature to Kelvin.

$21^{\circ}C+273=294^{\circ}K$
Rearrange to solve for moles:

$\frac{PV}{RT}=n$

Plug in our values and solve

$\frac{1.03*16.5}{.0821*294}=n$

.704moles=n

Then, we need to multiply by the molar mass of helium; recall that helium is not a diatomic gas.

$.704 moles He * \frac{4g He}{mol} = 2.81 g$

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Example Question #11 : Ideal Gas Law

What is the density of hydrogen gas at $-10 ^{\circ}C$ and .95 atm ?

$R=.0821 \frac{(atm*L)}{(K*mol)}$

Possible Answers:

$88 \frac{mg}{L}$

$88 \frac{g}{L}$

$44\frac{mg}{L}$

$88 \frac{g}{ml}$

Correct answer:

$88 \frac{mg}{L}$

Explanation:

We will use the equation

PV=nRT

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atm*L)}{(K*mol)}$

is the temperature in Kelvin

We need to convert celsius to Kelvin

$-10^{\circ}C+273=263^{\circ}K$

Then, we will assume 1mol to make our calculations easier.

Rearrange for volume:

$\frac{nRT}{P}=V$

Plug in our values and solve

$\frac{1*.0821*263}{.95}=V$

V=22.7 L

Again, assume 1mol and multiply by the molar mass of hydrogen gas, $\frac{2g}{mol}$ :

$1mol*\frac{2gH_2}{1molH_2}=2 g$

We divide mass by volume to get density.

$D=\frac{2g}{22.7L}$

$D=88\frac{mg}{L}$

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Example Question #11 : Ideal Gas Law

What is the mass of 6 L of oxygen gas at $-10^{\circ}C$ and .95 atm ?

$R=.0821 \frac{(atm*L)}{(K*mol)}$

Possible Answers:

8.45g

4.22g

.263g

25.22g

Correct answer:

8.45g

Explanation:

We will use the equation:

PV=nRT

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atm*L)}{(K*mol)}$

is the temperature in Kelvin

Rearrange to solve for moles:

$\frac{PV}{RT}=n$

Plug in our values and solve

$\frac{.95*6}{.0821*263}=n$

.263moles=n

Then, we need to multiply by the molar mass of oxygen gas, which is diatomic.

$.263 moles O_2 * \frac{32g O_2}{mol} = 8.45 g$

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Example Question #51 : Thermodynamics

What is the density of neon gas at $-5^{\circ} C$ and .97 atm ?

$R=.0821 \frac{(atm*L)}{(K*mol)}$

Possible Answers:

$.885 \frac{g}{L}$

$37 \frac{g}{L}$

$29\frac{g}{L}$

$88\frac{g}{ml}$

Correct answer:

$.885 \frac{g}{L}$

Explanation:

We will use the equation:

PV=nRT

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atm*L)}{(K*mol)}$

is the temperature in Kelvin

First we need to convert our temperature to Kelvin:

273+(-5)=268K

We will assume 1mol to make our calculations easier.

Rearrange for volume:

$\frac{nRT}{P}=V$

Plug in our values and solve.

$\frac{1mol*.0821 \frac{(atm*L)}{(K*mol)}268^{\circ}K}{.97atm}=V$

22.7 L=V

Again, assume 1mol and multiply by the molar mass of neon $\frac{20.2g}{mol}$ :

$1molNe*\frac{20.2gNe}{molNe}=20.2 gNe$

We divide mass by volume to find density.

$D=\frac{M}{V}$

$D=\frac{20.2g}{22.7L}$

$.885\frac{g}{L}=D$

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Example Question #53 : Thermodynamics

What is the volume of 1 kg of gaseous water at $125^{\circ}C$ and 1.02 atm ?

$R=.0821 \frac{(atm*L)}{(K*mol)}$

Possible Answers:

1780L

55.6 L

2555 L

111 L

Correct answer:

1780L

Explanation:

We will use the equation

PV=nRT

Where:

is the pressure in atm

is the volume in liters

is the number of moles of gas

$R=.0821 \frac{(atm*L)}{(K*mol)}$

is the temperature in Kelvin

We need to convert the temperature to Kelvin

125C+273=398K

Rearrange our original equation for volume:

$\frac{nRT}{P}=V$

We need to find the moles of nitrogen gas. We divide the mass by the molar mass of water, which is $18 \frac{g}{mol}$ .

$\frac{1000 g}{18 \frac{g}{mol}}=55.6mol$

Plug in our values to our rearranged original equation and solve.

$\frac{55.6*0821*398}{1.02}=V$

V=1780 L

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Example Question #54 : Thermodynamics

A balloon filled is filled with pure nitrogen gas. The balloon is determined to have a volume of .75 L on a day when the temperature is $20^{\circ}C$ , and the air pressure is .85 atm .

How many nitrogen molecules are present?

Possible Answers:

None of these

$4.57\times10^{22}$

38.4

$3.33\times10^{22}$

$1.57\times 10^{22}$

Correct answer:

$1.57\times 10^{22}$

Explanation:

We will use our ideal gas equation.

PV=nRT

Where is the pressure

is the volume

is the number of moles

is the gas constant

is the temperature in Kelvin

We rearrange the equation to solve for n

$n=\frac{PV}{RT}$

A common mistake is using the wrong gas constant, . We need to use

$.0821 \frac{L\cdot atm}{K\cdot mol}$

We convert our temperture from Celsius to Kelvin

$20^{\circ}C+273=293^{\circ}K$

We plug in our values

$n=\frac{.85*.75}{.0821*293}$

n=.0265 moles

We then need to multiply by Avogadros number to convert to number of molecules.

$.0265moles\cdot \frac{6.02\times10^{23}molecules}{mole}=n$

$1.57 \times 10^{22} molecules=n$

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Example Question #11 : Ideal Gas Law

Which of the following graphs shows an incorrect relationship for an ideal gas?

Possible Answers:

Varsity tutors ideal gas question incorrect answer 3

Varsity tutors ideal gas question incorrect answer 1

Varsity tutors ideal gas question incorrect answer 2

Varsity tutors ideal gas question correct answer

Correct answer:

Varsity tutors ideal gas question correct answer

Explanation:

This question is asking for us to identify which graph shows an incorrect relationship between volume, pressure, and temperature of an ideal gas. To distinguish between which graphs are correct, and which one isn't, we'll need to use the ideal gas law.

PV=nRT

It's also important to note that each graph in the answer choices shows a linear, positive relationship between the x and y-variables.

The only graph that shows an incorrect relationship is the graph that has as the y-axis and $\frac{V}{T}$ as the x-axis. From the ideal gas law shown above, we would expect a positive linear relationship of a graph showing vs. $\frac{T}{V}$ . If we instead have vs. $\frac{V}{T}$ , the graph would have a negative relationship.

All of the other graphs shown in the remaining answer choices show correct relationships. vs. $\frac{T}{V}$ gives a positive relationship. Likewise, vs. gives a positive relationship. And finally, vs. $\frac{T}{P}$ also gives a positive relationship.

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Example Question #56 : Thermodynamics

15moles of N_2 gas are heated from $-2^\circ C$ to $25^\circ C$ and the pressure increases from .95atm to 1.05atm . Determine the volume in the final state.

$R=.0821\frac{L*atm}{mole*K}$

Possible Answers:

165L

230L

350L

800L

450L

Correct answer:

350L

Explanation:

Only the final state data will be needed Convert Celsius to Kelvin:

273+25=298

Use the ideal gas law:

PV=nRT

Where

is the pressure in atm

is the volume in

is the number of moles

is the gas constant, $.0821\frac{L*atm}{mole*K}$

is the temperature in

Plug in values and solve:

1.05*V=15*.0821*298

350L=V

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AP Physics 2 : Thermodynamics

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #2 : Ideal Gas Law

Example Question #11 : Ideal Gas Law

Example Question #51 : Thermodynamics

Example Question #11 : Ideal Gas Law

Example Question #11 : Ideal Gas Law

Example Question #51 : Thermodynamics

Example Question #53 : Thermodynamics

Example Question #54 : Thermodynamics

Example Question #11 : Ideal Gas Law

Example Question #56 : Thermodynamics

All AP Physics 2 Resources

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