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AP Physics 2 : Thermodynamics

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6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Kinetic Molecular Theory

What is the ratio of the rate of effusion of carbon dioxide (CO_2) to nitrogen (N_2) ?

Possible Answers:

1.25

1.0

1.57

0.80

0.64

Correct answer:

0.80

Explanation:

We can use Graham's law to solve this problem, which states that the ratio of effusion between to gasses is the square root of the inverse of the gases' molecular masses:

$\frac{effusion_a}{effusion_b} = \sqrt{\frac{M_B}{M_A}}$

Therefore, in this problem we get:

$\frac{effusion_{CO_2}}{effusion_{N_2}} = \sqrt{\frac{M_{N_2}}{M_{CO_2}}}$

Now we just need to determine the molecular masses of the two gasses, starting with carbon dioxide:

$M_{CO_2} = M_C + 2\cdot M_O = 12.011+2(15.999) = 44$

$M_{N_2} = 2\cdot M_N = 2(14.007) = 28$

Plugging these in:

$\frac{effusion_{CO_2}}{effusion_{N_2}} = \sqrt{\frac{28}{44}} = 0.80$

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Example Question #1 : Properties Of Ideal Gases

Which of the following is not a property of an ideal gas?

Possible Answers:

The gas particles have strong intermolecular forces acting on them

All of these answers are properties of ideal gases

The gas particles have elastic collisions

The gas particles have no volume

The gas particles have almost no mass

Correct answer:

The gas particles have strong intermolecular forces acting on them

Explanation:

Ideal gases have particles that are considered to be point masses. This means that all of their (extremely small) mass is contained in a single, infinitesimal point of zero volume. These particles experience elastic collisions, which means they lose no kinetic energy when they collide with other particles. If the particles experienced strong intermolecular forces, then they couldn't have elastic collisions, so they wouldn't be ideal gases anymore. Ideal gases don't experience any intermolecular forces.

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Example Question #2 : Properties Of Ideal Gases

A real gas becomes more like an ideal gas at __________ temperatures and __________ pressures.

Possible Answers:

lower . . . higher

higher . . . lower

lower . . . lower

Real gases can never be made to act more like ideal gases

higher . . . higher

Correct answer:

higher . . . lower

Explanation:

When the temperature is higher, the kinetic energy of particles is higher. Because their energy is higher, they bounce more energetically, letting them ignore most intermolecular forces (which is something ideal gases don't experience).

An important assumption for ideal gases is that the volume of the particles is negligible compared to the volume of their container. For real gases, as pressure increases, the particles get closer together, and their volume gets less negligible when compared to the volume of the container. This is why real gases behave more ideally at lower pressures.

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Example Question #3 : Properties Of Ideal Gases

What is the temperature in Kelvin for 1mol of gas at 5atm and a volume of 10L?

$1{ atm}= 101.3{ kPa}= 101325 \frac{{N}}{{m}^2}$

$R= 8.314{ }\frac{{L}\cdot {kPa}}{{mol} \cdot {K}}$

Possible Answers:

151{ K}

713{ K}

1214{ K}

228{ K}

609{ K}

Correct answer:

609{ K}

Explanation:

Use the ideal gas law equation.

PV=nRT

Manipulate the equation, substitute the givens, and solve for temperature. Write the gas constants necessary to solve the problem.

$T=\frac{PV}{nR}$

$1{ atm}= 101.3{ kPa}= 101325 \frac{{N}}{{m}^2}$

$R= 8.314{ }\frac{{L}\cdot {kPa}}{{mol} \cdot {K}}$

$T=\frac{PV}{nR}= \frac{(5{ atm})(\frac{101.3{ kPa}}{1{ atm}})(10{ L})}{(1{ mol})(8.314{ }\frac{{L}\cdot {kPa}}{{mol} \cdot {K}})}$

T=609.213 { K}

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Example Question #2 : Properties Of Ideal Gases

Which of the following is the correct formula to find the internal energy of an ideal gas?

Possible Answers:

$U=\frac{1}{2}kT^2$

$U=\frac{1}{2}NkT$

$U=\frac{3}{2}NkT$

$U=\sqrt{\frac{3kT}{M}}$

$U=\frac{3}{2}kT$

Correct answer:

$U=\frac{3}{2}NkT$

Explanation:

An ideal gas has no molecular interactions besides perfectly elastic collisions. The energy is the combined kinetic energies of ideal gas molecules since there is no potential energy. For point molecules of the monatomic gas, the equation can be written as:

$U=\frac{3}{2}NkT$

Note that $K_{{avg}}=\frac{3}{2}kT$ is the equation for average kinetic energy also in Joules.

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Example Question #5 : Properties Of Ideal Gases

We have 10L of oxygen gas at $19^\circ C$ . We increase the temperature to $38 ^\circ C$ and keep the pressure the same. What is the new volume?

Possible Answers:

20 L

7.5 L

6.8 L

10.65 L

Correct answer:

10.65 L

Explanation:

We will use the relationship:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Rearrange for the final volume:

$\frac{V_1 T_2}{T_1}={V_2}$

Convert our temperatures to kelvin.

$19^{\circ}C+273=292^{\circ}K$

$38^{\circ}C+273=311^{\circ}K$

Plug in our values and solve.

$\frac{V_1 T_2}{T_1}={V_2}$

$\frac{10L* 292^{\circ}K}{311^{\circ}K}={V_2}$

10.65 L=V_2

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Example Question #1 : Properties Of Ideal Gases

We have 15 L of CO_2 gas at $10 ^\circ C$ . We increase the temperature to $30 ^\circ C$ , while keeping the pressure constant. What is the new volume?

Possible Answers:

5 L

45 L

16.1 L

6.6 L

Correct answer:

16.1 L

Explanation:

We will use the relationship:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Rearrange to solve for the final volume:

$\frac{V_1 T_2}{T_1}={V_2}$

Convert our temperatures to kelvin:

$10^{\circ}C+273=283^{\circ}K$

$30^{\circ}C+273=303^{\circ}K$

Plug in our values and solve.

$\frac{15L* 303^{\circ}K}{283^{\circ}K}={V_2}$

16.1 L=V_2

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Example Question #2 : Properties Of Ideal Gases

We have 13 L of helium gas at 1.3 atm . We decrease the pressure to .85 atm , while keeping the temperature the same. What is the new volume?

Possible Answers:

13 L

8.5 L

19.9 L

22.4 L

Correct answer:

19.9 L

Explanation:

We will use the relationship:

P_1V_1=P_2V_2

Rearrange to solve for the final volume:

$\frac{P_1V_1}{P_2}=V_2$

Plug in our values and solve.

$\frac{1.3atm*13L}{.85atm}=V_2$

19.9 L=V_2

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Example Question #1 : Properties Of Ideal Gases

We have 6 L of neon gas at .5 atm . We increase the pressure to 1.5 atm while keeping the temperature constant. What is the new volume?

Possible Answers:

18 L

3 L

2 L

6 L

Correct answer:

2 L

Explanation:

We will use the relationship:

P_1V_1=P_2V_2

Rearrange to solve for the final volume:

$\frac{P_1V_1}{P_2}=V_2$

Plug in our values and solve.

$\frac{.5atm*6L}{1.5atm}=V_2$

2L=V_2

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Example Question #1 : Properties Of Ideal Gases

We have 16 L of neon gas. We increase the temperature from $10 ^\circ C$ to $50 ^\circ C$ , and increase the pressure from 1 atm to 5 atm . What will be the new volume?

Possible Answers:

8 L

16 L

80 L

3.65 L

Correct answer:

3.65 L

Explanation:

First, we will convert celsius to Kelvin.

$10^{\circ}C+273=283^{\circ}K$

$50^{\circ}C+273=323^{\circ}K$

Then, we will use the relationship:

$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$

Solve for volume final volume:

$\frac{P_1 V_1 T_2}{T_1 P_2}=V_2$

Plug in our values and solve.

$\frac{1atm* 16L*323^{\circ}K}{283^{\circ}K *5atm}=V_2$

3.65 L=V_2

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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 #1 : Kinetic Molecular Theory

Example Question #1 : Properties Of Ideal Gases

Example Question #2 : Properties Of Ideal Gases

Example Question #3 : Properties Of Ideal Gases

Example Question #2 : Properties Of Ideal Gases

Example Question #5 : Properties Of Ideal Gases

Example Question #1 : Properties Of Ideal Gases

Example Question #2 : Properties Of Ideal Gases

Example Question #1 : Properties Of Ideal Gases

Example Question #1 : Properties Of Ideal Gases

All AP Physics 2 Resources

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