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AP Physics 2 : Kinetic Molecular Theory

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Example Questions

Example Question #11 : Thermodynamics

What is the average molecular kinetic energy of a gas at a temperature of $27\ ^{\circ}C$ ?

$k = 1.38*10^{-23}\ \frac{\textup{J}}{\textup{K}}$

Possible Answers:

$6.21*10^{-21}\textup{ J}$

$4.6*10^{-23}\textup{ J}$

$9.8*10^{-21}\textup{ J}$

$3.5*10^{-22}\textup{ J}$

$7.2*10^{-22}\textup{ J}$

Correct answer:

$6.21*10^{-21}\textup{ J}$

Explanation:

We know that the expression for average kinetic energy is:

$K_{avg}=\frac{1}{2}mv_{avg}^2$

Where average velocity of a gas is:

$v_{avg} = \sqrt{\frac{3kT}{m}}$

Plugging this into our expression for kinetic energy, we get:

$K_{avg} = \frac{1}{2}m\left ( \frac{3kT}{m}\right ) = \frac{3}{2}kT$

Plugging in our values, we get:

$K_{avg} = \frac{3}{2}\left ( 1.38x10^{-23}\right )(300K)$

$K_{avg} = 6.21*10^{-21}J$

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Example Question #13 : Thermodynamics

A mixture of methane (CH_4) and oxygen (O_2) is held at a constant temperature of $150\ ^{\circ}C$ and a pressure of $0.8\textup{ atm}$ . What is the ratio of the average velocity of methane to that of oxygen?

Possible Answers:

$\sqrt{\frac{1}{2}}$

$\sqrt{2}$

$\frac{1}{2}$

Correct answer:

$\sqrt{2}$

Explanation:

For gases at equal temperatures, we know that they both have the same average kinetic energy:

$K_{CH_{4}} = K_{O_{2}}$

$\frac{1}{2}M_{CH_{4}}V_{CH_{4}}^2 = \frac{1}{2}M_{O_2}V_{O_2}^2$

Note that is the molecular weight of each molecule, and is the average velocity of each molecule. Then rearranging for the ratio of velocities:

$\frac{V_{CH_{4}}^2}{V_{O_2}^2} = \frac{M_{O_2}}{M_{CH_{4}}}$

Then taking the square root of each side:

$\frac{V_{CH_{4}}}{V_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{CH_{4}}}}$

Now all we need is the molecular weight of each molecule:

$M_{O_{2}} = 2\cdot M_O = 2(16) = 32$

$M_{CH_{4}} = M_C+4\cdot M_H = 12 + 4(1) = 16$

Plugging these in, we get:

$\frac{V_{CH_{4}}}{V_{O_2}} = \sqrt{\frac{32}{16}}=\sqrt{2}$

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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:

1.0

1.57

0.64

1.25

0.80

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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AP Physics 2 : Kinetic Molecular Theory

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #11 : Thermodynamics

Example Question #13 : Thermodynamics

Example Question #1 : Kinetic Molecular Theory

All AP Physics 2 Resources

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