Kinetic Molecular Theory - AP Physics 2

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Question

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

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

Answer

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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Question

A mixture of methane and oxygen 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?

Answer

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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Question

What is the ratio of the rate of effusion of carbon dioxide to nitrogen ?

Answer

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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