Behaviour of Perfect Gas and Kinetic Theory Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A gas has a molar specific heat at constant volume of \(20.8 \, \text{J mol}^{-1} \text{K}^{-1}\). What is its \(C_p\)? (\(R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}\))

\(C_p = C_v + R\).

\(C_p = 20.8 + 8.31 = 29.11 \, \text{J mol}^{-1} \text{K}^{-1} \approx 29.1 \, \text{J mol}^{-1} \text{K}^{-1}\).

25.0 J mol⁻¹ K⁻¹
29.1 J mol⁻¹ K⁻¹
33.2 J mol⁻¹ K⁻¹
37.5 J mol⁻¹ K⁻¹
2

A gas mixture has equal masses of helium and oxygen. What is the ratio of their number of molecules?

Let mass = \(m\). \(N = \frac{m}{M} \times N_A\).

For He (\(M = 4\)): \(N_{\text{He}} = \frac{m}{4} \times N_A\).

For \(O_2\) (\(M = 32\)): \(N_{\text{O}_2} = \frac{m}{32} \times N_A\).

Ratio \(\frac{N_{\text{He}}}{N_{\text{O}_2}} = \frac{\frac{m}{4}}{\frac{m}{32}} = \frac{32}{4} = 8:1\).

1:8
4:1
8:1
1:4
3

A gas has a volume of 22.4 litres at STP. How many moles are present if the temperature is raised to 546 K at constant pressure?

At STP, 22.4 litres = 1 mole.

Charles’ law: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), but moles remain constant at constant \(P\).

Initial \(\mu = 1 \, \text{mol}\), remains 1 mole as \(V\) adjusts with \(T\).

0.5 mol
1.0 mol
2.0 mol
4.0 mol
2

A gas has a \(C_v\) of \(27.0 \, \text{J mol}^{-1} \text{K}^{-1}\). What is its \(C_p\)? (\(R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}\))

\(C_p = C_v + R\).

\(C_p = 27.0 + 8.31 = 35.31 \, \text{J mol}^{-1} \text{K}^{-1} \approx 35.3 \, \text{J mol}^{-1} \text{K}^{-1}\).

31.2 J mol⁻¹ K⁻¹
33.2 J mol⁻¹ K⁻¹
35.3 J mol⁻¹ K⁻¹
37.4 J mol⁻¹ K⁻¹
3

What is the collision frequency of a gas molecule with a mean free path of \(1.2 \times 10^{-7} \, \text{m}\) and average speed of 480 m/s?

Collision frequency = \(\frac{}{l}\).

\(\frac{480}{1.2 \times 10^{-7}} = 4.0 \times 10^9 \, \text{s}^{-1}\).

\(2.0 \times 10^9 \, \text{s}^{-1}\)
\(3.0 \times 10^9 \, \text{s}^{-1}\)
\(4.0 \times 10^9 \, \text{s}^{-1}\)
\(5.0 \times 10^9 \, \text{s}^{-1}\)
3

The rms speed of a gas is 550 m/s at 275 K. At what temperature will the rms speed be 1100 m/s?

\(v_{\text{rms}} \propto \sqrt{T}\), \(\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}\).

\(\frac{1100}{550} = \sqrt{\frac{T_2}{275}}\), \(2 = \sqrt{\frac{T_2}{275}}\).

Square both sides: \(4 = \frac{T_2}{275}\), \(T_2 = 1100 \, \text{K}\).

550 K
825 K
1100 K
1375 K
3

A gas has a \(C_v\) of \(17.1 \, \text{J mol}^{-1} \text{K}^{-1}\). What is its \(C_p\)? (\(R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}\))

\(C_p = C_v + R\).

\(C_p = 17.1 + 8.31 = 25.41 \, \text{J mol}^{-1} \text{K}^{-1} \approx 25.4 \, \text{J mol}^{-1} \text{K}^{-1}\).

20.8 J mol⁻¹ K⁻¹
25.4 J mol⁻¹ K⁻¹
29.1 J mol⁻¹ K⁻¹
33.2 J mol⁻¹ K⁻¹
2

What is the temperature at which the rms speed of nitrogen molecules is 1032 m/s? (Molecular mass of \(N_2 = 28 \, \text{u}\), \(k_B = 1.38 \times 10^{-23} \, \text{J K}^{-1}\))

\(v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}\), \(m = \frac{28 \times 10^{-3}}{6.02 \times 10^{23}} = 4.65 \times 10^{-26} \, \text{kg}\).

\(1032^2 = \frac{3 \times 1.38 \times 10^{-23} \times T}{4.65 \times 10^{-26}}\), \(T = \frac{1.065 \times 10^6 \times 4.65 \times 10^{-26}}{4.14 \times 10^{-23}} \approx 1196 \, \text{K}\).

900 K
1200 K
1500 K
1800 K
2

The mean free path of a gas is \(1.2 \times 10^{-7} \, \text{m}\) with a number density of \(4.0 \times 10^{25} \, \text{m}^{-3}\). What is the molecular diameter?

\(l = \frac{1}{\sqrt{2} n \pi d^2}\), \(d^2 = \frac{1}{\sqrt{2} n \pi l}\).

\(d^2 = \frac{1}{1.414 \times 4.0 \times 10^{25} \times 3.14 \times 1.2 \times 10^{-7}} = \frac{1}{2.13 \times 10^{-19}} \approx 4.69 \times 10^{-20}\).

\(d = \sqrt{4.69 \times 10^{-20}} \approx 2.17 \times 10^{-10} \, \text{m}\).

\(1.5 \times 10^{-10} \, \text{m}\)
\(2.0 \times 10^{-10} \, \text{m}\)
\(2.17 \times 10^{-10} \, \text{m}\)
\(2.5 \times 10^{-10} \, \text{m}\)
3

What is the average translational kinetic energy of a nitrogen molecule (\(N_2\)) at 300 K? (\(k_B = 1.38 \times 10^{-23} \, \text{J K}^{-1}\))

Average translational KE per molecule = \(\frac{3}{2} k_B T\).

Substitute: \(\frac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \, \text{J} \).

\(4.14 \times 10^{-21} \, \text{J}\)
\(6.21 \times 10^{-21} \, \text{J}\)
\(8.28 \times 10^{-21} \, \text{J}\)
\(12.42 \times 10^{-21} \, \text{J}\)
2

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