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  4. Kinetic Theory Chapter-Wise Test 3

Kinetic Theory Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the volume of 0.3 moles of an ideal gas at 2.5 atm and 127°C? (\(R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}\))

\(PV = \mu R T\), \(V = \frac{\mu R T}{P}\).

\(T = 127 + 273 = 400 \, \text{K}\), \(P = 2.5 \times 1.01 \times 10^5 = 2.525 \times 10^5 \, \text{Pa}\).

\(V = \frac{0.3 \times 8.31 \times 400}{2.525 \times 10^5} = 3.95 \times 10^{-3} \, \text{m}^3 \approx 3.95 \, \text{litres}\).

3.0 litres
3.95 litres
4.5 litres
5.0 litres
2

A gas occupies 44.8 litres at STP. How many moles are present? (Molar volume at STP = 22.4 litres)

Number of moles (\(\mu\)) = \(\frac{\text{Volume}}{\text{Molar volume}}\).

\(\mu = \frac{44.8}{22.4} = 2.0 \, \text{mol}\).

1.0 mol
1.5 mol
2.0 mol
2.5 mol
3

A solid has a molar specific heat capacity of \(24.9 \, \text{J mol}^{-1} \text{K}^{-1}\). How many degrees of freedom per atom does it have? (\(R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}\))

\(C = f \times \frac{R}{2}\), \(24.9 = f \times \frac{8.31}{2}\).

\(f = \frac{24.9 \times 2}{8.31} \approx 5.99 \approx 6\).

4
5
6
7
3

The molar specific heat at constant pressure for a polyatomic gas with 2 vibrational modes is: (\(R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}\))

Polyatomic gas: 3 translational + 3 rotational + 2 vibrational modes.

Total degrees of freedom = \(3 + 3 + 2 \times 2 = 10\).

\(C_v = 5R\), \(C_p = C_v + R = 6R = 6 \times 8.31 = 49.86 \, \text{J mol}^{-1} \text{K}^{-1} \).

33.24 J mol⁻¹ K⁻¹
41.55 J mol⁻¹ K⁻¹
49.86 J mol⁻¹ K⁻¹
58.17 J mol⁻¹ K⁻¹
3

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

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

\(d^2 = \frac{1}{1.414 \times 3 \times 10^{25} \times 3.14 \times 3 \times 10^{-7}} = \frac{1}{4.0 \times 10^{-19}} = 2.5 \times 10^{-20}\).

\(d = \sqrt{2.5 \times 10^{-20}} \approx 1.58 \times 10^{-10} \, \text{m}\).

\(1.0 \times 10^{-10} \, \text{m}\)
\(1.58 \times 10^{-10} \, \text{m}\)
\(2.0 \times 10^{-10} \, \text{m}\)
\(2.5 \times 10^{-10} \, \text{m}\)
2

The rms speed of a gas is 350 m/s at 175 K. At what temperature will the rms speed be 700 m/s?

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

\(\frac{700}{350} = \sqrt{\frac{T_2}{175}}\), \(2 = \sqrt{\frac{T_2}{175}}\).

Square both sides: \(4 = \frac{T_2}{175}\), \(T_2 = 700 \, \text{K}\).

350 K
525 K
700 K
875 K
3

A gas at 2 atm and 300 K occupies 20 litres. If the pressure is increased to 4 atm at constant temperature, what is the new volume?

Boyle’s law: \(P_1 V_1 = P_2 V_2\).

\(P_1 = 2 \, \text{atm}\), \(V_1 = 20 \, \text{litres}\), \(P_2 = 4 \, \text{atm}\).

\(V_2 = \frac{P_1 V_1}{P_2} = \frac{2 \times 20}{4} = 10 \, \text{litres}\).

5 litres
10 litres
15 litres
20 litres
2

A gas at 3 atm and 300 K has a volume of 10 litres. If the temperature rises to 900 K at constant pressure, what is the new volume?

Charles’ law: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\).

\(V_1 = 10 \, \text{litres}\), \(T_1 = 300 \, \text{K}\), \(T_2 = 900 \, \text{K}\).

\(V_2 = V_1 \times \frac{T_2}{T_1} = 10 \times \frac{900}{300} = 30 \, \text{litres}\).

15 litres
20 litres
25 litres
30 litres
4

A mixture of 0.4 moles of helium and 0.6 moles of nitrogen is at 400 K in a 25-litre container. What is the total pressure? (\(R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}\))

\(PV = \mu R T\), \(P = \frac{\mu R T}{V}\).

Total moles = \(0.4 + 0.6 = 1.0\), \(V = 25 \times 10^{-3} \, \text{m}^3\).

\(P = \frac{1.0 \times 8.31 \times 400}{25 \times 10^{-3}} = 1.3284 \times 10^5 \, \text{Pa} \approx 1.33 \, \text{atm}\).

1.00 atm
1.33 atm
1.66 atm
2.00 atm
2

What is the time between collisions for a gas molecule with a mean free path of \(1.5 \times 10^{-7} \, \text{m}\) and average speed of 450 m/s?

\(\tau = \frac{l}{} = \frac{1.5 \times 10^{-7}}{450} = 3.33 \times 10^{-10} \, \text{s}\).

\(2.0 \times 10^{-10} \, \text{s}\)
\(3.33 \times 10^{-10} \, \text{s}\)
\(4.5 \times 10^{-10} \, \text{s}\)
\(6.0 \times 10^{-10} \, \text{s}\)
2

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