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Kinetic Theory of Gases MCQs for NEET — Physics Questions with Answers

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N molecules each of mass m of gas A and 2N molecules each of mass 2m of gas B are contained in the same vessel at temperature T. The mean square of the velocity of molecules of gas B is $v^{2}$ and the mean square of x component of the velocity of molecules of gas A is $w^{2}$ . The ratio is $\frac{w^{2}}{v^{2}}$ is

$1$

$2$

$\frac{1}{3}$

$\frac{2}{3}$

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Explanation

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A gas is filled in a cylinder, its temperature is increased by 20% on Kelvin scale and volume is reduced by 10%. How much percentage of the gas will leak out

30%

40%

15%

25%

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Explanation

$4 . Let initial conditions = V . T and final conditions = V' . T' By Charle' s law V \propto T [P remains constant] \frac{V}{T} = \frac{V'}{T'} \Rightarrow \frac{V}{T} = \frac{V'}{1.2 T'} \Rightarrow V' = 1.2 V But as per question, volume is reduced by 10 % means V = 0.9 V So percentage of volume leaked out = \frac{(1.2 - 0.9) V}{1.2 V} \times 100 = 25 %$

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The air density at Mount Everest is less than that at the sea level. It is found by mountaineers that for one trip lasting a few hours, the extra oxygen needed by them corresponds to 30,000 cc at sea level (pressure 1 atmosphere, temperature 27°C). Assuming that the temperature around Mount Everest is –73°C and that the oxygen cylinder has capacity of 5.2 litre, the pressure at which $O_{2}$ be filled (at site) in cylinder is

3.86 atm

5.00 atm

5.77 atm

1 atm

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Explanation

$1 . At sea level PV = μRT \Rightarrow 1 \times (30 litre) = μ \times (0.0821) \times 300 \Rightarrow Number of moles μ = 1.22 (Remember R = 0.0821 \frac{atm - litre}{mole - K}) On Everest P \times 5.2 = 1.22 \times 0.0821 \times 200 \Rightarrow P = 3.86 atm$

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$\frac{1}{2}$ mole of helium gas is contained in a container at S.T.P. The heat energy needed to double the pressure of the gas, keeping the volume constant (specific heat of the gas $= 3 J {gm}^{- 1} K^{- 1}$ ) is

3276 J

1638 J

819 J

409.5 J

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Explanation

$2 . Molecular mass of He; M = 4 gm \Rightarrow Molar value of C_{V} = {Mc}_{V} = 4 \times 3 = 12 \frac{J}{mole - kelvin} At constant volume P \propto T therefore on doubling the pressure temperature also doubles i . e ., T_{2} = 2 T_{1} \Rightarrow ∆ T = T_{2} - T_{1} = 273 K Also {(∆ Q)}_{V} = {μC}_{V} ∆ T = \frac{1}{2} \times 12 \times 273 = 1638 J$

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The equation of state of a gas is given by $(P + \frac{{aT}^{2}}{V}) V^{c} = (RT + b),$ where a, b, c and R are constants. The isotherms can be represented by $P = {AV}^{m} - {BV}^{n},$ where A and B depend only on temperature and

$m = - c and n = - 1$

$m = c and n = 1$

$m = - c and n = 1$

$m = c and n = - 1$

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Explanation

$1 . (P + \frac{{aT}^{2}}{V}) V^{c} = (RT + b) \Rightarrow P = (RT + b) V^{- c} - ({aT}^{2}) V^{- 1} Comparing this equation with P = {AV}^{m} - {BV}^{n} we get m = - c and n = - 1$

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70 calories of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 30°C to 35°C. The amount of heat required to raise the temperature of same gas through the same range (30°C to 35°C) at constant volume (R = 2 cal/mol/K)

30 cal

50 cal

70 cal

90 cal

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Explanation

$2 . {(∆ Q)}_{P} = {μC}_{P} ∆ T \Rightarrow 2 \times C_{P} \times (35 - 30) \Rightarrow C_{P} = 7 \frac{cal}{mole ‐ K} ∵ C_{P} - C_{V} = R \Rightarrow C_{V} = C_{P} - R = 7 - 2 = 5 \frac{cal}{mole ‐ kevin} ∴ {(∆ Q)}_{V} = {μC}_{V} ∆ T = 2 \times 5 \times (35 - 30) = 50 cal .$

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A closed compartment containing gas is moving with some acceleration in horizontal direction. Neglect effect of gravity. Then the pressure in the compartment is

Same everywhere

Lower in the front side

Lower in the rear side

Lower in the upper side

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Three closed vessels A, B and C are at the same temperature T and contain gases which obey the Maxwellian distribution of velocities. Vessel A contains only $O_{2}, B$ only $N_{2}$ and C a mixture of equal quantities of $O_{2}$ and $N_{2}$ . If the average speed of the $O_{2}$ molecules in vessel A is $V_{1},$ that of the $N_{2}$ molecules in vessel B is $V_{2},$ the average speed of the $O_{2}$ molecules in vessel C is

$(V_{1} + V_{2}) / 2$

$V_{1}$

${(V_{1} V_{2})}^{1 / 2}$

$\sqrt{3 kT / M}$

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Explanation

2. Average speed of gas molecules is $\sqrt{\frac{8 kT}{π m}} .$ It depends on temperature and molecules mass. So the

average speed of $O_{2}$ will be same in (A) and (C).

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A box containing N molecules of a perfect gas at temperature $T_{1}$ and pressure $P_{1}$ . The number of molecules in the box is doubled keeping the total kinetic energy of the gas same as before. If the new pressure is $P_{2}$ and temperature $T_{2}$ , then

$P_{2} = P_{1}, T_{2} = T_{1}$

$P_{2} = P_{1}, T_{2} = \frac{T_{1}}{2}$

$P_{2} = 2 P_{1}, T_{2} = T_{1}$

$P_{2} = 2 P_{1}, T_{2} = \frac{T_{1}}{2}$

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Explanation

$2 . E_{Total} = \frac{f}{2} NkT \Rightarrow T \propto \frac{1}{N} [∵ E_{Total}, f, k = constant] \Rightarrow \frac{T_{2}}{T_{1}} = \frac{N_{1}}{N_{2}} = \frac{1}{2} \Rightarrow T_{2} = \frac{T_{1}}{2} Also from PV = NkT \Rightarrow P \propto NT \Rightarrow \frac{P_{1}}{P_{2}} = \frac{N_{1}}{N_{2}} . \frac{T_{1}}{T_{2}} = \frac{1}{2} \times \frac{2}{1} = \frac{1}{1} .$

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Two containers of equal volume contain the same gas at pressures $P_{1}$ and $P_{2}$ and absolute temperatures $T_{1}$ and $T_{2}$ respectively. On joining the vessels, the gas reaches a common pressure P and common temperature T. The ratio P/T is equal to

$\frac{P_{1}}{T_{1}} + \frac{P_{2}}{T_{2}}$

$\frac{P_{1} T_{1} + P_{2} T_{2}}{{(T_{1} + T_{2})}^{2}}$

$\frac{P_{1} T_{2} + P_{2} T_{1}}{{(T_{1} + T_{2})}^{2}}$

$\frac{P_{1}}{2 T_{1}} + \frac{P_{2}}{2 T_{2}}$

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