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Physics MCQs for NEET — Practice Questions with Answers

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The temperature of 5 moles of a gas which was held at constant volume was changed from $100 ° C$ to $120 ° C$ . The change in internal energy was found to be 80 Joules. The total heat capacity of the gas at constant volume will be equal to

8 J/K

0.8 J/K

4.0 J/K

0.4 J/K

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Explanation

$3 . {(∆ Q)}_{V} = {μC}_{V} ∆ T = w ∆ T (W = Heat capacity) \Rightarrow W = \frac{{(∆ Q)}_{V}}{∆ T} = \frac{∆ U}{∆ T} [∵ {(∆ Q)}_{V} = ∆ U] \Rightarrow W = \frac{80}{(120 - 100)} = 4 J / K .$

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The temperature at which the r.m.s. speed of hydrogen molecules is equal to escape velocity on earth surface, will be

1060 K

5030 K

8270 K

10063 K

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Explanation

$4 . Escape velocity from the earth' s surface is 11.2 km / \sec . So, v_{rms} = v_{escape} = \sqrt{\frac{3 RT}{M}} \Rightarrow T = \frac{{(V_{escape})}^{2} \times M}{3 R} = \frac{{(11.2 \times 10^{3})}^{2} \times (2 \times 10^{- 3})}{3 \times 8.31} = 10063 K$

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Inside a cylinder having insulating walls and closed at ends is a movable piston, which divides the cylinder into two compartments. On one side of the piston is a mass m of a gas and on the other side a mass 2 m of the same gas. What fraction of volume of the cylinder will be occupied by the larger mass of the gas when the piston is in equilibrium ? Consider that the movable piston is conducting so that the temperature is the same throughout

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{2}{3}$

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The diameter of oxygen molecules is $2.94 \times 10^{- 10} m .$ The Vander Waal's gas constant 'b' in $m^{3} / mol$ will be

$3.2$

$16$

$32 \times 10^{- 4}$

$32 \times 10^{- 6}$

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Explanation

$4 . Vander Waal' s gas constant b = 4 \times total volume of all the molecules of the gas in the enclosure or b = 4 \times N \times \frac{4}{3} π {(\frac{d}{2})}^{3} = \frac{2}{3} {πNd}^{3} = \frac{2}{3} \times 3.14 \times 6.02 \times 10^{23} \times {(2.94 \times 10^{\times - 10})}^{3} = 32 \times 10^{- 6} \frac{m^{3}}{mol}$

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The temperature of the mixture of one mole of helium and one mole of hydrogen is increased from $0 ° C$ to $100 ° C$ at constant pressure. The amount of heat delivered will be

600 cal

1200 cal

1800 cal

3600 cal

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Explanation

$2 . {(C_{P})}_{mix} = \frac{μ_{1} C_{P_{1}} + μ_{2} C_{P_{2}}}{μ_{1} + μ_{2}} (C_{P_{1}} (He) = \frac{5}{2} R and C_{P_{2}} (H_{2}) = \frac{7}{2} R) = \frac{1 \times \frac{5}{2} R + 1 \times \frac{7}{2} R}{1 + 1} = 3 R = 3 \times 2 = 6 cal / mol ° C ∴ Amount of heat needed to raise the temperature from 0 ° C to 100 ° C {(∆ Q)}_{P} = {μC}_{P} ∆ T = 2 \times 6 \times 100 = 1200 cal$

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A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per $O_{2}$ molecule to that per $N_{2}$ molecule is

1 : 1

1 : 2

2 : 1

Depends on the moments of inertia of the two molecules

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Explanation

1. Average kinetic energy per molecule per degree of freedom $= \frac{1}{2} kT$ . Since both the gases are

diatomic and at same temperature (300 K), both will have the same number of rotational degree of

freedom i.e. two. Therefore, both the gases will have the same average rotational kinetic energy per

molecule $(= 2 \times \frac{1}{2} kT = kT)$ . Thus $\frac{E_{1}}{E_{2}} = \frac{1}{1}$ .

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A gas mixture consists of 2 mole of oxygen and 4 mole of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

4 RT

15 RT

9 RT

11 RT

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Explanation

$4 . Total internal energy of system = U_{oxygen} + U_{argon} = μ_{1} \frac{f_{1}}{2} RT + μ_{2} \frac{f_{2}}{2} RT = 2 \frac{5}{2} RT + 4 \frac{3}{2} RT = 5 RT + 6 RT = 11 RT [As f_{1} = 5 (for oxygen) and f_{2} = 3 (for argon)]$

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A jar contains a gas and few drops of water at T K. The pressure in the jar is 830 mm of mercury. The temperature of jar is reduced by 1%. The saturated vapour pressure of water at the two temperatures are 30 mm and 25 mm of mercury. Then the new pressure in the jar will be

917 mm of Hg

717 mm of Hg

817 mm of Hg

None of these

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Explanation

$3 . At T K, pressure of gas (P) in the jar pressure = Total pressure - saturated vapour pressure \Rightarrow P = (830 - 30) = 800 mm of Hg New temperature T' = (T - \frac{T}{100}) = \frac{99 T}{100} Using Charle' s law \frac{P}{T} = \frac{P'}{T} \Rightarrow P' = \frac{PT'}{T} = \frac{800 \times 99 T}{100 T} = 792 mm of Hg Saturated vapour pressure at T' = 25 mm of Hg ∴ Total pressure in the jar = Actual pressure of gas + Saturated vapour pressure = 792 + 25 = 817 mm of Hg .$

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Molar specific heat of oxygen at constant pressure $C_{P} = 7.2 cal / mol ° C and R = 8.3 joule / mol / K .$ At constant volume, 5 mol of oxygen is heated from $10 ° C to 20 ° C,$ the quantity of heat required is approximately

$25 cal$

$50 cal$

$250 cal$

$500 cal$

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Explanation

$3 . Given R = 8.3 \frac{Joule}{mol K} = \frac{8.3}{4.2} \frac{cal}{mol K} = 2 \frac{cal}{mol K} C_{V} = C_{P} - R = (7.2 - 2) \approx 5 cal / mol l ° C At constant volume required heat {(∆ Q)}_{V} = {μC}_{V} ∆ T = 5 \times 5 \times 10 = 250 cal$

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One mole of an ideal gas requires 207 J heat to raise the temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10 K, the heat required is

198.7 J

29 J

215.3 J

124 J

(Given the gas constant $R = 8.3 J / mol ‐ K$ )

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Explanation

$4 . {(∆ Q)}_{P} = {μC}_{P} ∆ T \Rightarrow 207 = 1 \times C_{P} \times 10 \Rightarrow C_{P} = 20.7 \frac{Joule}{mol ‐ K} . Also C_{P} - C_{V} = R \Rightarrow C_{V} = C_{P} - R = 20.7 - 8.3 = 12.4 \frac{Joule}{mole ‐ K} So, {(∆ Q)}_{V} = {μC}_{V} ∆ T = 1 \times 12.4 \times 10 = 124 J .$

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