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Physics MCQs for NEET — Practice Questions with Answers

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An ideal gas is found to obey an additional law $V P^{2}$ = constant. The gas is initially at temperature T and volume V. When it expands to a volume 2 V, the temperature becomes:

T $\sqrt{2}$

2T $\sqrt{2}$

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Explanation

(a)

$V P^{2} = c o n s \tan t, P V = R T ∴ V \frac{R^{2} T^{2}}{V^{2}} = c o n s \tan t ∴ \frac{T^{2}}{V} = c o n s \tan t \frac{T^{2}}{V} = \frac{T'^{2}}{2 V} = T'^{2} = 2 T^{2} = T' = \sqrt{2} T$

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The root-mean-square speed of hydrogen molecules at 300 K is 1930 m/s. Then the root mean square speed of oxygen molecules at 900 K will be: [MH CET 2002; Odisha JEE 2010]

1930 $\sqrt{3}$ m/s

836 m/s

643 m/s

$\frac{1930}{\sqrt{3}}$ m/s

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Explanation

(b)

$V_{r m s} = \sqrt{\frac{3 R T}{M}} \Rightarrow \frac{{(V_{r m s})}_{O_{2}}}{{(V_{r m s})}_{H_{2}}} = \sqrt{\frac{T_{O_{2}}}{T_{H_{2}}} \times \frac{M_{H_{2}}}{M_{O_{2}}}} = \sqrt{\frac{900}{300} \times \frac{2}{32}} {(V_{r m s})}_{O_{2}} = 836 m / s$

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What is the mass of 2 liters of nitrogen at 22.4 atmospheric pressure and 273 K?

28 g

14 $\times 22.4$ g

56 g

None of these

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Explanation

(c)

22.4 litres of nitrogen at 1 atmosphere or 1 litre at 22.4 atmospheric pressure has a mass of 1 gm mole or 28 g. Therefore, 2 litres at 22.4 atmospheres have a mass of 56 gm.

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Mean kinetic energy per degree of freedom of gas molecules is: [ MP PET 1995; RPET 1999, 2001, 03; Haryana CEET 2000; DCE 2009]

$\frac{3}{2} k T$

$\frac{1}{2} k T$

$\frac{3}{2} R T$

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Explanation

(c)

According to the law of equipartition of energy, kinetic energy per degree of freedom of a gas molecule is $\frac{1}{2} k T$ .

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The root mean square speed of oxygen molecules (O2) at a certain absolute temperature is v. If the temperature is doubled and oxygen gas dissociates into oxygen atom, the rms speed would be :

$v$

$\sqrt{2} v$

$2 v$

$2 \sqrt{2} v$

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Explanation

(c)

$V_{rms} = \sqrt{\frac{3 RT}{M}} V = \sqrt{\frac{3 RT}{M}} V' = \sqrt{\frac{3 R x 2 T}{M / 2}} = \sqrt{\frac{3 RT}{M}} x 2 V' = 2 v$

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If P is the pressure of the gas then the KE per unit volume of the gas is:

$\frac{P}{2}$

$P$

$\frac{3 P}{2}$

$2 P$

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Explanation

(c)

$P = \frac{1}{3} \frac{mn v^{2}}{V} \frac{(K . V .)}{V} = \frac{3 P}{2}$

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$n_{1}$ mole of monoatomic gas is mixed with $n_{2}$ mole of diatomic gas such that $γ_{m i x} = 1.5$

$n_{1} = 2 n_{2}$

$n_{2} = 2 n_{1}$

$n_{1} = n_{2}$

$n_{1} = 3 n_{2}$

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Explanation

(c)

For a mixture,

$γ_{m i x} = \frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}}$

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$Molar heat capacity of the process P = \frac{a}{T} for a monoatomic gas,' a' being positive constant i s -$

$\frac{7}{2} R$

$\frac{5}{2} R$

$\frac{3}{2} R$

None of these

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Explanation

$(1) PT = a P \frac{PV}{nR} = a P^{2} V = constant {PV}^{1 / 2} = constant {PV}^{x} = constant . . . . P o l y t r o p i c p r o c e s s w i t h x = \frac{1}{2} C = \frac{R}{γ - 1} + \frac{R}{1 - x} \frac{R}{(\frac{5}{3} - 1)} + \frac{R}{1 - \frac{1}{2}} = \frac{7}{2} R$

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Two chambers containin m₁g and m₂g of a gas at pressure P₁ and P₂ respectively are put in contact with each other. If temperature remains constant, the common pressure reached will be

$\frac{P_{1} P_{2} (m_{1} + m_{2})}{(P_{2} m_{1} + m_{2} P_{1})}$

$\frac{(m_{1} + m_{2})}{(P_{2} m_{1} + m_{2} P_{1})}$

$\frac{P_{1} P_{2}}{(P_{2} m_{1} + m_{2} P_{1})}$

$\frac{P_{1} P_{2}}{(m_{1} + m_{2})}$

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Explanation

$(1) According to Boyle' s law \frac{P}{ρ} = const . ∴ V_{1} = \frac{m_{1}}{ρ_{1}} = \frac{{km}_{1}}{P_{1}} and V_{2} = \frac{{km}_{2}}{P_{2}} Total volume = k (\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{1}}) Let mixture has common pressure P and common density ρ ρ = \frac{(m_{1} + m_{2})}{k (\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}})} \Rightarrow P = kρ = \frac{(m_{1} + m_{2})}{(\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}})} = \frac{P_{1} P_{2} (m_{1} + m_{2})}{(P_{2} m_{1} + m_{2} P_{1})}$

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A rocket is propelled by a gas which is initially at a temperature of 4000 K. The temperature of the gas falls to 1000 K as it leaves the exhaust nozzle. The gas which will acquire the largest momentum while leaving the nozzle is:

Hydrogen

Helium

Nitrogen

Argon

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Explanation

Heavier gas will acquire largest momentum i.e. Argon.

$E_{g a s} = \frac{p_{g a s}^{2}}{2 M} M o r e t h e m o l e c u l a r m a s s M, m o r e i s t h e m o m e n t u m p_{g a s}^{} A r g o n h a s h i g h e s t M, s o l a r g e s t m o m e n t u m p_{g a s}^{}$

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