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Physics NEET Questions — Free MCQ Practice

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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

A.

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

B.

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

C.

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

D.

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

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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

A.

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

B.

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

C.

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

D.

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

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The root mean square speed of the molecules of a diatomic gas is v. When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is

A.

$\sqrt{2} v$

B.

$v$

C.

$2 v$

D.

$4 v$

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Explanation

$3 . v_{rms} = \sqrt{\frac{3 RT}{M}} . According to problem T will becomes 2 T and M will becomes M / 2 so the value of v_{rms} will increase by \sqrt{4} = 2 times i . e . new root mean square velocity will be 2 v .$

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A closed vessel contains 8gm of oxygen and 7gm of nitrogen. The total pressure is 10 atm at a given temperature. If now oxygen is absorbed by introducing a suitable absorbent the pressure of the remaining gas in atm will be

A.

2

B.

10

C.

4

D.

5

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Explanation

$4 . From Dalton' s law final pressure of the mixture of nitrogen and oxygen P_{mix} = P_{1} + P_{2} = \frac{μ_{1} RT}{V} + \frac{μ_{2} RT}{V} = \frac{m_{1}}{M_{1}} \frac{RT}{V} + \frac{m_{2}}{M_{2}} \frac{RT}{V} = \frac{8}{32} \frac{RT}{V} + \frac{7}{28} \frac{RT}{V} = \frac{RT}{2 V} \Rightarrow 10 = \frac{RT}{2 V} . . . . . . . . (i) When oxygen is absorbed then for nitrogen let pressure is P = \frac{7}{28} \frac{RT}{V} \Rightarrow P = \frac{RT}{4 V} . . . . . . . . . . (ii) From equation (i) and (ii) we get pressure of the nitrogen P = 5 atm .$

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${CO}_{2} (O = C = O)$ is a triatomic gas. Mean kinetic energy of one gram gas will be (If N-Avogadro's number, k-Boltzmann's constant and molecular weight of ${CO}_{2} = 44$ , Degree of freedom f = 7)

A.

$(3 / 88) NkT$

B.

$(5 / 88) NkT$

C.

$(6 / 88) NkT$

D.

$(7 / 88) NkT$

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Explanation

$4 . Mean kinetic energy for μ mole gas = μ . \frac{f}{2} RT ∴ E = μ \frac{7}{2} RT = (\frac{m}{M}) \frac{7}{2} NkT = \frac{1}{44} (\frac{7}{2}) NkT = \frac{7}{88} NkT [As f = 7 and M = 44 for {CO}_{2}]$

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40 calories of heat is needed to raise the temperature of 1 mole of an ideal monoatomic gas from 20°C to 30°C at a constant pressure. The amount of heat required to raise its temperature over the same interval at a constant volume $(R = 2 calorie {mole}^{- 1} K^{- 1})$ is

A.

20 calorie

B.

40 calorie

C.

60 calorie

D.

80 calorie

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Explanation

$1 . At constant pressure {(∆ Q)}_{P} = {μC}_{P} ∆ T = 1 \times C_{P} \times (30 - 20) = 40 \Rightarrow C_{P} = 4 \frac{cal}{mole \times kelvin} ∴ C_{V} = C_{P} - R = 4 - 2 = 2 \frac{cal}{mole \times kelvin} Now {(∆ Q)}_{V} = {μC}_{V} ∆ T = 1 \times 2 \times (30 - 20) = 20 cal$

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The pressure and volume of saturated water vapour are P and V respectively. It is compressed isothermally thereby volume becomes V/2, the final pressure will be

A.

More than 2P

B.

P

C.

2P

D.

4P

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Explanation

2. Saturated water vapour do not obey gas laws.

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If the intermolecular forces vanish away, the volume occupied by the molecules contained in 4.5 kg water at standard temperature and pressure will be

A.

$5.6 m^{3}$

B.

$4.5 m^{3}$

C.

$11.2 litre$

D.

$11.2 m^{3}$

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Explanation

$1 . When inter molecular forces vanish away the water will act as an ideal gas . The volume of one mole of water at S . T . P . = 22.4 litre We know that 1 mole of water = 18 gm So volume of 4.5 kg of water = \frac{22.4 \times 4.5 \times 10^{3}}{18} = 5.6 \times 10^{3} litre = 5.6 m^{3}$

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When an air bubble of radius ‘r’ rises from the bottom to the surface of a lake, its radius becomes 5r/4 (the pressure of the atmosphere is equal to the 10 m height of water column). If the temperature is constant and the surface tension is neglected, the depth of the lake is

A.

3.53 m

B.

6.53 m

C.

9.53 m

D.

12.53 m

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Explanation

$3 . According to Boyle' s law {(P_{1} V_{1})}_{At top of the lake} = {(P_{2} V_{2})}_{At the bottom of the lake} \Rightarrow P_{1} V_{1} = (P_{1} + h) V_{2} \Rightarrow 10 \times \frac{4}{3} π {(\frac{5 π}{4})}^{3} = (10 + h) \times \frac{4}{3} {πr}^{3} \Rightarrow h = \frac{610}{64} = 9.53 m$

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At standard temperature and pressure the density of a gas is $1.3 kg / m^{3}$ and the speed of the sound in gas is 330 m/sec. Then the degree of freedom of the gas will be

A.

3

B.

4

C.

5

D.

6

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Explanation

$3 . We know that v = \sqrt{\frac{γP}{ρ}} γ = \frac{v^{2} ρ}{P} = \frac{{(330)}^{2} \times 1.3}{1.015 \times 10^{5}} = 1.4 = 1 + \frac{2}{n} \Rightarrow n = 5$

Physics MCQs for NEET — Practice Questions with Answers

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