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Physics NEET Questions — Free MCQ Practice

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One mole of ideal monoatomic gas $(γ = 5 / 3)$ is mixed with one mole of diatomic gas $(γ = 7 / 5)$ . What is $γ$ for the mixture? $γ$ denotes the ratio of specific heat at constant pressure, to that at constant volume

A.

3/2

B.

23/15

C.

35/23

D.

4/3

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Explanation

1.

$γ_{mix} = \frac{\frac{μ_{1} γ_{1}}{γ_{1} - 1} + \frac{μ_{2} γ_{2}}{γ_{2} - 1}}{\frac{μ_{1}}{γ_{1} - 1} + \frac{μ_{2}}{γ_{2} - 1}} = \frac{\frac{1 \times \frac{5}{3}}{(\frac{5}{3} - 1)} + \frac{1 \times \frac{7}{5}}{(\frac{7}{5} - 1)}}{\frac{1}{(\frac{5}{3} - 1)} + \frac{1}{(\frac{7}{5} - 1)}} = \frac{3}{2} = 1.5$

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A gaseous mixture contains equal number of hydrogen and nitrogen molecules. Specific heat measurements on this mixture at temperatures below 100 K would indicate that the value of $γ$ (ratio of specific heats) for this mixture is

A.

3/2

B.

4/3

C.

5/3

D.

7/5

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Explanation

3. Below 100 K only translational degree of freedom is considered. Hence

$γ_{mixture} = \frac{μ_{1} C_{p, 1} + μ_{2} C_{p, 2}}{μ_{1} C_{v, 1} + μ_{2} C_{v, 2}} where μ_{1} and μ_{2} are moles of samples A and B γ_{mixture} = \frac{\frac{μ_{1} γ_{1}}{γ_{1} - 1} + \frac{μ_{2} γ_{2}}{γ_{2} - 1}}{\frac{μ_{1}}{γ_{1} - 1} + \frac{μ_{2}}{γ_{2} - 1}}$

according to question,

$μ_{1} = μ_{2} and γ_{1} = γ_{2} = 1 + \frac{2}{3} = \frac{5}{3}$

$\Rightarrow γ_{mix} = γ_{1} = \frac{5}{3}$

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One mole of monoatomic gas and three moles of diatomic gas are put together in a container. The molar specific heat $(in J K^{- 1} {mol}^{- 1})$ at constant volume is $(R = 8.3 J K^{- 1} {mol}^{- 1})$

A.

18.7

B.

18.9

C.

19.2

D.

None of the above

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Explanation

$1 . C_{V_{mix}} = \frac{μ_{1} C_{V_{1}} + μ_{2} C_{V_{2}}}{μ_{1} + μ_{2}} = \frac{1 \times \frac{3}{2} R + 3 \times \frac{5}{2} R}{1 + 3} = \frac{9}{4} R = \frac{9}{4} \times 8.3 = 18.7$

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The number of translational degrees of freedom for a diatomic gas is

A.

2

B.

3

C.

5

D.

6

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Explanation

2. Number of translational degrees of freedom (3) are same for all types of gases.

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For a gas if ratio of specific heats at constant pressure and volume is $γ$ then value of degrees of freedom is

A.

$\frac{3 γ - 1}{2 γ - 1}$

B.

$\frac{2}{γ - 1}$

C.

$\frac{9}{2} (γ - 1)$

D.

$\frac{25}{2} (γ - 1)$

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Explanation

$2 . γ = 1 + \frac{2}{f}, \Rightarrow γ - 1 = \frac{2}{f} \Rightarrow \frac{f}{2} = \frac{1}{γ - 1} \Rightarrow f = \frac{2}{γ - 1}$

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If a gas has n degrees of freedom ratio of specific heats of gas is

A.

$\frac{1 + n}{2}$

B.

$1 + \frac{1}{n}$

C.

$1 + \frac{n}{2}$

D.

$1 + \frac{2}{n}$

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Explanation

$4 . γ = \frac{C_{P}}{C_{V}} = \frac{\frac{f}{2} R + R}{\frac{f}{2} R} = 1 + \frac{2}{f} = 1 + \frac{2}{n}$

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A gaseous mixture consists of 16g of helium and 16g of oxygen. The ratio $\frac{C_{P}}{C_{V}}$ of the mixture is

A.

1.4

B.

1.54

C.

1.59

D.

1.62

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Explanation

$4 . μ_{1} = moles of helium = \frac{16}{4} = 4 μ_{2} = moles of oxygen = \frac{16}{32} = \frac{1}{2} \frac{\frac{4 \times 5 / 3}{\frac{5}{3} - 1} + \frac{1 / 2 \times 7 / 5}{\frac{7}{5} - 1}}{\frac{4}{\frac{5}{3} - 1} + \frac{1 / 2}{\frac{7}{5} - 1}} = 1.62$

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The pressure exerted by the gas on the walls of the container because

A.

It loses kinetic energy

B.

It sticks with the walls

C.

On collision with the walls there is a change in momentum

D.

It is accelerated towards the walls

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Explanation

3. Pressure, $P = \frac{F}{A} = \frac{1}{A} . \frac{∆ p}{∆ t} (∆ p = Change in momentum)$

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Gas at a pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to

A.

$4 P_{0}$

B.

$2 P_{0}$

C.

$P_{0}$

D.

$\frac{P_{0}}{2}$

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Explanation

$2 . v_{rms} = \sqrt{\frac{3 P}{ρ}} = \sqrt{\frac{3 PV}{m}} \Rightarrow v_{rms} \propto \sqrt{\frac{P}{M}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{P_{1}}{P_{2}} \times \frac{m_{2}}{m_{1}}} \Rightarrow \frac{v}{2 v} = \sqrt{\frac{P_{0}}{P_{2}} \times \frac{m / 2}{m}} \Rightarrow P_{2} = 2 P_{0}$

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A box contains n molecules of a gas. How will the pressure of the gas be effected, if the number of molecules is made 2n?

A.

Pressure will decrease

B.

Pressure will remain unchanged

C.

Pressure will be doubled

D.

Pressure will become three times

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Explanation

(c)

$PV = nRT n = Molecules \times N_{A} \Rightarrow \frac{P_{2}}{P_{1}} = \frac{N_{2}}{N_{1}} = \frac{2}{1} \Rightarrow P_{2} = 2 P_{1}$

Physics MCQs for NEET — Practice Questions with Answers

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