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Thermodynamics MCQs for NEET — Physics Questions with Answers

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The isothermal bulk modulus of a perfect gas at normal pressure is -

$1 .013 \times 10^{5} N / m^{2}$

$1 .013 \times 10^{6} N / m^{2}$

$1.013 \times 10^{- 11} N / m^{2}$

$1 .013 \times 10^{11} N / m^{2}$

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Explanation

$E_{θ} = P = 1 .013 \times 10^{5} N / m^{2}$

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In an isothermal change, an ideal gas obeys -

Boyle's law

Charle's law

Gaylussac law

None of the above

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Explanation

In isothermal process, compressibility E_θ = ρ.

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An ideal gas A and a real gas B have their volumes increased from V to 2 V under isothermal conditions. The increase in internal energy

Will be same in both A and B

Will be zero in both the gases

Of B will be more than that of A

Of A will be more than that of B

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Explanation

No change in the internal energy of ideal gas but for real gas, internal energy increases because work is done against intermolecular forces.

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The specific heat of a gas in an isothermal process is -

Infinite

Zero

Negative

Remains constant

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Explanation

In isothermal process temperature remains constant. i.e., $Δ T = 0$ .

Hence according to $C = \frac{Q}{m Δ T}$ ⇒ $C_{i s o} = \infty$

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The latent heat of vaporisation of water is 2240 J/gm. If the work done in the process of expansion of 1 g is 168 J, then increase in internal energy is

2408 J

2240 J

2072 J

1904 J

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Explanation

$Δ Q = Δ U + Δ W$

$\Rightarrow Δ U = Δ Q - Δ W = 2240 - 168 = 2072 J$

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A cylinder fitted with a piston contains 0.2 moles of air at temperature 27°C. The piston is pushed so slowly that the air within the cylinder remains in thermal equilibrium with the surroundings. Find the approximate work done by the system if the final volume is twice the initial volume

543 J

345 J

453 J

600 J

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Explanation

$W = μ R T \log_{e} (\frac{V_{2}}{V_{1}})$

$= 0 .2 \times 8 .3 \times \log_{e} 2 \times (27 + 273)$

$= 0 .2 \times 8 .3 \times 300 \times 0 .693 = 345 J$

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If a cylinder containing a gas at high pressure explodes, the gas undergoes -

Reversible adiabatic change and fall of temperature

Reversible adiabatic change and rise of temperature

Irreversible adiabatic change and fall of temperature

Irreversible adiabatic change and rise of temperature

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Explanation

Gas cylinder suddenly explodes is an irreversible adiabatic change and work done against expansion reduces the temperature.

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The work done in an adiabatic change in a gas depends only on

Change is pressure

Change is volume

Change in temperature

None of the above

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Explanation

Work done in adiabatic change $= \frac{μ R (T_{1} - T_{2})}{γ - 1}$

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In adiabatic expansion

ΔU = 0

ΔU = negative

ΔU = positive

ΔW = zero

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Explanation

In case of adiabatic expansion ΔW = positive and ΔQ = 0

from First law-

$Δ Q = Δ U + Δ W$ ⇒ $Δ U = - Δ W$ i.e., ΔU will be negative.

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A monoatomic gas $(γ = 5 / 3)$ is suddenly compressed to $\frac{1}{8}$ of its original volume adiabatically, then the pressure of the gas will change to -

$\frac{24}{5}$

$\frac{40}{3}$

32 times its initial pressure

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Explanation

$P V^{γ}$ = constant

⇒ $\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ} \Rightarrow P_{2} = {(8)}^{5 / 3} P_{1} = 32 P_{1}$

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