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Physics MCQs for NEET — Practice Questions with Answers

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Air in a cylinder is suddenly compressed by a piston, which is then maintained at the same position. With the passage of time

The pressure decreases

The pressure increases

The pressure remains the same

The pressure may increase or decrease depending upon the nature of the gas

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Explanation

Due to sudden compression (adiabatic) ,the temperature of the system increases to a very high value. This causes the flow of heat from system to the surroundings, thus decreasing the temperature. This decrease in temperature results in decrease in pressure.

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The adiabatic Bulk modulus of a perfect gas at pressure P is given by

P/2

γ P

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Explanation

Adiabatic Bulk modulus $E_{ϕ} = γ P$

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An adiabatic process occurs at constant

Temperature

Pressure

Heat

Temperature and pressure

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Explanation

In adiabatic process, no heat transfers between system and surrounding.

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A polyatomic gas $(γ = \frac{4}{3})$ is compressed to $\frac{1}{8}$ of its volume adiabatically. If its initial pressure is P₀, its new pressure will be -

8 P₀

16 P₀

6 P₀

2 P₀

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Explanation

$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$

⇒ $P_{2} = P_{1} {(\frac{V_{1}}{V_{2}})}^{γ} = P_{0} {(8)}^{4 / 3} = 16 P_{0} .$

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In an adiabatic expansion of a gas initial and final temperatures are T₁ and T₂ respectively, then the change in internal energy of the gas is -

$\frac{R}{γ - 1} (T_{2} - T_{1})$

$\frac{R}{γ - 1} (T_{1} - T_{2})$

R(T₁ – T₂)

Zero

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Explanation

$Δ U = - Δ W = - \frac{R (T_{1} - T_{2})}{(γ - 1)}$

$= \frac{R (T_{2} - T_{1})}{γ - 1}$

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A cycle tyre bursts suddenly. This represents an

Isothermal process

Isobaric process

Isochoric process

Adiabatic process

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Explanation

The process is very fast, so the gas fails to gain or lose heat. Hence this process in adiabatic

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One mole of helium is adiabatically expanded from its initial state $(P_{i}, V_{i}, T_{i})$ to its final state $(P_{f}, V_{f}, T_{f})$ . The decrease in the internal energy associated with this expansion is equal to

$C_{V} (T_{i} - T_{f})$

$C_{P} (T_{i} - T_{f})$

$\frac{1}{2} (C_{P} + C_{V}) (T i - T_{f})$

$(C_{P} - C_{V}) (T_{i} - T_{f})$

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Explanation

$Δ U = μ C_{V} Δ T = 1 \times C_{V} (T_{f} - T_{i}) = - C_{V} (T_{i} - T_{f})$

⇒ |ΔU| = C_V (T_i – T_f)

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A diatomic gas initially at 18°C is compressed adiabatically to one-eighth of its original volume. The temperature after compression will be

10°C

887°C

668 K

144°C

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Explanation

$T V^{γ - 1} =$ constant

$\Rightarrow T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1} = (273 + 18) {(\frac{V}{V / 8})}^{0 .4} = 668 K$

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During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio C_p/C_v for the gas is

$\frac{3}{2}$

$\frac{4}{3}$

$\frac{5}{3}$

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Explanation

Given $P \propto T^{3}$ , but we know for an adiabatic process, the pressure $P \propto T^{γ / γ - 1}$

So $\frac{γ}{γ - 1} = 3 \Rightarrow γ = \frac{3}{2} \Rightarrow \frac{C_{P}}{C_{V}} = \frac{3}{2}$

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One mole of an ideal gas at an initial temperature of T K does 6 R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be -

(T + 2.4)K

(T – 2.4)K

(T + 4)K

(T – 4)K

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Explanation

$W = \frac{R (T_{i} - T_{f})}{γ - 1}$

⇒ $6 R = \frac{R (T - T_{f})}{(\frac{5}{3} - 1)}$ ⇒ $T_{f} = (T - 4) K .$

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