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Physics NEET Questions — Free MCQ Practice

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

A.

Change is pressure

B.

Change is volume

C.

Change in temperature

D.

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

A.

ΔU = 0

B.

ΔU = negative

C.

ΔU = positive

D.

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 -

A.

$\frac{24}{5}$

B.

8

C.

$\frac{40}{3}$

D.

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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The pressure and density of a diatomic gas $(γ = 7 / 5)$ change adiabatically from (P, d) to (P', d'). If $\frac{d'}{d} = 32$ , then $\frac{P'}{P}$ should be

A.

1/128

B.

32

C.

128

D.

None of the above

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Explanation

Volume of the gas $V = \frac{m}{d}$ and using $P V^{γ}$ = constant

We get $\frac{P'}{P} = {(\frac{V}{V'})}^{γ} = {(\frac{d'}{d})}^{γ} = {(32)}^{7 / 5} = 128$

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An ideal gas at 27°C is compressed adiabatically to $\frac{8}{27}$ of its original volume. If $γ = \frac{5}{3}$ , then the rise in temperature is

A.

450 K

B.

375 K

C.

225 K

D.

405 K

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Explanation

$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} \Rightarrow T_{2} = 300 {(\frac{27}{8})}^{\frac{5}{3} - 1} = 300 {(\frac{27}{8})}^{\frac{2}{3}}$

$=300 {\{{(\frac{27}{8})}^{1 / 3}\}}^{2} = 300 {(\frac{3}{2})}^{2} = 675 K$

$\Rightarrow Δ T =675-300=375 K$

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Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically. Work done is

A.

More in the adiabatic process

C.

Neither of them

D.

Equal in both processes

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Explanation

For the expansion of an ideal gas, the work done is greater in the isothermal process compared to the adiabatic process. In the isothermal process, heat is continuously absorbed from the surroundings, allowing for more expansion and thus more work done.

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Which is the correct statement ?

A.

For an isothermal change PV = constant

B.

In an isothermal process the change in internal energy must be equal to the work done

C.

For an adiabatic change $\frac{P_{2}}{P_{1}} = {(\frac{V_{2}}{V_{1}})}^{γ}$ , where γ is the ratio of specific heats

D.

In an adiabatic process work done must be equal to the heat entering the system

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Explanation

Since PV = RT and T = constant;

∴ PV = constant.

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The slopes of isothermal and adiabatic curves are related as -

A.

Isothermal curve slope = adiabatic curve slope

B.

Isothermal curve slope = γ × adiabatic curve slope

C.

Adiabatic curve slope = γ × isothermal curve slope

D.

Adiabatic curve slope = $\frac{1}{2} \times$ isothermal curve slope

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Explanation

For Isothermal process PV = constant

$\Rightarrow (\frac{d P}{d V}) = \frac{- P}{V} =$ Slope of Isothermal curve

For adiabatic $P V^{γ} =$ constant

$\Rightarrow \frac{d P}{d V} = \frac{- γ P}{V} =$ Slop of adiabatic curve slope

Clearly, ${(\frac{d P}{d V})}_{adiabatic} = γ {(\frac{d P}{d V})}_{Isothermal}$

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During the adiabatic expansion of 2 moles of a gas, the internal energy of the gas is found to decrease by 2 joules, the work done during the process by the gas will be equal to -

A.

1 J

B.

–1 J

C.

2 J

D.

– 2 J

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Explanation

$d Q = 0 = - 2 + d W \Rightarrow d W = 2 J$

⇒ Work done by the gas = 2 J

● ● ● ● ●

If $γ$ denotes the ratio of two specific heats of a gas, the ratio of slopes of adiabatic and isothermal PV curves at their point of intersection is

A.

$\frac{1}{γ}$

B.

$γ$

C.

$γ - 1$

D.

$γ + 1$

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Explanation

Slope of adiabatic curve = $γ$ × (Slope of isothermal curve)

$Isothermal proccess : PV = constant Differentiating, we get PdV + VdP = 0 \Rightarrow Slope of isothermal curve {(\frac{dP}{dV})}_{iso} = \frac{- P}{V} . . . . . . . . . (1) Adiabtic process : {PV}^{γ} = constant Differentiating, we get {PγV}^{γ - 1} dV + V^{γ} dP = 0 \Rightarrow Slope of adiabatic curve {(\frac{dP}{dV})}_{adi} = \frac{- γP}{V} . . . . . . . . . . . . (2) \Rightarrow Ratio of slopes \frac{{(\frac{dP}{dV})}_{adi}}{{(\frac{dP}{dV})}_{iso}} = \frac{(- γP / V)}{- P / V} = γ$

Physics MCQs for NEET — Practice Questions with Answers

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