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Two Carnot engines A and B are operated in succession. The first one, A receives heat from a source at T₁ = 800 K and rejects to sink at T₂K. The second engine B receives heat rejected by the first engine and rejects to another sink at T₃ = 300 K. If the work outputs of two engines are equal, then the value of T₂ is -

100K

300K

550K

700K

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Explanation

$η_{A} = \frac{T_{1} - T_{2}}{T_{1}} = \frac{W_{A}}{Q_{1}}$ ⇒ $η_{B} = \frac{T_{2} - T_{3}}{T_{2}} = \frac{W_{B}}{Q_{2}}$

∴ $\frac{Q_{1}}{Q_{2}} = \frac{T_{1}}{T_{2}} \times \frac{T_{2} - T_{3}}{T_{1} - T_{2}} = \frac{T_{1}}{T_{2}}$

∴ W_A = W_B

∴ $T_{2} = \frac{T_{1} + T_{3}}{2} = \frac{800 + 300}{2} = 550 K$

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When an ideal monoatomic gas is heated at constant pressure, fraction of heat energy supplied which increases the internal energy of gas, is

$\frac{2}{5}$

$\frac{3}{5}$

$\frac{3}{7}$

$\frac{3}{4}$

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Explanation

For monoatomic gas

$γ = \frac{C_{P}}{C_{V}} = \frac{5}{3}$ we know $Δ Q = n C_{P} Δ T$

and $Δ U = n C_{V} Δ T$ ⇒ $\frac{Δ U}{Δ Q} = \frac{C_{V}}{C_{P}} = \frac{3}{5}$

i.e. fraction of heat energy to increase the internal energy be 3/5.

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When an ideal gas (γ = 5/3) is heated under constant pressure, then what percentage of given heat energy will be utilised in doing external work ?

40 %

30 %

60 %

20 %

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Explanation

$Δ Q = Δ U + Δ W \Rightarrow \frac{Δ W}{Δ Q} = 1 - \frac{Δ U}{Δ Q} = 1 - \frac{n C_{V} d T}{n C_{P} d T}$

$\Rightarrow \frac{Δ W}{Δ Q} = 1 - \frac{C_{V}}{C_{P}} = 1 - \frac{3}{5} = \frac{2}{5} = 0 .4$

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Which one of the following gases possesses the largest internal energy?

2 moles of helium occupying 1m³at 300 K

56 kg of nitrogen at $10^{5}$ Nm^–2 and 300 K

8 grams of oxygen at 8 atm and 300 K

6 × 10²⁶ molecules of argon occupying 40 m³ at 900 K

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Explanation

$U = \frac{n fR T}{2}$

⇒ ${(U)}_{N} = \frac{56 \times 10^{3}}{28} \times \frac{5}{2} R \times 300$

and ${(U)}_{A r} = \frac{6 \times 10^{26}}{6 \times 10^{23}} \times \frac{3}{2} R \times 900 \Rightarrow {(U)}_{A r} < {(U)}_{N}$

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Two samples A and B of a gas initially at the same pressure and temperature are compressed from volume V to V/2 (A isothermally and B adiabatically). The final pressure of A is

Greater than the final pressure of B

Equal to the final pressure of B

Less than the final pressure of B

Twice the final pressure of B

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Explanation

A is compressed isothermally, hence

$P_{1} V = P_{2} \frac{V}{2} \Rightarrow P_{2} = 2 P_{1}$

and B is compressed adiabatically, hence

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

Since $γ > 1$ , hence $P_{2}' > P_{2}$ or $P_{2} < {P^{'}}_{2}$

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Initial pressure and volume of a gas are P and V respectively. First it is expanded isothermally to volume 4V and then compressed adiabatically to volume V. The final pressure of gas will be [Given : $γ = 1.5]$ -

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Explanation

In isothermal process $P_{1} V_{1} = P_{2} V_{2}$

or $P V = P_{2} \times 4 V$

∴ $P_{2} = \frac{P}{4}$

In adiabatic process

$P_{2} {V_{2}}^{γ} = P_{3} {V_{3}}^{γ}$

⇒ $\frac{P}{4} \times {(4 V)}^{1 .5} = P_{2} V^{1 .5}$

⇒ $P_{3} = 2 P$

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A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by 62°C, the efficiency of the engine is doubled. The temperatures of the source and sink are -

80°C, 37°C

95°C, 28°C

90°C, 37°C

99°C, 37°C

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Explanation

Initially $η = (1 - \frac{T_{2}}{T_{1}}) = \frac{W}{Q} = \frac{1}{6}$ ...(i)

Finally $η' = (1 - \frac{T_{2}'}{T_{1}}) = (1 - \frac{(T_{2} - 62)}{T_{1}}) = 1 - \frac{T_{2}}{T_{1}} + \frac{62}{T_{1}}$

$= η + \frac{62}{T_{1}}$ ....(ii)

It is given that $η' = 2 η .$ Hence solving equation (i) and (ii)

⇒ $T_{1} = 372 K = 99 ° C$ and $T_{2} = 310 K = 37 ° C$

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An ideal gas expands in such a manner that its pressure and volume can be related by equation $P V^{5 / 3} = c o n s \tan t$ . During this process, the gas is

Heated

Cooled

Neither heated nor cooled

First heated and then cooled

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Explanation

$P V^{5 / 3} =$ constant represents adiabatic equation.

So during the expansion of ideal gas internal energy of gas decreases and temperature falls.

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P-V diagram of a diatomic gas is a straight line passing through origin. The molar heat capacity of the gas in the process will be -

4 R

2.5 R

3 R

$\frac{4 R}{3}$

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Explanation

P-V diagram of the gas is a straight line passing through origin. Hence $P \propto V$ or $P V^{- 1} =$ constant

Molar heat capacity in the process $P V^{x} = constant is$

$C = \frac{R}{γ - 1} + \frac{R}{1 - x}$ ; Here $γ = 1 .4$ (For diatomic gas)

⇒ $C = \frac{R}{1 .4 - 1} + \frac{R}{1 + 1}$ ⇒ $C = 3 R$

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Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is

30 K

18 K

50 K

42 K

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Explanation

In both cylinders A and B the gases are diatomic (γ = 1.4). Piston A is free to move i.e. it is isobaric process. Piston B is fixed i.e. it is isochoric process. If same amount of heat ΔQ is given to both then

${(Δ Q)}_{isobaric} = {(Δ Q)}_{isochoric}$ ⇒ $μ C_{p} {(Δ T)}_{A} = μ C_{v} {(Δ T)}_{B}$

⇒ ${(Δ T)}_{B} = \frac{C_{p}}{C_{v}} {(Δ T)}_{A} = γ {(Δ T)}_{A} = 1 .4 \times 30 = 42 K .$

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.