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One mole of a perfect gas in a cylinder fitted with a piston has a pressure P, volume V and temperature 273 K. If the temperature is increased by 1 K keeping pressure constant, the increase in volume is

$\frac{2 V}{273}$

$\frac{V}{91}$

$\frac{V}{273}$

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Explanation

For isobaric process $\frac{V_{2}}{V_{1}} = \frac{T_{2}}{T_{1}} \Rightarrow V_{2} = V \times \frac{274}{273}$

Increase $= \frac{274 V}{273} - V = \frac{V}{273}$

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Unit mass of a liquid with volume V₁ is completely changed into a gas of volume V₂ at a constant external pressure P and temperature T. If the latent heat of evaporation for the given mass is L, then the increase in the internal energy of the system is -

Zero

$P (V_{2} - V_{1})$

$L - P (V_{2} - V_{1})$

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Explanation

$Δ Q = Δ U + P Δ V$

⇒ mL = ΔU + P(V₂ – V₁)

⇒ ΔU = L – P (V₂ – V₁) (âˆµ m = 1)

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A gas expands 0.25m³ at constant pressure 10³ N/m², the work done is -

2.5 ergs

250 J

250 W

250 N

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Explanation

$Δ W = P Δ V = 10^{3} \times 0 .25 = 250 J$

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If 300 ml of a gas at 27°C is cooled to 7°C at constant pressure, then its final volume will be -

540 ml

350 ml

280 ml

135 ml

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Explanation

V ∝ T at constant pressure

⇒ $\frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}}$

⇒ $V_{2} = \frac{V_{1} T_{2}}{T_{1}} = \frac{300 \times 280}{300} = 280 m l .$

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Which of the following is correct in terms of increasing work done for the same initial volume and final volume ?

Adiabatic < Isothermal < Isobaric

Isobaric < Adiabatic < Isothermal

Adiabatic < Isobaric < Isothermal

None of these

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A sample of gas expands from volume V₁ to V₂. The amount of work done by the gas is greatest when the expansion is

Isothermal

Isobaric

Adiabatic

Equal in all cases

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In a reversible isochoric change -

ΔW = 0

ΔQ = 0

ΔT = 0

ΔU = 0

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Explanation

$Δ V = 0 \Rightarrow P Δ V = 0 \Rightarrow Δ W = 0$

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Entropy of a thermodynamic system does not change when this system is used for -

Conduction of heat from a hot reservoir to a cold reservoir

Conversion of heat into work isobarically

Conversion of heat into internal energy isochorically

Conversion of work into heat isochorically

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Explanation

Entropy does not change when there is no heat exchange in a process.

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Which is incorrect ?

In an isobaric process, Δp = 0

In an isochoric process, ΔW = 0

In an isothermal process, ΔT = 0

In an isothermal process, ΔQ = 0

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Explanation

In isothermal process ΔQ ≠ 0.

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A Carnot engine working between 300 K and 600K has work output of 800 J per cycle. What is amount of heat energy supplied to the engine from source per cycle ?

1800 J/cycle

1000 J/cycle

2000 J/cycle

1600 J/cycle

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Explanation

$η = \frac{T_{1} - T_{2}}{T_{1}} = \frac{W}{Q}$ ⇒ $Q = (\frac{T_{1}}{T_{1} - T_{2}}) W$

$= \frac{600}{(600 - 300)} \times 800$ =1600 J

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Yes. You can attempt every NEET 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.