JEE Advanced 2021 · Paper 1 · Chemistry Q14

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ChemistryMultiple correct+4 / -2Question 33 of 57

An ideal gas undergoes a reversible isothermal expansion from state I to state II followed by a reversible adiabatic expansion from state II to state III. The correct plot(s) representing the changes from state I to state III is(are)

( $p$ : pressure, $V$ : volume, $T$ : temperature, $H$ : enthalpy, $S$ : entropy)

p-V plot (shown)

p (vertical) against V (horizontal): a falling curve from I to II, then a steeper falling curve from II to III.

p-T plot (shown)

p (vertical) against T (horizontal): a vertical drop at constant T from I down to II, then a curve from II down and to the left (lower p and T) to III.

H-S plot (shown)

H (vertical) against S (horizontal): a horizontal segment from I to II (S increasing at constant H), then a vertical rise from II up to III at constant S.

T-S plot (shown)

T (vertical) against S (horizontal): a horizontal segment from I to II (S increasing at constant T), then a vertical drop from II down to III at constant S.

Answer and worked solution

Official answer: A, B, D

What each step does.

Isothermal step I→II: $T$ and $H$ are constant (ideal gas), $p$ falls and $S$ rises.
Reversible adiabatic step II→III: $S$ is constant, and $T$ , $H$ and $p$ all fall.

Check each plot.

(A) $p$ - $V$ : a falling isotherm, then a steeper adiabat. Correct.
(B) $p$ - $T$ : a vertical drop at constant $T$ , then a curve to lower $p$ and $T$ . Correct.
(C) $H$ - $S$ : it shows $H$ increasing on the adiabatic step, but $H$ decreases there. Wrong.
(D) $T$ - $S$ : horizontal at constant $T$ , then a vertical drop at constant $S$ . Correct.

(A), (B), (D)

Topics:Entropy First law and enthalpy

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JEE Advanced 2021 · Paper 1, Chemistry question 14