JEE Advanced 2021 Paper 1 Q14 Chemistry Thermodynamics & Thermochemistry First Law & Internal Energy Hard

JEE Advanced 2021 Paper 1 · Q14 · First Law & Internal Energy

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)

  1. A. [Figure: option A — p-V plot. State I at high pressure / low volume. A smooth concave curve (isotherm) from I down-right to II at intermediate p and larger V. A second steeper concave curve (adiabat) from II down-right to III at low p and even larger V. Both segments show p decreasing as V increases.]
  2. B. [Figure: option B — p-T plot. State I at high p, fixed T. Vertical arrow down from I to II at the same T (the isothermal expansion). Then a curved arrow from II down-left to III at lower p and lower T (the adiabatic expansion cools the gas).]
  3. C. [Figure: option C — H-S plot. State I at low H. Horizontal arrow from I to II at the same H but higher S (isothermal expansion of an ideal gas is isoenthalpic and entropy-increasing). Then a vertical arrow up from II to III at the same S but higher H. (The vertical adiabatic step is shown going UP in H, which is incorrect for an adiabatic expansion of an ideal gas.)]
  4. D. [Figure: option D — T-S plot. State I at high T. Horizontal arrow from I to II at the same T but higher S (isothermal expansion). Then a vertical arrow from II downward to III at the same S but lower T (reversible adiabatic = isentropic, with cooling on expansion).
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, D

Solution

Reversible isothermal expansion of an ideal gas: T constant, V increases, p decreases, $\Delta U = 0$, $\Delta H = 0$, $\Delta S > 0$.

Reversible adiabatic expansion of an ideal gas: q = 0, isentropic ($\Delta S = 0$), V increases, p decreases, T decreases (and so $\Delta H < 0$).

(A) p-V: both steps go down-right (p falls, V rises). Correct.

(B) p-T: in the isothermal step T is constant and p drops vertically; in the adiabatic step both p and T drop further. The graph in (B) matches this. Correct.

(C) H-S: in the isothermal step $\Delta H = 0$ (horizontal); in the adiabatic step $\Delta S = 0$ (vertical), and H must DECREASE because T decreases. The figure shows H increasing in the adiabatic step, so (C) is wrong.

(D) T-S: in the isothermal step T is constant ($\Delta T = 0$, S increases — horizontal arrow); in the adiabatic step S is constant and T decreases (vertical arrow downward). Correct.

Hence (A), (B) and (D) are correct.

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