JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Chemical Thermodynamics

A reversible cyclic process for an ideal gas is shown below. Here,

P, V

, and

T

are pressure, volume and temperature, respectively. The thermodynamic parameters

q, w, H

and

U

are heat, work, enthalpy and internal energy, respectively.

The correct option(s) is (are)

Answer: (B), (C)

Solution

From the P-V diagram:

(A) For isochoric process AC:

q_{A C} = Δ U_{B C} = n c_{V} (T_{2} - T_{1})

✓
For isothermal process AB:

w_{A B} = n R T_{1} ln \frac{V _{2}}{V _{1}} \neq = P_{2} (V_{2} - V_{1})

✗

(B) For isobaric process BC:

w_{B C} = - P_{2} (V_{2} - V_{1})

(work done by gas is negative for compression, but the magnitude is correct)
Also,

q_{B C} = Δ H_{A C} = n c_{p} (T_{2} - T_{1})

✓

(C) For process CA (cooling):

Δ H_{C A} = n c_{p} (T_{1} - T_{2}) < 0

and

Δ U_{C A} = n c_{V} (T_{1} - T_{2}) < 0

Since

c_{p} > c_{V}

, we have

∣Δ H_{C A} ∣ > ∣Δ U_{C A} ∣

, so

Δ H_{C A} < Δ U_{C A}

✓

(D) This is opposite to (C), so incorrect.

Correct answers: (B) and (C).

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · One or more correct options