JEE Advanced 2022 — Paper 1 — PHY-III solution

JEE Advanced2022Paper 1PHY-III

List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.

List-I	List-II
(I) $1 0^{- 3}$ kg of water at 100 $^{\circ}$ C is converted to steam at the same temperature, at a pressure of $1 0^{5}$ Pa. The volume of the system changes from $1 0^{- 6}$ m $^{3}$ to $1 0^{- 3}$ m $^{3}$ in the process. Latent heat of water = 2250 kJ/kg.	(P) 2 kJ
(II) 0.2 moles of a rigid diatomic ideal gas with volume $V$ at temperature 500 K undergoes an isobaric expansion to volume $3 V$ . Assume $R = 8.0$ J mol $^{- 1}$ K $^{- 1}$ .	(Q) 7 kJ
(III) One mole of a monoatomic ideal gas is compressed adiabatically from volume $V = \frac{1}{3}$ m $^{3}$ and pressure 2 kPa to volume $\frac{V}{8}$ .	(R) 4 kJ
(IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given 9 kJ of heat and undergoes isobaric expansion.	(S) 5 kJ
	(T) 3 kJ

Which one of the following options is correct?

A
I $\to$ T, II $\to$ R, III $\to$ S, IV $\to$ Q
B
I $\to$ S, II $\to$ P, III $\to$ T, IV $\to$ P
C
I $\to$ P, II $\to$ R, III $\to$ T, IV $\to$ Q
D
I $\to$ Q, II $\to$ R, III $\to$ S, IV $\to$ T

Solution

(I) First law: $Δ U = Q - W$ . Here $Q = m L = 1 0^{- 3} \times 2250 = 2.25$ kJ; $W = P Δ V = 1 0^{5} (1 0^{- 3} - 1 0^{- 6}) \approx 0.1$ kJ. $Δ U \approx 2.15$ kJ $\approx$ 2 kJ. I $\to$ P.

(II) Rigid diatomic: $C_{V} = \frac{5 R}{2}$ . At constant pressure, $V \propto T$ , so $T_{2} = 3 T_{1} = 1500$ K, $Δ T = 1000$ K. $Δ U = n C_{V} Δ T = 0.2 \times \frac{5 \times 8}{2} \times 1000 = 4$ kJ. II $\to$ R.

(III) Adiabatic, $γ = 5/3$ . $P_{2} = P_{1} (V_{1} / V_{2})^{γ} = 2 \times 8^{5/3} = 2 \times 32 = 64$ kPa. $Δ U = \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1}) = \frac{3}{2} (64 \times \frac{1}{3 \times 8} - 2 \times \frac{1}{3}) = \frac{3}{2} (\frac{8}{3} - \frac{2}{3}) = 3$ kJ. III $\to$ T.

(IV) Diatomic with vibration active: $C_{V} = \frac{7 R}{2}$ , $C_{P} = \frac{9 R}{2}$ . For isobaric process, $\frac{Δ U}{Q} = \frac{C _{V}}{C _{P}} = \frac{7}{9}$ . $Δ U = \frac{7}{9} \times 9 = 7$ kJ. IV $\to$ Q.

The correct matching is option (C).

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