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. [2022]

Question

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. [2022]

	List - I		List - II
(I)	$10^{- 3} kg$ of water at $100 ° C$ is converted to steam at the same temperature, at a pressure of $10^{5} Pa$ . The volume of the system changes from $10^{- 6} m^{3}$ to $10^{- 3} m^{3}$ 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 monatomic 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?

Smart Achievers · Accepted Answer

(3)

$(i) By first law of thermodynamics,$

$Δ U = Δ Q - Δ W = M L_{v} - P Δ V$

$= 10^{- 3} \times 2250 \times 10^{3} - 10^{5} \times (10^{- 3} - 10^{- 6})$

$= 2250 - 100 ≃ 2150 J = 2.15 kJ$ $So (I) \to (P)$

$(i i) P = \frac{n R T}{V} = \frac{0.2 \times 8 \times 500}{V} = \frac{800}{V} Pa$

$Δ U = \frac{f}{2} P Δ V = \frac{5}{2} \times \frac{800}{V} \times 2 V = 4000 J = 4 kJ$

$So (II) \to (R)$

$(i i i) P V^{γ} = constant \Rightarrow P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$

$\Rightarrow 2 V^{γ} = P_{2} {(\frac{V}{8})}^{γ} \Rightarrow P_{2} = 2 \times 8^{γ} = 2 \times 8^{5 / 3} = 64 kPa$

$So, Δ U = \frac{f}{2} (P_{2} V_{2} - P_{1} V_{1})$ $= \frac{3}{2} (64 \times \frac{1}{24} - 2 \times \frac{1}{3}) \times 10^{3}$ $= 3 kJ$

$So (III) \to (T)$

$(i v) Here f = 7$

$So, Δ U = n C_{v} Δ T = \frac{f}{2} n R Δ T = \frac{7}{2} n R Δ T$

$and, Δ Q = n C_{p} Δ T = (\frac{f}{2} + 1) n R Δ T = \frac{9}{2} n R Δ T = \frac{9}{2} \times \frac{2}{7} Δ U$

$Δ U = \frac{7}{9} Δ Q$

$So, Δ U = \frac{7}{9} Δ Q = \frac{7}{9} \times 9 = 7 kJ$

$So (IV) \to (Q)$

Solution

1 (I) → (T); (II) → (R); (III) → (S); (IV) → (Q)

2 (I) → (S); (II) → (P); (III) → (T); (IV) → (P)

3 (I) → (P); (II) → (R); (III) → (T); (IV) → (Q)

4 (I) → (Q); (II) → (R); (III) → (S); (IV) → (T)

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