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) 10 - 3   kg of water 100   ° C is converted…

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.

	List-I		List-II
(I)	$10^{- 3} kg$ of water $100 ° C$ 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 $= 2250 kJ {kg}^{- 1}$	(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?

Aicharya · Accepted Answer

(I)
According to the first law of thermodynamics,
$∆ Q = ∆ U + W \Rightarrow ∆ U = ∆ Q - W$
$\Rightarrow U = M L - P Δ V$
$= 10^{- 3} \times 2250 - 10^{2} kP \times (10^{- 3} - 10^{- 6}) m^{3}$
$= 2.25 kJ - 0.1 kJ$
$= 2.15 kJ$
Therefore, I-P
(II)
For isobaric process,
$\frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}} \Rightarrow T_{2} = 3 \times 500 = 1500 K$
Now the change in the internal energy will be,
$∆ U = n C_{V} ∆ T = 0.2 \times \frac{5}{2} \times 8 \times 1000 = 4 kJ$
Therefore, II-R
(III)
For adiabatic expansion of monoatomic gas $(γ = \frac{5}{3})$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ} \Rightarrow (2 kPa) \times V_{0}^{\frac{5}{3}} = P_{2} \times {(\frac{V_{0}}{8})}^{\frac{5}{3}}$
$\Rightarrow P_{2} = 64 kPa$
Now the change in the internal energy will be,
$Δ U = n C_{v} Δ T$
$= \frac{3 n R Δ T}{2} = \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1}) = \frac{3}{2} \times (64 \times \frac{1}{3 \times 8} - 2 \times \frac{1}{3})$
$= \frac{3}{2} \times (\frac{8}{3} - \frac{2}{3}) = 3 kJ$
Therefore, III-T
(IV)
For diatomic gas whose molecules vibrate,
$C_{V} = \frac{6}{2} R$ and $C_{P} = \frac{8}{2} R$
$\Rightarrow Δ U = n C_{V} Δ T = 3 n R Δ T$
$Δ Q = n C_{p} Δ T = 4 n R Δ T$
$\frac{Δ U}{Δ Q} = \frac{3}{4}$
$Δ U = \frac{3}{4} \times 9 = 6.75 ≃ 7 kJ$
Therefore, IV-Q

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List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either…

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