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A monoatomic ideal gas of 2 moles was found to be undergoing a thermodynamic polytropic process whose volume coefficient is varying with temperature as shown.
$(R = \frac{25}{3} J / m o l . K) .$ Match the two columns.
List-I List-II
P) Work done by gas in joules, for every 1K rise in temperature, if $\tan θ = 3$ 1) 0
Q) Internal energy rise of the gas in joules, for every 1K rise in temperature, if $\tan θ = 2$ 2) 25
R) Heat absorbed by the gas in joules, for every 1K rise in temperature, if $\tan θ = 1$ 3) $\frac{125}{3}$
S) Value of $θ$ in degrees for which the internal energy rise of the gas for every 1 K rise in temperature is 25 J 4) $\frac{50}{9}$

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$P \to 2, 3; Q \to 1, 3, 4; R \to 2; S \to 3, 4$

$P \to 2, 4; Q \to 1, 2; R \to 4; S \to 1, 3, 4$

$P \to 4; Q \to 2; R \to 3; S \to 1, 2, 3, 4$

$P \to 4; Q \to 3, 4; R \to 2, 3, 4; S \to 1, 2$

answer is A.

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Detailed Solution

$(P) \tan θ = \frac{1}{T α} \Rightarrow 3 = \frac{V d T}{T d V} \Rightarrow \frac{3 d V}{V} = \frac{d T}{T} \ln V^{3} = \ln T + C \Rightarrow T V^{- 3} = constant$ .

This is polytropic process with $x - 1 = - 3 \Rightarrow x = - 2$

$W = \frac{n R}{x - 1} (T_{1} - T_{2}) = \frac{2 (\frac{25}{3})}{- 3} (- 1) = + \frac{50}{9} J (Q) Δ U = n C_{V} Δ T = \frac{2 (\frac{25}{3})}{(\frac{5}{3} - 1)} (1) = 25 J$

$(R) Δ Q_{P} = n C_{P} Δ T = (2) [\frac{5 R}{2}] Δ T = 5 \times \frac{25}{3} (1) = \frac{125}{3} J$

$(S) Δ U = 25 J$ for any polytropic co-efficient i.e., for any value of $' θ' .$

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