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An ideal gas whose adiabatic exponent equals to $\frac{7}{5}$ is expanded according to the law P = 2V.The initial volume of the gas is equal to $V_{o} = 1$ unit. As a result of expansion the volume increases 4 times. (Take R = units)
Column-I Column-II
i. Work done by the gas p. 25 units
ii. Increment in internal energy of the gas q. 45 units
iii. Heat supplied to the gas r. 75 units
iv. Molar heat capacity of the gas in the process s. 15 units
t. 55 units
Now much the given columns and select the correct option from the codes given below.
Codes

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i -t, ii - s, iii-q, iv-r

i - p, ii - q, iii - r, iv - s

i-s, ii-r, iii-q, iv - p

i-q, ii - r, iii -s, iv - t

answer is B.

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Detailed Solution

$W = \int PdV = \int_{V_{0}}^{4 V_{0}} 2 VdV = {(V^{2})}_{V_{0}}^{4 V_{0}} = 15 {V_{0}}^{2} = 15 units$

From PV = nRT, $2 V^{2} = nRT$

$\Rightarrow 2 ({V_{2}}^{2} - {V_{1}}^{2}) = nR (∆ T) nR ∆ T = 30 {V_{0}}^{2}$

$∆ U = {nC}_{v} ∆ T = \frac{nR}{γ - 1} ∆ T = \frac{30 {V_{0}}^{2}}{γ - 1} = \frac{30 {(1)}^{2}}{\frac{7}{5} - 1} = \frac{30}{2} (5)$

= 75 units

$Q = W + ∆ U = 15 + 30 = 45 units$

Molar heat capacity:

$C = C_{v} + \frac{R}{1 - x} = \frac{5}{2} R + \frac{R}{1 - (- 1)} = \frac{5}{2} R + \frac{R}{2} = 3 R$

= $3 \times \frac{25}{3} = 25 units$

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