A process $1 \to 2$ using diatomic gas is shown on the P-V diagram below. $P_{2} = 2 P_{1} = 10^{6} N / m^{2}, V_{2} = 4 V_{1} = 0.4 m^{3}$ . The molar heat capacity of the gas in this process will be

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$\frac{22 R}{7}$

$\frac{25 R}{13}$

$\frac{35 R}{11}$

$\frac{35 R}{12}$

answer is D.

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Detailed Solution

$Δ w = (\frac{p_{1} + p_{2}}{2}) (v_{2} - v_{1}) = \frac{3}{4} \times 10^{6} \times 0.3 = \frac{9}{4} \times 10^{5} T_{2} = \frac{10^{6} \times 0.4}{n R} T_{1} = \frac{0.5 \times 10^{6} \times 0.1}{n R} Δ U = n (\frac{5 R}{2}) \frac{(105) (3.5)}{n R} = \frac{35}{4} \times 10^{5} Δ θ = 11 \times 10^{5} = \frac{n c}{n R} {10^{5}} {3.5} c = \frac{22 R}{7}$