ThermodynamicsMedium
Question
10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_{1}$ to $P_{2}$ is $\alpha$ Joule $\left( P_{1} = 21.7\text{ }Pa \right.\ $ and $\left. \ P_{2} = 30\text{ }Pa,C_{v} = 21\text{ }J/K.mol,R = 8.3\text{ }J/mol.K \right)$. The value of $\alpha$ is $\_\_\_\_$ .
Options
A.24
B.15
C.21
D.28
Solution
$\Delta Q = {nC}_{v}\Delta T$ (isochoric)
$${= \frac{C_{v}}{R} \cdot nR\Delta T = \frac{C_{v}}{R}\left( P_{2} - P_{1} \right)V }{= \frac{21}{8.3} \times (30 - 21.7) \times 1 = 21\text{ }J}$$
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