A diatomic ideal gas undergoes a thermodynami | vedclass

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(A) For process $AB$,the volume is constant $(V = V_0)$,so it is an isochoric process.Work done $W_{AB} = 0$.The heat given is $\Delta Q = \Delta U =

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$2.5$

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(A) For process $AB$,the volume is constant $(V = V_0)$,so it is an isochoric process.Work done $W_{AB} = 0$.The heat given is $\Delta Q = \Delta U = n C_v \Delta T$.For a diatomic gas,$C_v = \frac{5}{2} R$.Using the ideal gas law $PV = nRT$,we have $n \Delta T = \frac{\Delta(PV)}{R} = \frac{P_B V_B - P_A V_A}{R}$.Given $P_A = P_0, V_A = V_0$ and $P_B = 2P_0, V_B = V_0$.So,$n \Delta T = \frac{(2P_0)(V_0) - (P_0)(V_0)}{R} = \frac{P_0 V_0}{R}$.Substituting this into the heat equation:$\Delta Q = n \left( \frac{5}{2} R \right) \Delta T = \frac{5}{2} R (n \Delta T) = \frac{5}{2} R \left( \frac{P_0 V_0}{R} \right) = 2.5 P_0 V_0$.

$A$ diatomic ideal gas undergoes a thermodynamic change according to the $P-V$ diagram shown in the figure. The heat given to the gas in process $AB$ is: (in $, P_0 V_0$)

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