The initial pressure and volume of a given ma | vedclass

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(B) For an adiabatic process, the relationship between temperature and volume is given by $T V^{\gamma-1} = 	ext{constant}$.Let the temperature after

Vedclass · Accepted Answer

$(2^{\gamma-1}-1) \frac{p_{0} V_{0}}{\gamma-1}$

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(B) For an adiabatic process, the relationship between temperature and volume is given by $T V^{\gamma-1} = 	ext{constant}$.Let the temperature after adiabatic compression be $T$. Then, $T_{0} V_{0}^{\gamma-1} = T \left(\frac{V_{0}}{2}ight)^{\gamma-1}$.Solving for $T$, we get $T = T_{0} 2^{\gamma-1}$.When the gas is allowed to come to thermal equilibrium with the surroundings at constant volume $\frac{V_{0}}{2}$, the heat released $\Delta Q$ is equal to the change in internal energy: $\Delta Q = n C_{V} \Delta T$.Using $C_{V} = \frac{R}{\gamma-1}$ and the ideal gas law $n R T_{0} = p_{0} V_{0}$, we have:$\Delta Q = n \left(\frac{R}{\gamma-1}ight) (T - T_{0}) = \frac{n R T_{0}}{\gamma-1} (2^{\gamma-1} - 1)$.Substituting $n R T_{0} = p_{0} V_{0}$, the heat released is $\frac{p_{0} V_{0}}{\gamma-1} (2^{\gamma-1} - 1)$.

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