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(B) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}$.Using the ideal gas equation

Vedclass · Accepted Answer

$T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}}$

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(B) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}$.Using the ideal gas equation for $1$ mole,$PV = RT$,we can write $V = \frac{RT}{P}$.Substituting this into the adiabatic equation: $P \left( \frac{RT}{P} ight)^{\gamma} = 	ext{constant}$.This simplifies to $P^{1-\gamma} T^{\gamma} = 	ext{constant}$.Therefore,$P_1^{1-\gamma} T_1^{\gamma} = P_2^{1-\gamma} T_2^{\gamma}$.Rearranging for $T_2$: $\left( \frac{T_2}{T_1} ight)^{\gamma} = \left( \frac{P_1}{P_2} ight)^{1-\gamma} = \left( \frac{P_2}{P_1} ight)^{\gamma - 1}$.Taking the $\gamma$-th root on both sides: $T_2 = T_1 \left( \frac{P_2}{P_1} ight)^{\frac{\gamma - 1}{\gamma}}$.

One mole of an ideal gas $(C_p/C_v = \gamma)$ at absolute temperature $T_1$ is adiabatically compressed from an initial pressure $P_1$ to a final pressure $P_2$. The resulting temperature $T_2$ of the gas is given by

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