A diatomic gas undergoes an adiabatic change. | vedclass

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Question

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

Vedclass · Accepted Answer

$3.5$

Vedclass · Answer

(C) For an adiabatic process, the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}$.From the ideal gas equation, $PV = RT$, we have $V = \frac{RT}{P}$.Substituting $V$ in the adiabatic equation:$P \left( \frac{RT}{P} ight)^{\gamma} = 	ext{constant}$$P \cdot \frac{T^{\gamma}}{P^{\gamma}} = 	ext{constant}'$$P^{1-\gamma} T^{\gamma} = 	ext{constant}'$$P^{1-\gamma} = \frac{	ext{constant}'}{T^{\gamma}}$$P = 	ext{constant}'' \cdot T^{\frac{\gamma}{\gamma-1}}$Comparing this with $P \propto T^{x}$, we get $x = \frac{\gamma}{\gamma-1}$.For a diatomic gas, the adiabatic exponent $\gamma = 1.4$ (or $\frac{7}{5}$).Substituting the value of $\gamma$:$x = \frac{1.4}{1.4 - 1} = \frac{1.4}{0.4} = 3.5$.Therefore, the correct option is $C$.

$A$ diatomic gas undergoes an adiabatic change. Its pressure $P$ and temperature $T$ are related as $P \propto T^{x}$, where $x$ is:

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