A diatomic gas undergoes adiabatic change. Its pressure 'P' and temperature 'T' are related as $\mathrm{P} \propto \mathrm{T}^{\mathrm{x}}$, where $\mathrm{x}$ is

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Aicharya · Accepted Answer

$\begin{array}{ll}	ext {Adiabatic change } ightarrow \mathrm{PV}^{\gamma}=\mathrm{k} & \mathrm{PV}=\mathrm{RT} \ \mathrm{P} \frac{\mathrm{R}^{\gamma} \mathrm{T}^{\gamma}}{\mathrm{p}^{\gamma}}=\mathrm{k} & \mathrm{V}^{\gamma}=\left(\frac{\mathrm{RT}}{\mathrm{P}}ight)^{\gamma}\end{array}$ $\mathrm{P}^{1-\gamma} \mathrm{T}^{\gamma}=\mathrm{k}^{\prime}$ $\mathrm{P}^{1-\gamma}=\frac{\mathrm{k}^{\prime}}{\mathrm{T}^{\gamma}}$ $P=\frac{k^{n}}{T^{\gamma / 1-\gamma}}=k^{n} T\left(\frac{\gamma}{1-\gamma}ight)$ $	herefore \quad \mathrm{x}=\frac{\gamma}{\gamma-1}$ For diatomic gas $\gamma=1.4$ $	herefore \quad x=\frac{1.4}{0.4}=3.5$

A diatomic gas undergoes adiabatic change. Its pressure 'P' and temperature 'T' are related as $\mathrm{P}…

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