The P-V diagram for two gases during an adiab | vedclass

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(A) For an adiabatic process, the equation is $PV^{\gamma} = 	ext{constant}$.Differentiating with respect to $V$, we get $\frac{dP}{dV} V^{\gamma} +

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$O_2$ and $He$

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(A) For an adiabatic process, the equation is $PV^{\gamma} = 	ext{constant}$.Differentiating with respect to $V$, we get $\frac{dP}{dV} V^{\gamma} + P \gamma V^{\gamma-1} = 0$.Thus, the slope $m = \frac{dP}{dV} = -\gamma \frac{P}{V}$.From the figure, the magnitude of the slope of curve $B$ is greater than the magnitude of the slope of curve $A$ $(|m_B| > |m_A|)$.Therefore, $\gamma_B > \gamma_A$.For monoatomic gases (like $He$, $Ar$), $\gamma = 1.67$.For diatomic gases (like $O_2$), $\gamma = 1.4$.Since $\gamma_B > \gamma_A$, curve $B$ represents a monoatomic gas and curve $A$ represents a diatomic gas.Thus, $A$ corresponds to $O_2$ and $B$ corresponds to $He$ (or $Ar$).Comparing with the options, $A$ is $O_2$ and $B$ is $He$ fits the condition.

The $P-V$ diagram for two gases during an adiabatic process is shown in the figure. Curves $A$ and $B$ correspond to:

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