The P-V diagram for two adiabatic processes i | vedclass

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(B) For an adiabatic process, the equation is $PV^{\gamma} = 	ext{constant}$.The slope of the adiabatic curve is given by $\frac{dP}{dV} = -\gamma \f

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$O_2$ and $He$

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(B) For an adiabatic process, the equation is $PV^{\gamma} = 	ext{constant}$.The slope of the adiabatic curve is given by $\frac{dP}{dV} = -\gamma \frac{P}{V}$.Thus, the magnitude of the slope is directly proportional to the adiabatic index $\gamma$ (i.e., $|	ext{slope}| \propto \gamma$).For a monoatomic gas (like $He$), $\gamma = 1.66$, and for a diatomic gas (like $O_2$), $\gamma = 1.4$.Since $\gamma_{monoatomic} > \gamma_{diatomic}$, the slope of the adiabatic curve for a monoatomic gas is steeper than that for a diatomic gas.From the given $P-V$ graph, curve $2$ is steeper than curve $1$.Therefore, curve $2$ corresponds to the monoatomic gas $(He)$ and curve $1$ corresponds to the diatomic gas $(O_2)$.

The $P-V$ diagram for two adiabatic processes is given. Curves $1$ and $2$ correspond to:

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