The figure shows an adiabatic curve on a log | vedclass

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(A) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = K$,where $K$ is a constant.Taking t

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Monatomic

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(A) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = K$,where $K$ is a constant.Taking the logarithm on both sides,we get $\log T + (\gamma-1) \log V = \log K$.This equation is of the form $y = mx + c$,where $y = \log T$,$x = \log V$,and the slope $m = -(\gamma-1)$.From the given graph,the coordinates of points $A$ and $B$ are $(1, 4)$ and $(4, 2)$ respectively.The slope $m$ is calculated as $\frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 4}{4 - 1} = \frac{-2}{3}$.Equating the slopes,we have $-(\gamma-1) = -\frac{2}{3}$,which implies $\gamma - 1 = \frac{2}{3}$.Therefore,$\gamma = 1 + \frac{2}{3} = \frac{5}{3}$.Since $\gamma = \frac{5}{3}$ for a monatomic gas,the gas is monatomic.

The figure shows an adiabatic curve on a $\log T$ and $\log V$ scale for an ideal gas. The gas is ............

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