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Q. Figure shows, the adiabatic curve on a log $T$ and $\log V$ scale performed on ideal gas. The gas isPhysics Question Image

Thermodynamics

Solution:

$T V^{\gamma-1}=K$
$\log T+(\gamma-1) \log V=0$
$\log T=-(\gamma-1) \log V$
$y=-(\gamma-1) x$
$\Rightarrow -(\gamma-1)=-\frac{2}{3}$
$\frac{y}{x}=-(\gamma-1)=$ slope $=\frac{2-4}{4-1}$
$\gamma=\frac{5}{3}$
$\therefore $ Monoatomic.