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Q.
The figure shows graphs of pressure versus density for an ideal gas at two temperatures $T_1$ and $ T_2$, then
Odisha JEEOdisha JEE 2010Kinetic Theory
Solution:
$pV = nRT = \frac{m}{M}RT$
$\Rightarrow \, \, \, \, \, \, \frac{P}{m / V} = \frac{RT}{M} \Rightarrow \frac{p}{
ho} = \frac{RT}{M}$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \frac{p}{
\rho} \propto T$
$\therefore $ for a given gas M is constant
Temperature is directly proportional to the slope of p - $
\rho$ graph. So, $T_1 > T_2$ .
