Question

The figure shows graphs of pressure versus density for an ideal gas at two temperatures $T_1$ and $T_2$, then

• A. $T_1 > T_2$
• B. $T_1 = T_2$
• C. $T_1 < T_2$
• D. Nothing can be predicted

Answer 1

Answer: (A)

$T_1 > T_2$

Answer 2

$pV = nRT = \frac{m}{M}RT$
$\Rightarrow \, \, \, \, \, \, \frac{P}{m / V} = \frac{RT}{M} \Rightarrow \frac{p}{ ho} = \frac{RT}{M}$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \frac{p}{ \rho} \propto T$
$\therefore$ for a given gas M is constant
Temperature is directly proportional to the slope of p - $\rho$ graph. So, $T_1 > T_2$ .