Question

The figure shows pressure versus density graph for an ideal gas at two temperatures $T_1$ and $T_2$

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

Answer 1

The correct option is A $T_1$ > $T_2$

${\left(\frac{dP}{d\rho }\right)}_{T_1}>{\left(\frac{dP}{d\rho }\right)}_{T_2}$

This means, the curve for $T_1$ is more steeper than that of $T_2$

$\therefore T_1>T_2$