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Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius $r_A$ and $r_B$, respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas A and B are displaced by $16\text{ cm}$ and $9\text{ cm}$, respectively. If the change in their internal energy is the same, then the ratio $r_A / r_B$ is equal to:
A
$\frac{4}{3}$
B
$\frac{3}{4}$
C
$\frac{2}{\sqrt{3}}$
D
$\frac{3}{2}$
Explanation
Since $Q$ and $\Delta U$ are equal, work done $W = P \Delta V$ is identical. $\pi r_A^2 (16) = \pi r_B^2 (9) \implies r_A / r_B = 3/4$.
Detailed Solution
From the first law of thermodynamics: $Q = \Delta U + W$. Since equal heat $Q_A = Q_B$ is supplied and internal energy change is identical $\Delta U_A = \Delta U_B$, the work done by both gases must be equal: $W_A = W_B$. At constant pressure $P$: $W = P \Delta V \implies P \Delta V_A = P \Delta V_B \implies \Delta V_A = \Delta V_B$. For a cylindrical piston of radius $r$ displaced by $d$: $\Delta V = (\pi r^2) d$. Thus: $\pi r_A^2 (16\text{ cm}) = \pi r_B^2 (9\text{ cm}) \implies \frac{r_A^2}{r_B^2} = \frac{9}{16} \implies \frac{r_A}{r_B} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.