Question

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 $cm$ and 9 $cm$, respectively. If the change in their internal energy is the same, then the ratio

$\frac{r_A}{r_B}$ is equal to

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (B)

$\frac{3}{4}$

Answer 2

The core idea is based on the First Law of Thermodynamics ($Q = \Delta U + W$). Given that equal heat ($Q$) is supplied and the change in internal energy ($\Delta U$) is the same for both gases, it implies that the work done ($W$) by both gases must be equal ($W_A = W_B$). Since the process occurs at constant pressure ($P$), the work done is $W = P \Delta V$. As $P$ is also the same for both, it follows that the change in volume ($\Delta V$) for both gases must be equal ($\Delta V_A = \Delta V_B$). The change in volume for a cylinder with a movable piston is given by $\Delta V = \text{Area} \times \text{displacement} = \pi r^2 h$. Equating the changes in volume, $\pi r_A^2 h_A = \pi r_B^2 h_B$, which simplifies to $r_A^2 h_A = r_B^2 h_B$. Rearranging this gives $\frac{r_A}{r_B} = \sqrt{\frac{h_B}{h_A}}$. Plugging in the given displacements $h_A = 16 \, \text{cm}$ and $h_B = 9 \, \text{cm}$, we get $\frac{r_A}{r_B} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.

Answer: The ratio $\frac{r_A}{r_B}$ is equal to $\frac{3}{4}$. The correct option is (2).