JEE MAIN - Physics (2023 - 10th April Morning Shift - No. 6)

Consider two containers A and B containing monoatomic gases at the same Pressure (P), Volume (V) and Temperature (T). The gas in A is compressed isothermally to $$\frac{1}{8}$$ of its original volume while the gas in B is compressed adiabatically to $$\frac{1}{8}$$ of its original volume. The ratio of final pressure of gas in B to that of gas in A is
$$\frac{1}{8}$$
8$$^\frac{3}{2}$$
4
8

Explanation

The final pressure of gas in container A after isothermal compression can be found using the equation of state for an ideal gas, $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature. For an isothermal process, the temperature $T$ is constant, so the equation becomes $P_1V_1 = P_2V_2$.

The final volume is $\frac{1}{8}$ of the initial volume, so $P_2 = P_1 \times \frac{V_1}{V_2} = P_1 \times 8 = 8P_1$.

The final pressure of gas in container B after adiabatic compression can be found using the adiabatic equation for an ideal gas, $PV^\gamma = \text{constant}$, where $\gamma$ is the ratio of the heat capacities, which is $\frac{5}{3}$ for a monoatomic gas.

Since $V_2 = \frac{V_1}{8}$, we have $P_2 = P_1 \times \left(\frac{V_1}{V_2}\right) ^ \gamma = P_1 \times 8^\frac{5}{3} = P_1 \times 2^5 = 32P_1$.

The ratio of the final pressure of gas in B to that of gas in A is therefore $\frac{32P_1}{8P_1} = 4$.

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