Select the correct statement.
- The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules.
- The temperature of gas is $-73^{\circ} \mathrm{C}$. When the gas is heated to $527^{\circ} \mathrm{C}$, the r.m.s. speed of the molecules is doubled.
- The temperature of gas is $-100^{\circ} \mathrm{C}$. When the gas is heated to $+627^{\circ} \mathrm{C}$, the r.m.s. speed of the molecules is four times.
- The product of pressure and volume of an ideal gas will be equal to half the translational kinetic energy.
Solution
Option A: The product of pressure and volume for an ideal gas is given by $PV = nRT$.
The translational kinetic energy is $K_{\text{trans}} = \frac{3}{2}nRT = \frac{3}{2}PV$.
Thus $PV = \frac{2}{3}K_{\text{trans}}$, not equal to $K_{\text{trans}}$ itself. Option A is incorrect.
Option B: Convert temperatures to Kelvin: $T_1 = -73 + 273 = 200$ K, $T_2 = 527 + 273 = 800$ K.
Since $v_{\text{rms}} \propto \sqrt{T}$, we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = 2$.
The r.m.s. speed doubles, making Option B correct.
Option C: Convert temperatures: $T_1 = -100 + 273 = 173$ K, $T_2 = 627 + 273 = 900$ K.
The ratio $\frac{v_2}{v_1} = \sqrt{\frac{900}{173}} \approx 2.28 \neq 4$.
Option C is incorrect.
Option D: As shown previously, $PV = \frac{2}{3}K_{\text{trans}} \neq \frac{1}{2}K_{\text{trans}}$.
Option D is incorrect.
The only correct statement is Option B.
Asked in: MHT CET 2025 (23 April Shift 2)