Q.35. Two closed vessels of same volume are joined through a narrow tube and both vessels are filled

Q: What is the correct answer for JEE Main Physics Q35?

The correct answer is A. Sol. Let the volume of each vessel be $V$. Initial state: Total volume $V_{\text{total}} = 2V$. Initial pressure $P_1 = 90 \text{ kPa}$. Initial temperature $T_1 = 400 \text{ K}$. Using the ideal gas law $PV=nRT$, the initial total number of moles is $n_{\text{total}} = \frac{P_1 V_{\text{total}}}{RT_1} = \frac{(90 \times 10^3)(2V)}{R(400)}$. Final state: Let the final pressure be $P_f$. One vesse

Q: Q.35. Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure 90 kPa and temperature 400 K. Keeping the temperature of one vessel constant…

The correct answer is A. Sol. Let the volume of each vessel be $V$. Initial state: Total volume $V_{\text{total}} = 2V$. Initial pressure $P_1 = 90 \text{ kPa}$. Initial temperature $T_1 = 400 \text{ K}$. Using the ideal gas law $PV=nRT$, the initial total number of moles is $n_{\text{total}} = \frac{P_1 V_{\text{total}}}{RT_1} = \frac{(90 \times 10^3)(2V)}{R(400)}$. Final state: Let the final pressure be $P_f$. One vesse

Q: What are the options for this JEE Main Physics question?

A. 100 · B. 120 · C. 90 · D. 105

Question

Q.35. Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure 90 kPa and temperature 400 K. Keeping the temperature of one vessel constant at 400 K the second vessel temperature is raised to 500 K. The final pressure in the vessels is ________ kPa.

FirstInTest · Accepted Answer

Sol. Let the volume of each vessel be $V$. Initial state: Total volume $V_{	ext{total}} = 2V$. Initial pressure $P_1 = 90 	ext{ kPa}$. Initial temperature $T_1 = 400 	ext{ K}$. Using the ideal gas law $PV=nRT$, the initial total number of moles is $n_{	ext{total}} = \frac{P_1 V_{	ext{total}}}{RT_1} = \frac{(90 	imes 10^3)(2V)}{R(400)}$. Final state: Let the final pressure be $P_f$. One vessel has volume $V$ and temperature $T_A = 400 	ext{ K}$. The other vessel has volume $V$ and temperature $T_B = 500 	ext{ K}$. The total number of moles remains constant. $n_{	ext{total}} = n_A + n_B$. $n_A = \frac{P_f V}{RT_A} = \frac{P_f V}{R(400)}$. $n_B = \frac{P_f V}{RT_B} = \frac{P_f V}{R(500)}$. So, $\frac{(90 	imes 10^3)(2V)}{R(400)} = \frac{P_f V}{R(400)} + \frac{P_f V}{R(500)}$. Cancel $V/R$ from all terms: $\frac{90 	imes 10^3 	imes 2}{400} = \frac{P_f}{400} + \frac{P_f}{500}$. $\frac{180 	imes 10^3}{400} = P_f \left(\frac{1}{400} + \frac{1}{500}ight)$. $450 = P_f \left(\frac{500+400}{400 	imes 500}ight) = P_f \left(\frac{900}{200000}ight)$. $P_f = 450 	imes \frac{200000}{900} = 0.5 	imes 200000 = 100000 	ext{ Pa} = 100 	ext{ kPa}$.

Answer

100

Answer

120

Answer

90

Answer

105

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