Physics Q35 – JEE-MAIN 2026

Home Physics ThermodynamicsQuestion #522

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

Clue 1 of 3: Conservation of Moles

The total number of moles of air in the combined system remains constant throughout the process.

🥷

Ninja Strategy•Range Estimation and Elimination

Estimate the expected range of the final pressure based on temperature changes to eliminate options that are too low or too high, then perform the full calculation if needed.

🧠 Full Solution Path

Step 1: Calculate Initial Total Moles✦ Active

Let the volume of each vessel be $V$ . Initially, both vessels are at pressure $P_{i} = 90 kPa$ and temperature $T_{i} = 400 K$ . Using the ideal gas law $P V = n R T$ , the initial moles in each vessel are $n_{1} = n_{2} = \frac{P_{i} V}{R T_{i}}$ . The total initial moles are $n_{t o t a l} = n_{1} + n_{2} = \frac{2 P_{i} V}{R T_{i}}$ .

n_{t o t a l} = \frac{2 \times 90 \times V}{R \times 400} = \frac{180 V}{400 R} = \frac{9 V}{20 R}

Step 2: Set Up Final Moles and Pressure○ Expand

In the final state, the temperature of the first vessel is $T_{1}^{'} = 400 K$ and the second vessel is $T_{2}^{'} = 500 K$ . Let the final pressure be $P_{f}$ . The moles in each vessel are $n_{1}^{'} = \frac{P_{f} V}{R T_{1}^{'}}$ and $n_{2}^{'} = \frac{P_{f} V}{R T_{2}^{'}}$ . The total final moles are $n_{t o t a l}^{'} = n_{1}^{'} + n_{2}^{'}$ .

n_{t o t a l}^{'} = \frac{P_{f} V}{R \times 400} + \frac{P_{f} V}{R \times 500} = \frac{P_{f} V}{R} (\frac{1}{400} + \frac{1}{500})

Step 3: Apply Conservation of Moles and Solve for Final Pressure○ Expand

By conservation of moles, $n_{t o t a l} = n_{t o t a l}^{'}$ . Equating the expressions from Step 1 and Step 2:

\frac{9 V}{20 R} = \frac{P_{f} V}{R} (\frac{1}{400} + \frac{1}{500})

\frac{9}{20} = P_{f} (\frac{500 + 400}{400 \times 500})

\frac{9}{20} = P_{f} (\frac{900}{200000})

\frac{9}{20} = P_{f} (\frac{9}{2000})

P_{f} = \frac{9}{20} \times \frac{2000}{9} = \frac{2000}{20} = 100 kPa

💡 Teacher's Secret Hint

Ensure to cancel common terms like $V$ and $R$ early to simplify calculations.

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Physics Question 35 – JEE-MAIN 2026

Clue 1 of 3: Conservation of Moles

🧠 Full Solution Path