There are two vessels filled with an ideal gas where volume of one is double the volume of other. The large vessel contains the gas at 8 kPa at 1000 K while the smaller vessel contains the gas at 7 kPa at 500 K . If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at 600 K , at steady state the pressure in the vessels will be (in kPa ).

Heat and Thermodynamics PYQ — JEE Main 2025 · 4 Apr · Shift 2 · Q64 with Solution

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Correct answer: D

Let the smaller vessel have volume $V$ .

Then the larger vessel has volume $2V$ .
Initial moles in each vessel using the ideal gas law:

$n=\frac{PV}{RT}$
- For the large vessel: $n_1=\frac{(8)(2V)}{R(1000)}=\frac{16V}{1000R}=\frac{2V}{125R}$
- For the small vessel: $n_2=\frac{(7)(V)}{R(500)}=\frac{7V}{500R}$
Total number of moles initially:

$n_{\text{total}}=n_1+n_2$ $n_{\text{total}}=\frac{16V}{1000R}+\frac{7V}{500R}$

Taking common denominator $1000R$ :

$n_{\text{total}}=\frac{16V}{1000R}+\frac{14V}{1000R}=\frac{30V}{1000R}=\frac{3V}{100R}$
After connecting the vessels, the gas reaches steady state at common temperature $600\,\text{K}$ and common pressure $P$ .

Total volume available:

$V_{\text{total}}=V+2V=3V$

Apply ideal gas law to the whole system:

$P(3V)=n_{\text{total}}R(600)$

Substitute $n_{\text{total}}=\dfrac{3V}{100R}$ :

$P(3V)=\frac{3V}{100R}\cdot R \cdot 600$

$P(3V)=18V$

$P=\frac{18V}{3V}=6\,\text{kPa}$
Option check:
- A: $24$ ❌
- B: $4.4$ ❌
- C: $18$ ❌
- D: $6$ ✅

Therefore, the steady-state pressure is:

$\boxed{6\,\text{kPa}}$