There are two vessels filled with an ideal gas where volume of one is double the volume of the other. The large vessel c

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There are two vessels filled with an ideal gas where volume of one is double the volume of the other. The large vessel contains the gas at 8 kPa at 1000 K while the smaller vessel contains the gas at…

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Given: Volume of the larger vessel is double that of the smaller vessel. Initial states are $P_1 = 8 \, \text{kPa},\ T_1 = 1000 \, \text{K}$ for the larger vessel and $P_2 = 7 \, \text{kPa},\ T_2 = 500 \, \text{K}$ for the smaller vessel. Final common temperature is $T_f = 600 \, \text{K}$ .

Find: Final common pressure in the connected vessels.

Use the ideal gas law:

PV = nRT

So the moles in the two vessels are

n_1 = \frac{P_1V_1}{RT_1}, \qquad n_2 = \frac{P_2V_2}{RT_2}

Given $V_1 = 2V_2$ , substitute into the expressions:

n_1 = \frac{8 \times 2V_2}{R \times 1000} = \frac{16V_2}{1000R}

n_2 = \frac{7V_2}{R \times 500} = \frac{14V_2}{1000R}

Hence total moles are

n_{\text{total}} = n_1 + n_2 = \frac{16V_2}{1000R} + \frac{14V_2}{1000R} = \frac{30V_2}{1000R}

After connection, the total volume is

V_1 + V_2 = 2V_2 + V_2 = 3V_2

At the final state,

P_f(3V_2) = n_{\text{total}}R(600)

Substituting $n_{\text{total}}$ ,

P_f(3V_2) = \frac{30V_2}{1000R} \cdot R \cdot 600

Therefore,

P_f = \frac{30 \times 600}{1000 \times 3} = 6 \, \text{kPa}

Therefore, the final pressure in both vessels is $6 \, \text{kPa}$ . The correct option is B.

The second approach shown in the source also concludes the same result, although it interchanges the vessel labels while substituting volumes. The final answer still comes out to $6 \, \text{kPa}$ .

Given: Two rigid vessels are connected, and the total amount of gas remains constant.

Find: Final common pressure at $600 \, \text{K}$ .

For an ideal gas in separate vessels, total moles are proportional to the sum of $\frac{PV}{T}$ for each part:

n_{\text{total}}R = \frac{P_1V_1}{T_1} + \frac{P_2V_2}{T_2}

At the final state,

P_f(V_1+V_2) = n_{\text{total}}RT_f

So,

P_f = \frac{T_f\left(\frac{P_1V_1}{T_1} + \frac{P_2V_2}{T_2}\right)}{V_1+V_2}

Take the smaller vessel volume as $V$ . Then the larger vessel has volume $2V$ . Hence,

P_f = \frac{600\left(\frac{8 \cdot 2V}{1000} + \frac{7 \cdot V}{500}\right)}{3V}

= \frac{600\left(\frac{16V}{1000} + \frac{14V}{1000}\right)}{3V}

= \frac{600 \cdot 30}{1000 \cdot 3} = 6 \, \text{kPa}

This works because the gas redistributes, but the total number of moles remains unchanged.

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