Select the correct observation for a 8.21 litre container, filled with 2 moles of He at 300K)

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It has pressure 6 atm.

If it is an open rigid container, its pressure increases to 8 atm on heating to 400K.

If it is closed non-rigid (like thin skin balloon), its volume increases to 16.42 lit on heating to 600K.

When connected with another similar empty container maintained at 150K while maintaining original container at 300K, pressure reduced to $\frac{2}{3} atm$

answer is A, C.

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Detailed Solution

Given,

Volume $= 8.21 L$

$\begin{array}{l} n_{re} = 2 moles \\ T = 300 K \end{array}$

From ideal gas equation,

$\begin{array}{l} p v = n R T \\ P \times 8.21 = 2 \times 0.0821 \times 300 \\ P = 6 atm \end{array}$

Open container

$P = Pgas + Patm .$

$Pages = \frac{2 \times 0.821 \times 400}{8.21} = 8 atm$

so, pressure will be more than 8 atm
closed non rigid (P= constant)

$\begin{array}{l} \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}, \frac{8.21}{300} = \frac{V_{2}}{600} \\ v_{2} = 16.42 \\ \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}, \frac{6}{300} = \frac{P_{2}}{150} \\ P_{2} = 3 atm, P_{1} = 6 a + m \end{array}$