JEE Main · ChemistryMedium

$Two flasks of equal volume are connected by a narrow tube of negligible volume. One flask contains 1 mole of H_{2} at 300 K and the other contains 1 mole of He at 600 K. If the connecting valve is opened, what is the final pressure in terms of the initial pressure P of the H_{2} flask?$

Show correct answer & step-by-step solution

Correct answer: C — $$1.25 P$$

Solution

$Initial state: P_{1} = P, V_{1} = V, n_{1} = 1, T_{1} = 300 K. For the second flask, P_{2} = \frac{n _{2} R T _{2}}{V} = \frac{1 \cdot R \cdot 600}{V} = 2 P .$
$Total moles n_{t o t a l} = 1 + 1 = 2 .$
$Final pressure P_{f} = \frac{n _{t o t a l} R T _{f}}{2 V}, where T_{f} is the average temperature 450 K.$
$P_{f} = \frac{2 \cdot R \cdot 450}{2 V} = \frac{450 R}{V} = 1.25 \cdot \frac{300 R}{V} = 1.25 P .$
$Hence the answer is (C).$