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Two glass bulbs A (of 100 mL capacity), and B (of 150 mL capacity) containing same gas are connected by a small tube of negligible volume. At particular temperature the pressure in A was found to be 20 times more than that in bulb B. The stopcock is opened Without changing the temperature. The pressure in A will :

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drop by 75%

drop by 57%

will remain same

drop by 25%

answer is B.

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Detailed Solution

It is known that;

$\begin{array}{l} P_{1} V_{1} + P_{2} V_{2} = P (V_{1} + V_{2}) \\ \Rightarrow 20 x + 100 + x \times 150 = P \times 250 \\ P = 8.6 x \end{array}$

Pressure drop in A= 11.4.x

$∴ %$ drop $= 57 %$

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