Question

Two glass bulbs $A$ ( of $100ml$ volume) and $B$ (of $150ml$ volume) 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 temperature. The percentage decreases in pressure will be:

• A. $57\%$
• B. $22\%$
• C. $23\%$
• D. $45\%$

Answer 1

Answer: (A)

Before opening the stopcock:

$P_A\times 100=n_A\times R\times T\dots \dots (1)$

$P_B\times 150=n_B\times R\times T\dots \dots (2)$

After opening the stop cock-

$P\times 250=\left(n_A+n_B\right)\times R\times T$

$250P=\left(\frac{100P_A}{RT}+\frac{150P_B}{RT}\right)\times RT$

$250P=100P_A+150P_B$

Given $:P_A=20P_B$

$250P=100P_A+150\times \frac{P_A}{20}$

$P=\frac{107.5}{250}{P}_A=0.43P_A$

Percentage decrease in pressure

$=\frac{P_A-0.43P_A}{P_A}\times 100$

$=57\%$