Question

Two glass bulbs $A$ ( of $100 \,ml$ volume) and $B$ (of $150 \,ml$ 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:

$P_{A} \times 100=n_{A} \times R \times T \ldots \ldots(1)$

$P_{B} \times 150=n_{B} \times R \times T \ldots \ldots(2)$

After opening the stop cock-

$P \times 250=\left( n _{ A }+ n _{ B }\right) \times R \times T$

$250 P =\left(\frac{100 P _{ A }}{ RT }+\frac{150 P _{ B }}{ RT }\right) \times RT $

$250 P =100 P _{ A }+150 P _{ B }$

Given $: P_{A}=20 P_{B}$

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

$P=\frac{107.5}{250} P_{A}=0.43 P_{A}$

Percentage decrease in pressure

$=\frac{P_{A}-0.43 P_{A}}{P_{A}} \times 100$

$=57 \%$