Question

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

• A. drop by 75%
• B. drop by 57%
• C. drop by 25%
• D. will remain same

Answer 1

The correct option is B drop by 57%

Given: $V_A=100mL$, $V_B=150mL$,
Let $P_B=P$ then $P_A=20P$
and $V_f=V_A+V_B=100mL+150mL=250mL$
As we know, $n_T=n_A+n_B$
so, $\frac{P_f{V}_f}{RT}=\frac{P_A{V}_A}{RT}+\frac{P_B{V}_B}{RT}$
i.e $P_f{V}_f=P_A{V}_A+P_B{V}_B$
$\Rightarrow P_f\times 250mL=20P\times 100+P\times 150$
$P_f=\frac{2150P}{250}$
Decrease in $P$ of $A=20P-\frac{215}{25}\times P$
% drop $=\frac{285}{25}\times P\times \frac{1}{20P}\times 100=57\%$
Thus, the pressure of A will drop by $57$%.