Question

Two glass bulbs $A$ (of $100mL$ capacity), and $B$ (of $150mL$ 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 :

• A. Drop by $75$%
• B. Drop $57$%
• C. Drop by $25$%
• D. Will remain same

Answer 1

Answer: (B)

Drop $57$%

Let the original pressure in B be P.

$\therefore$ Pressure in A = 20 P.

Given that :

Volume in A = 100 mL

Volume in B = 150 mL

After opening the stopcock,

total volume $=(100+150)mL=250mL$

According to Boyle's law :

$P_f{V}_f=P_i{V}_i$

Let the partial pressure of gas in A be $P_A$ :

$P_A\times 250=100\times 20P$

$\Rightarrow P_A=\frac{100\times 20P}{250}=8P$

Let Partial pressure of gas in B be $P_B$ :

$P_B\times 250=150\times P$

$P_B=\frac{150P}{250}=0.6P$

$\therefore$ Total pressure $=8P+0.6P=8.6P$

Now,

Drop in pressure in A $=20P-8.6P=11.4P$

$\therefore \%$ drop in pressure $=\frac{11.4P}{20P}\times 100=57\%$

Therefore, the correct option is B.