Question

Two glass bulbs $A$ and $B$ at same temperature are connected by a very small tube having a stop-cock. Bulb $A$ has a volume of $100{cm}^3$ and contained the gas while bulb $B$ was empty. On opening the stop cock, the pressure fell down 20% The volume of the bulb $B$ is:

• A. $100c{m}^3$
• B. $200c{m}^3$
• C. $250c{m}^3$
• D. $400c{m}^3$

Answer 1

Answer: (B)

$200c{m}^3$

let the initial pressure in bulb $A=P$ and final pressure $=P_2$

Final pressuer is 40 % of the initial pressure. So $P_2=\left(40/100\right)P$

Volume in $A=100$ ml

Let initially volume of $B=V$

After opening the cock final volume will be $(100+V)$ mL

According to Boyle"s law

$P_1{V}_1=P_2{V}_2$

$P\times 100=\left(40/100\right)P\times (100+V)$

$60\times 100/40=V$

$V=150$ mL