Question

Two glass bulb $A$ and $B$ are connected by a very small tube (of negligible volume) having stopcock. Bulb $A$ has a volume of $100c{m}^3$ and contains certain gas while bulb $B$ is empty. On opening the stopcock, the pressure in $A$ fell down by 60%. The volume of bulb $B$ must be:

• A. 200 mL
• B. 150 mL
• C. 250 mL
• D. 100 mL

Answer 1

Answer: (B)

150 mL

Thus $P_2=0.4P_1$.

The expression for the boyle's law is $P_1{V}_1=P_2{V}_2$.

Substitute values in the above expression:

$P_1\times \left(100c{m}^3\right)=0.4P_1{V}_2$.

Thus $V_2=250c{m}^3$.

$V_2$ is the total volume and includes the volume of bulb and bulb $B$.

Hence, the volume of bulb $B$ is $250-100=150mL$.