Question

Two glass bulbs A and B are connected by a very small tube having a stop-cock. Bulb A has a volume of 100 $c{m}^3$ and contained the gas; while bulb B was empty. On opening the stop-cock, the pressure fell down to 40%. The volume of the bulb B must be

• A. $75\text{ c}{\text{m}}^3$
• B. $125\text{ c}{\text{m}}^3$
• C. $150\text{ c}{\text{m}}^3$
• D. $250\text{ c}{\text{m}}^3$

Answer 1

Answer: (C)

$P_1=1\text{ atm}$

$P_2=\text{1 atm}\times \frac{40}{100}=0.40\text{ atm}$

$V_1=100\text{ c}{\text{m}}^3$

$V_2=?$

At constant temperature, $P_1{V}_1=P_2{V}_2$

So $V_2=\frac{P_1{V}_1}{P_2}$

$-\frac{1\text{ atm}\times \text{100 c}{\text{m}}^3}{0.40\text{ atm}}$

$=250\text{ c}{\text{m}}^3$

Hence, the volume of bulb $B=(250-100){\mathrm{cm}}^3$

$=150{\text{cm}}^{\text{3}}$