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 $cm^{3}$ 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 $B=(250-100) \mathrm{cm}^3$

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