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 40%. The volume of the bulb B must be ?

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

Answer 1

Answer: (C)

150 $c{m}^3$

Let the initial pressure in bulb A be P.

Given that the pressure fell down $40\%$ on opening the stop cock

$\therefore$ Final pressure $=\frac{40}{100}P=\frac{2}{5}P$

Initial voume of bulb A $=100{cm}^3$

Let initial volume of bulb B be V.

$\therefore$ total volume after opening the stopcock $=100+V$

By using Boyle's law,

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

$P\times 100=\frac{2}{5}P\times (100+V)$

$\Rightarrow V=150{cm}^3$

Hence, the volume of bulb B must be $150{cm}^3$.