Question

A bulb of unknown volume $V$ contains an ideal gas at $2$ atm pressure. It was connected to another evacuated bulb of volume $0.5$ litre through a stopcock. When the stopcock was opened, the pressure in each bulb became $0.5$ atm. The V is:

• A. $17l$
• B. $1.7l$
• C. $0.17l$
• D. $0.34l$

Answer 1

Answer: (C)

$0.17l$

Initial volume $=V$

Initial Pressure $\left(P_i\right)=2$ $atm$

Final volume $=V+0.5$ $L$

Final pressure $=0.5$ $atm$

According to Boyle's law,

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

$\therefore 2\times V=0.5(V+0.5)$

$⟹4V=V+0.5$

$\therefore 3V=0.5$

$⟹V=0.17$ $L$

Hence, option $C$ is correct.