Question

A bulb of unknown volume ${}^{\prime }{V}^{\prime }$ contains an ideal gas at $1atm$ pressure. This bulb was connected to another evacuated bulb of volume $0.5L$ through a stopcock. When stopcock was opened, the pressure at each bulb became $530mmofHg$ while the temperature remained constant. Find $V$ in Litres.

• A. $11.52L$
• B. $1.152L$
• C. $1.52L$
• D. $2.152L$

Answer 1

The correct option is B $1.152L$

$1atm=760mmofHg$
$1mmofHg=\frac{1}{760}atm$
$530mmofHg=\frac{530}{760}atm$
Applying Boyle's law:
$P_1{V}_1=P_2{V}_2$ (at constant temperature)
$1\times V=\frac{530}{760}\times (V+0.5)$
$V=1.152L$