Question

A bubble of volume $V_1$ is present in the bottom of a pond at $15^{\circ }C$ and $1.5atm$ pressure. When it comes at the surface, it observes a pressure of 1 atm at $25^{\circ }C$ and has volume $V_2$. The value of $V_2/V_1$ is

• A. $15.5$
• B. $0.155$
• C. $155.0$
• D. $1.55$

Answer 1

Answer: (D)

$p_1=1.5$ atm , $T_1=15^{\circ }C=(15+273)K=288K$
$p_2=1$ atm , $T_2=25^{\circ }C=(25+273)=298K$
According to combined gas law,
$\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}$
$\Rightarrow \frac{p_1{T}_2}{T_1{p}_2}=\frac{V_2}{V_1}$
$\Rightarrow \frac{V_2}{V_1}=\frac{1.5\times 298}{288\times 1}=1.55$