Question

A bubble occupies volume $V_1$ at the bottom of a pond at ${20}^{\circ }\text{C}$ and 1.5 atm pressure. When it comes at the surface, it observes a pressure of 1 atm at ${25}^{\circ }\text{C}$ and occupies volume $V_2$. Find $\left(\frac{V_2}{V_1}\right)$

• A. 15.3
• B. 0.153
• C. 153
• D. 1.53

Answer 1

The correct option is D 1.53

Given $P_1$ = 1.5 atm $T_1$ = 20${}^{\circ }\text{C}$ = 293 K
$P_2$ = 1 atm, $T_2$ = 25${}^{\circ }\text{C}$ = 298 K
From, ideal gas equation,
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$\frac{V_2}{V_1}=\frac{P_1{T}_2}{P_2{T}_1}$
$=\frac{1.5\times 298}{1\times 293}=1.53$