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Ideal Gas Equation

A bubble of v...

Question

A bubble of volume $V_{1}$ is at the bottom of a pond at $15^{\circ} C$ and $1.5 a t m$ pressure. When it comes at the surface, it observes a pressure of $1 a t m$ at $25^{\circ} C$ and has volume $V_{2}$ ; given $(\frac{V_{2}}{V_{1}})$ .

15.5

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0.155

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155.0

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1.55

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Solution

The correct option is D $1.55$
Applying the formula,
$P_{1} = 1.5 a t m, T_{1} = 15 + 273 = 288 K$
$P_{2} = 1 a t m, T_{2} = 25 + 293 = 298 K$
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{V_{2}}{V_{1}} = \frac{P_{1} T_{2}}{P_{2} T_{1}}$
$= \frac{1.5 \times 288}{1 \times 298} = 1.5$ .

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