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Byju's Answer

Standard XII

Physics

Ideal Gas Equation

At the bottom...

Question

At the bottom of a lake where the temperature is ${7}^{0}\,C$ the pressure is $2.8$ atmosphere. An air bubble of radius $1 c m$ at the bottom rises to the surface. Where the temperature is ${27}^{0}\, C$. The radius of the air bubble at the surface is

3^{1 / 3}

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4^{1 / 3}

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5^{1 / 3}

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6^{1 / 3}

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Solution

The correct option is A $3^{1 / 3}$
We know that, by gas law
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{2.8 {\times V}_{1}}{280} = \frac{1 \times V_{2}}{300}$
$V_{2} = 3 V_{1}$
$R \propto V^{\frac{1}{3}}$
$∴ R_{2} = {3^{\frac{1}{3}} R}_{1}$

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At the bottom of a lake where the temperature is ${7}^{0}\,C$ the pressure is 2.8 atmosphere. An air bubble of radius 1cm at the bottom rises to the surface. Where the temperature is ${27}^{0}\, C$. The radius of the air bubble at the surface is

The correct option is A 31/3 We know that, by gas lawP1V1T1=P2V2T22.8×V1280=1×V2300V2=3V1R∝V13∴R2=313R1

At the bottom of a lake where the temperature is ${7}^{0}\,C$ the pressure is $2.8$ atmosphere. An air bubble of radius $1 c m$ at the bottom rises to the surface. Where the temperature is ${27}^{0}\, C$. The radius of the air bubble at the surface is

The correct option is A $3^{1 / 3}$
We know that, by gas law
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{2.8 {\times V}_{1}}{280} = \frac{1 \times V_{2}}{300}$
$V_{2} = 3 V_{1}$
$R \propto V^{\frac{1}{3}}$
$∴ R_{2} = {3^{\frac{1}{3}} R}_{1}$