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Standard XII

Physics

Ideal Gas Equation

An air bubble...

Question

An air bubbles starts rising from the bottom of a lake. Its diameter is 3.6 mm at the bottom and 4mm at the surface. The depth of the lake is 2.5m and the temperature at the surface is 40 $^{0} C$ then temperature at the bottom is :

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Solution

The correct option is D 12 $^{0}$ C
The air inside the bubble must follow the Ideal Gas Equation.
Hence $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$⟹ \frac{(P_{0} + ρ g (2.5)) \times \frac{4}{3} π r_{1}^{3}}{T_{2}} = \frac{P_{0} \times \frac{4}{3} π r_{2}^{3}}{313}$
$⟹ T_{2} = 285 K = 12^{\circ} C$

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An air bubbles starts rising from the bottom of a lake. Its diameter is 3.6 mm at the bottom and 4mm at the surface. The depth of the lake is 2.5m and the temperature at the surface is 40 0C then temperature at the bottom is :

The correct option is D 120CThe air inside the bubble must follow the Ideal Gas Equation.Hence P1V1T1=P2V2T2⟹(P0+ρg(2.5))×43πr31T2=P0×43πr32313⟹T2=285K=12∘C

An air bubbles starts rising from the bottom of a lake. Its diameter is 3.6 mm at the bottom and 4mm at the surface. The depth of the lake is 2.5m and the temperature at the surface is 40 $^{0} C$ then temperature at the bottom is :