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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
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An air bubble is at the bottom of the lake at $40m$ deep where the temperature is $6^{\circ}C$. The bubble rises to the surface where the temperature is $20^{\circ}C$ If it's volume at the bottom is $50\;cm^3$, what is its volume at the surface?

Note:

Specific gravity of mercury is $13.6$
1 Atmospheric pressure = $76cm$ of Mercury

$\begin{array}{1 1} (A)246.30\;cm^3 \\ (B) 128.15\;cm^3 \\ (C) 256.30\;cm^3 \\ (D) 512.60\;cm^3 \end{array}$

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1 Answer

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  • Ideal gas law for a fixed amount of gas: $\large\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
  • The ratio of expansion of the volume is dependent on the ratio of the temperatures in $^{\circ}K$
  • $1$ Atmospheric pressure = $76cm Hg \rightarrow P_{atm} = 0.76 \times 13600 \times 9.8 = 1.01292 \times 10^5 P_a$
Let the subscript$_1$ refer to the bottom of the lake and the subscript$_2$ to the surface of the lake.
The ratio of expansion of the volume is dependent on the ratio of the temperatures in $^{\circ}K$
$\Rightarrow T_1 =0^{\circ}C = 6 + 273 = 279^{\circ}K$ and $T_2 = 20^{\circ}C = 20 + 273 = 293^{\circ}K$
The pressure at the bottom of the lake $P_1$ is equal to the sum of the atmospheric pressure at its surface and the pressure due to the depth of the lake.
$1$ Atmospheric pressure = $76cm Hg \rightarrow P_{atm} = 0.76 \times 13600 \times 9.8 = 1.01292 \times 10^5 P_a$
$P_1 = P_{atm} + \rho h g = 1.01292 \times 10^5 P_a + 1000 \times 40m \times 9.8 = 4.93 \times 10^5 P_a$
$P_2$ is nothing but the atmospheric pressure at the surface, i.e, $P_2 = 1.01292 \times 10^5 P_a$
Now, applying ideal gas law, $\large\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$\Rightarrow V_2 = \large\frac{P_1 V_1 T_2}{P_2 T_2}$
$\Rightarrow V_2 = \large \frac{ 4.93 \times 10^5 P_a \times 50 cm^3 \times 293}{1.01292 \times 10^5 P_a \times 279}$
$\Rightarrow V_2 = 256.30cm^3$

 

answered Jan 22, 2014 by balaji.thirumalai
 

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