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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 $40\;m$ deep where the temperature is $0^{\circ}C$. 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 = $76\;cm$ of Mercury

$\begin{array}{1 1} 261.93\;cm^3 \\ 2.6193\;cm^3 \\ 52.387\;cm^3 \\ 523.387\;cm^3\end{array}$

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1 Answer

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Toolbox:
  • Ideal gas law for a fixed amount of gas: P1V1T1=P2V2T2\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 K^{\circ}K
  • 11 Atmospheric pressure = 76cmHgPatm=0.76×13600×9.8=1.01292×105Pa76cm Hg \rightarrow P_{atm} = 0.76 \times 13600 \times 9.8 = 1.01292 \times 10^5 P_a
Let the subscript1_1 refer to the bottom of the lake and the subscript2_2 to the surface of the lake.
The ratio of expansion of the volume is dependent on the ratio of the temperatures in K^{\circ}K
T1=0C=0+273=273K\Rightarrow T_1 =0^{\circ}C = 0 + 273 = 273^{\circ}K and T2=20C=20+273=293KT_2 = 20^{\circ}C = 20 + 273 = 293^{\circ}K
The pressure at the bottom of the lake P1P_1 is equal to the sum of the atmospheric pressure at its surface and the pressure due to the depth of the lake.
11 Atmospheric pressure = 76cmHgPatm=0.76×13600×9.8=1.01292×105Pa76cm Hg \rightarrow P_{atm} = 0.76 \times 13600 \times 9.8 = 1.01292 \times 10^5 P_a
P1=Patm+ρhg=1.01292×105Pa+1000×40m×9.8=4.93×105PaP_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
P2P_2 is nothing but the atmospheric pressure at the surface, i.e, P2=1.01292×105PaP_2 = 1.01292 \times 10^5 P_a
Now, applying ideal gas law, P1V1T1=P2V2T2\large\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
V2=P1V1T2P2T2\Rightarrow V_2 = \large\frac{P_1 V_1 T_2}{P_2 T_2}
V2=4.93×105Pa×50cm3×2931.01292×105Pa×273\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 273}
V2=261.93cm3\Rightarrow V_2 = 261.93 cm^3
answered Jan 22, 2014 by balaji.thirumalai
edited Aug 8, 2014 by balaji.thirumalai
 

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