Respuesta :
Answer:
Absolute Pressure at depth h = [tex]1.312\times 10^5\ Pa.[/tex]
Explanation:
Given:
- Radius of the bubble at depth h, [tex]\rm r_i = 5\ mm = 5\times 10^{-3}\ m.[/tex]
- Radius of the bubble at the surface of the water, [tex]\rm r_f = 6.6\ mm = 6.6\times 10^{-3}\ m.[/tex]
- Acceleration due to gravity, [tex]\rm g = 9.8\ m/s^2.[/tex]
Assumptions:
- Atmospheric pressure, [tex]\rm P_{atm} = 1.01\times 10^5\ Pa.[/tex]
- Density of water, [tex]\rm \rho = 1000\ kg/m^3.[/tex]
- Pressure at depth h = [tex]\rm P_i.[/tex]
- Pressure at the surface of water = [tex]\rm P_f.[/tex]
According to Ideal Gas Law,
[tex]\rm PV=nkT[/tex]
where,
- P = pressure.
- V = volume.
- n = number of molecules.
- k = Boltzmann constant.
- T = absolute temperature.
For the given case, the temperature of the air bubble is constant. The number of molecules of the bubble also does not change, therefore,
PV = constant.
[tex]\rm P_iV_i=P_fV_f[/tex]
The pressure at the surface is equal to the atmospheric pressure, [tex]\rm P_f=P_{atm}.[/tex]
The pressure at depth h is equal to sum of atmospheric pressure and the pressure of the water upto depth h, [tex]\rm P_i=P_{atm}+\rho gh.[/tex]
Using these values,
[tex]\rm (P_{atm} + \rho gh)V_i = P_{atm}V_f\\\dfrac{(P_{atm} + \rho gh)}{P_{atm}}=\dfrac{V_f}{V_i}=\dfrac{\dfrac 43 \pi r_f^3}{\dfrac 43 \pi r_i^3}=\dfrac{r_f^3}{r_i^3}\\\dfrac{(1.01\times 10^5)+1000\times 9.8\times h}{1.01\times 10^5}=\dfrac{(6.6\times 10^{-3})^3}{(5\times 10^{-3})^3}=2.299\\(1.01\times 10^5)+9800\times h=2.299\times 1.01\times 10^5=2.322\times 10^5\\9800\ h= 2.322\times 10^5-1.01\times 10^5=1.312\times 10^5\\h=\dfrac{1.312\times 10^5}{9800}=13.39\ m.[/tex]
The absolute temperature at that depth is given by
[tex]\rm P=\rho gh=1000\times 9.8\times 13.39=1.312\times 10^5\ Pa.[/tex]
