InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
What is the density of ocean water at a depth, where the pressure is 80.0 atm, given that its density at the surface is `1.03 xx 10^(3) kg m^(-3)`? Compressibilty of water `= 45.8 xx 10^(-11) Pa^(-1)`. Given 1 atm. =`1.013 xx 10^(5) Pa`. |
|
Answer» Let the given depth be h. Pressure at the given depth, p = 80.0 atm = `80xx1.01xx10^(5)` Pa Density of water at the surface, `rho_(1)=1.03xx10^(3)"kg"" m"^(-3)` Let `rho_(2)` be the density of water at the depth h. Let `V_(1)` be the volume of water of mass m at the surface. Let `V_(2)` be the volume of water of mass m at the depth h. Let `DeltaV` be the change in volume. `DeltaV=V_(1)-V_(2)` `=m((1)/(rho_(1))-(1)/(rho_(2)))` `therefore"Volumetric strain"=(DeltaV)/(V_(1))` `=m((1)/(rho_(1))-(1)/(rho_(2)))xx(rho_(1))/(m)` `therefore (DeltaV)/(V_(1))=1-(rho_(1))/(rho_(2))" "...(i)` Bulk modulus, `B=(pV_(1))/(DeltaV)` `(DeltaV)/(V_(1))=(P)/(B)` Compressibity of water = `(DeltaV)/(V_(1))=(P)/(B)` `(1)/(B)=45.8xx10^(-11)"Pa"^(-1)` `therefore(DeltaV)/(V_(1))=80xx1.013xx10^(5)xx45.8xx10^(-11)=3.71xx10^(-3)" "...(ii)` For equations (i) and (ii), we get: `1-(rho_(1))/(rho_(2))=3.71xx10^(-3)` `rho_(2)=(1.03xx10^(3))/(1-(3.71xx10^(-3)))` `=1.034xx10^(3)"kg m"^(-3)` Therefore, the density of water at the given depth (h) is `1.034xx10^(3)"kg m"^(-3)`. |
|