What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 xx 10 ^(3)kgm ^(-3)?Compressibility of water 45.8 10 ^(-11) Pa ^(-1), [1 Pa =1 Nm ^(-2)]

What is the density of water at a depth where pressure is 80.0 atm, gi

1.	What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 xx 10 ^(3)kgm ^(-3)?Compressibility of water 45.8 10 ^(-11) Pa ^(-1), [1 Pa =1 Nm ^(-2)]
Answer» Solution :Let density of water at a depth is `rho.` and density of water at surface is `rho.` VOLUME of water of mass M and V at the surface and V. at a depth, `therefore V = (M)/(rho) and V. =(M)/(rho.)` `therefore` Change in volume `Delta =V-V.` `= M [(1)/(rho) - (1)/(rho.)]` `therefore (Delta V )/(V) = M [ (1)/(rho )- (1)/(rho ) ] xx (rho )/(M) = (rho )/(rho )-(rho )/(rho.) =1- (rho )/(rho.)` `therefore `Volume strain `(Delta V)/(V) =1 - (rho )/(rho.)` compressibility `K = (1)/("BULK MODULUS")` `= (Delta V )/(PV) = (1)/(P ) [1- (rho )/(rho.)]` `45.8 xx 10 ^(-11) = (1)/(80 xx 1.013 xx 10 ^(5)) [ (1- 1.03 xx 10 ^(3))/( rho .)]` `45.8 xx 10 ^(-11) xx 80 xx 1.013 xx 10 ^(5) =1 - ( 1.03 xx 10 ^(3))/(rho .)` `therefore (1.03 xx 10 ^(3))/( rho .) =1 - 45.8 xx 10 ^(-11) xx 80xx 1.013 xx 10 ^(5)` `therefore rho . =1- 3.712 xx 10 ^(-3) =0.996288` `therefore rho . = (1.03 xx 10 ^(3))/( 0.996288)` `therefore rho . = 1034 KG m ^(-3)` `therefore rho . =1.034 xx 10 ^(3) kg m ^(-3)`