As liquid water is compressed at constant temperature, its pressure increases ac

Question

As liquid water is compressed at constant temperature, its pressure increases according to the formula: p=[1+2.5x104(1-V/V0)] atm, where V0 is its volume under atmospheric pressure, 1.013 = 105 Pa.

a. If some water has a volume of 1 liter at atmosphere pressure, what will be its volume at the bottom of the ocean, where the pressure is 500 atm?

b. How much work is done by a liter of water that is brought to the surface from the ocean bottom? (Hint: THe pressure is not constant, so you will have to integrate pdV.)

c. Knowing that to change the temperature of water by 1oC requires a change in internal energy of 4186 J per kg, calculate the change in temperature of the water sample brought up from the bottom of the ocean. Assume that it is closed and insulated, so that no heat or particles enter or leave the sample as it is raised.

Explanation / Answer

a) p =[1+2.5x10^4(1-V/V0)] (p -1)/(2.5*10^4) = 1- V/V0 V = [1- (p -1)/(2.5*10^4)]*V0 So, we can use V = [1- (500 -1)/(2.5*10^4)]*1 = 0.9800 *1 lit = .98 liter b) work = -p*dv work done =- integral ([1+2.5x10^4(1-V/V0)]dv) =-(1+2.5*10^4)*integral(dv) + 2.5*10^4 * integral (Vdv) = -(1+2.5*10^4)*(v2-v1) + 2.5*10^4 *(v2^2 - v1^2)/2 where v2 and v1 are final and initial volumes = -(1+(2.5*(10^4)))*(.98-1) + (2.5*(10^4))*(.98^2 - 1^2)/2 atm liter = 5.02 *1.013*10^5 pascal *10^-3m^3joules = 508.526 joules c)Internal energy change = work done as q = 0 4186 joules =rise in 1 degree 508.526 joules = 508.526/4186 degrees = 0.12148

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