Why doesn't volume change account for the energy change in this problem?

Problem: A mass of 200 g of saturated liquid water is completely vaporized at a constant pressure of 100 kPa. Determine the amount of energy transferred to the water.

My approach: Assuming change in KE & PE go to zero, I set that

dE = dU.


Also, I knew that

du = dh - d(PV).


And since P is constant, I thought it would make sense to take out P and have:

du = dh - P * dV


And I could look up the enthalpy of vaporization (dh), and I could find the specific volume change from saturated liquid to saturated vapor. And so I multiplied dh and d(specific volume) by the mass, and got:

dE = dh - P*(dV)


However, when I looked up the solution, it said the amount of energy transferred to the water was only dh, change in enthalpy.

Why? I found on wiki that dP goes to zero, and the volume pressure work goes to zero...but I do not understand because why not just take out Pressure constant and calculate dV and include it here? What's the reason? Please help...I cannot find any answers in the text :(

$$\Delta U=Q-W$$ $$W=P\Delta V$$ $$\Delta U=Q-P\Delta V$$ $$\Delta U+P\Delta V=\Delta H=Q$$

• Thank you very much sir. But isn't dE = dU here? if I were to find dE, I think I would still subtract PdV from dH...
– js9
Feb 26, 2016 at 3:55
• $\Delta U$ (or $\Delta E$, if you prefer) is not equal to Q here because expansion work is done. Do you see anything wrong with any of the equations I wrote such that Q would not be equal to $\Delta H$? Your mistake was that the work W is not zero. Feb 26, 2016 at 4:02
• No sir, I do not see anything wrong with any of your equations. But, I was looking for dU (which is dE) and so I subtracted the work PdV (which is non-zero) from dH (=Q). But the answer says I cannot subtract PdV from dH because dE = dH. And this is what I do not get. From your equation as well, shouldn't dE = dU = dH - PdV?
– js9
Feb 26, 2016 at 4:10
• When they said,"Determine the amount of energy transferred," they were referring to Q. If they wanted $\Delta U$, they would have said, "determine the change in internal energy." Feb 26, 2016 at 4:14
• Oh! I see! Thank you so much sir! I see now. Thank you for your time and patience!
– js9
Feb 26, 2016 at 4:15