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A water vapour filling the space under the piston of a cylinder is compressed (or expanded) so that it remain saturated all the time,being just on the verge of condensation. Find the molar heat capacity of the vapour in this process as a function of temperature T assuming the vapour to be an ideal gas and neglecting the specific volume of water in comparison with that of vapour.

My attempt is shown in the figure that follows

I am mainly concerned with two points in this- 1. I think that the number of moles of gas in the container will not remain constant. So the dn term in my solution cannot be put to zero. 2. Does the solution take in account the heat given to water to turn it into vapour?

If there is any better solution than this please let me know. Thank you enter image description here

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    $\begingroup$ The number of. moles of vapor does not change. It fills the container initially, and stays at the verge of condensation (but does not condense) during the operation. In your analysis, you are using the symbol q for two different entities. $\endgroup$ – Chet Miller Mar 9 '18 at 14:40
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$$dq=du+Pdv=dh-vdP=C_PdT-\frac{RT}{P}dP$$ Clausius Clapeyron: $$\frac{dP}{P}=\frac{\Delta H_v}{RT^2}dT$$where $\Delta H_v$ is the molar heat of vaporization at temperature T. So, combining these equations, we have: $$dq=\left(C_p-\frac{\Delta H_v}{T}\right)dT$$

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