1
$\begingroup$

Let's take water for an example: the specific heat for water vapor is 1.99. This would mean that the molar specific heat is the molar mass times the specific heat (or $18.0*1.99 = 35.8$).

  • Is 35.8 J/mol*K at constant volume, $C_v$, or constant pressure, $C_p$?
  • Why can't you calculate the molar heat capacity directly for water vapor using $3R = 24.94 J/mol*K$ for constant volume and $4R = 33.26 J/mol*K$ for constant pressure? Why don't these values match up with the 35.8?
Improve this question
$\endgroup$
1
  • $\begingroup$ You are trying to apply statements about gases to the liquid water. For liquids and solids $C_p$ and $C_v$ do not differ strongly at usual pressures, as the volume expansion is very small (and this volume times atmospheric pressure do not correspond to much work). $\endgroup$
    – Sebastian Riese
    Commented Oct 11, 2015 at 23:00

1 Answer 1

Reset to default
3
$\begingroup$

Sebastian's comment is incorrect, since you're working with water vapour, not liquid water. You're on the right track: since water vapour is a gas at STP, we can approximate it with the ideal gas law near STP, and $C_v \neq C_p$ too.

But remember: water is $H_2 O$, which is a polyatomic molecule! So the equation for is $C_v = 7R/2 \approx 29.1$ and $C_p = 9R/2 \approx 37.3$.

$37.3 \approx 35.8$, which suggests that the heat capacity of water vapour is based on its molar heat capacity at constant pressure. Perhaps the Wikipedia article just assumes you want constant pressure instead of constant volume, but of course, it's best to specifically search for what you want, $C_p$ or $C_v$, since they are going to be different.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.