# Why is steam assumed to be an ideal gas here? [closed]

A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporisation at 100 o C. Δ vap

H ⊝ for water at 373K =40.66 kJ mol • Because you need to assume it's an ideal gas to convert $p \Delta V$ to $\Delta n R T$ (recall that, for an ideal gas, $pV=nRT$). The other assumption being made is that $\Delta V$ due to the loss of the liquid water is negligible. – theorist Nov 25 '20 at 6:33
• Liquid water to gas has very negligible volume change ? @theorist – srijan Sri Nov 25 '20 at 6:36
• No. You have two effects: (1) you are decreasing the moles of liquid water; (2) you are increasing the moles of gaseous water by an equal amount. Effect #1 causes a decrease in volume. Effect #2 causes an increase in volume. In doing the above calculation, you are ignoring effect #1, and only considering the change in volume due to effect #2. I.e., you are not subtracting, from $\Delta V$, the change in the volume of liquid water, because it is negligible compared to the gain in volume of gaseous water. Specifically, $| \Delta V_{liquid}| \approx 1/1000 | \Delta V_{gas}|$ – theorist Nov 25 '20 at 6:43
• Ohk.Thank you v much.👍🏻 – srijan Sri Nov 25 '20 at 6:48
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## 1 Answer

The critical temperature of water is 647 K, its critical pressure is 22100 kPa, and its equilibrium vapor pressure at 298 K is 3.169 kPa. So the reduced temperature is 0.461 and its reduced pressure is 0.000143. From generalized compressibility factor charts, at this reduced temperature and pressure, the compressibility factor z is virtually indistinguishable from a value of 1, characteristic of an ideal gas. Also, from the steam tables, the specific volume of water vapor under these conditions is 43.36 m^3/kg = 0.7805 m^3/mole. So, $$\frac{Pv}{RT}=\frac{(3160)(0.7805)}{(8.314)(298.15)}=0.995$$