# High vapor pressure cooking

On a Wikipedia article about latent heat, it can be found:

The large value of the enthalpy of condensation of water vapor is the reason that steam is a far more effective heating medium than boiling water, and is more hazardous.

Let's compare the amount of substance of $$\ce{H2O (g,v)}$$ at two different temperatures $$T$$:

A cooking-pot is partially filled in with water, and it has $$\pu{1 L}$$ of free volume. How many moles of water vapour will we find at $$\pu{100 ^\circ C}$$ and $$\pu{1 atm}$$? What about $$\pu{2 atm}$$ and $$\pu{125 ^\circ C}$$?

Once applied ideal gas law for each condition and dividing them I find:

$$\frac{2}{1} = \frac{398}{373}\times\frac{n_2}{n_1}$$

Indeed, the number of water molecules in vapor is almost doubled. It might justify that is not so much the temperature as the increase in vapor density the responsible for cooking faster. (Confirming what Wikipedia article states).

Is this reasoning right?

A $2$ liters cooking-pot is filled in with $1$ liter of liquid water: how many moles of water vapour will we find at $100\ ^\circ \mathrm{C}$ and $1\ \mathrm{atm}$? What about $2\ \mathrm{atm}$ and $125\ ^\circ \mathrm{C}$ (data taken from Wikipedia)?
The initial volume of liquid water is $V_{\mathrm l,0}=1.0000\ \mathrm l$. Assuming an initial temperature of $T_0=20.000\ \mathrm{^\circ C}$ and initial pressure of $p_0=1.0000\ \mathrm{atm}$ (not given in the question), the volume of liquid water corresponds to an amount of $n=55.409\ \mathrm{mol}$.
At a temperature of $T_1=100.00\ \mathrm{^\circ C}$ and pressure of $p_1=1.0000\ \mathrm{atm}$, the water is completely evaporated to steam (The boiling point of water at a pressure of $p_1=1.0000\ \mathrm{atm}$ is $T_{\mathrm b,1}=99.974\ \mathrm{^\circ C}$.); i.e. there is no equilibrium between liquid water and steam. Therefore, the answer to your question “how many moles of water vapour will we find” is the entire given amount of water of $n=55.409\ \mathrm{mol}$. At this temperature and pressure, however, the amount of $n=55.409\ \mathrm{mol}$ corresponds to a volume of $V_1=1670.3\ \mathrm l$, which doesn’t fit into the mentioned $2\ \mathrm l$ pot.
At a temperature of $T_2=125.00\ \mathrm{^\circ C}$ and pressure of $p_2=2.0000\ \mathrm{atm}$, the water is also completely evaporated to steam (The boiling point of water at a pressure of $p_2=2.0000\ \mathrm{atm}$ is $T_{\mathrm b,2}=120.63\ \mathrm{^\circ C}$.); i.e. again, there is no equilibrium between liquid water and steam. Therefore, the answer to your question “how many moles of water vapour will we find” is again the entire given amount of water of $n=55.409\ \mathrm{mol}$. At this new temperature and pressure, the amount of $n=55.409\ \mathrm{mol}$ corresponds to a volume of $V_2=884.21\ \mathrm l$, which still doesn’t fit into the mentioned $2\ \mathrm l$ pot.