## Boiling Point of Water At High Pressure

Calculate at what temperature water will boil if the outside pressure is 400 kPa. Under standard conditions the atmospheric pressure is 101 325 Pa and the boiling point of water is 373.15 K (100 °C).

Consider the density of water vapor to be constant and equal to 0.8 kg m−3.

• #### Hint

To solve this problem it is necessary to use the Clausius–Clapeyron relation in the following form:

$\frac{\mathrm{d}p}{\mathrm{d}T} = \frac{l_{w}}{T\left( \frac{1}{\rho _{v}}-\frac{1}{\rho _{w}}\right) }.$
• #### Analysis

We will use the Clausius–Clapeyron relation and we will solve it as a differential equation, calculating the definite integral from the initial pressure to the final pressure and between the two boiling points. We will get a relation between the boiling point and the pressure from this integral.

• #### Notation

 ps = 400 kPa = 4·105 Pa outside pressure p0 = 101 325 Pa standard atmospheric pressure T0 = 373.15 K boiling point of water under standard atmospheric pressure ρv = 0.8 kg m−3 density of water vapor Tw = ? boiling point of water under outside pressure

Found in Physics Reference Tables:

 lw = 2256 kJ kg−1 = = 2.256·106 J kg−1 specific heat of vaporization of water ρw = 1000 kg m−3 water density
• #### Solution

The boiling point of water increases with increasing pressure. This fact is used for example when pressure cooking. We will use the Clausius–Clapeyron relation to calculate this phenomenon numerically:

$\frac{\mathrm{d}p}{\mathrm{d}T} = \frac{l_{w}}{T\left( \frac{1}{\rho _{v}}-\frac{1}{\rho _{w}}\right) },$

where p is pressure, T is thermodynamic temperature, lw is specific heat of vaporization of water, ρv is density of water vapor and ρw is water density. This equation describes the phase transition, i.e. the relation between the phase transition point and pressure.

We cannot consider the right-hand side of the equation to be constant because the relation between boiling point and pressure is relatively large, due to the big difference between the two densities of the two phases. We will therefore treat it as a differential equation and we will solve it using the method of separation of variables

$\frac{\mathrm{d}p}{\mathrm{d}T} = \frac{l_{w}}{T\left( \frac{1}{\rho _{v}}-\frac{1}{\rho _{w}}\right) } \Rightarrow \mathrm{d}p = \frac{l_{w}}{\left( \frac{1}{\rho _{v}}-\frac{1}{\rho _{w}}\right) }\frac{\mathrm{d}T}{T}$

and then we will integrate from p0 (atmospheric pressure) to ps (outside pressure) with respect to pressure, and from T0 (B.P. of water under standard atmospheric pressure) to Tw (the B.P. of water we are looking for) with respect to temperature. We can consider the specific heat of vaporization of water and both the densities to be constant, therefore we can factor them out of the integrand.

$\int_{p_{0}}^{p_{s}}\, \mathrm{d}p =\frac{l_{w}}{\left( \frac{1}{\rho _{v}}-\frac{1}{\rho _{w}}\right) }\int_{T_{0}}^{T_{w}}{\frac{1}{T}}\, \mathrm{d}T$ $p_{s}-p_{0} = \frac{l_{w}}{\left( \frac{1}{\rho _{v}}-\frac{1}{\rho _{w}}\right) }\left[ \ln T\right] _{T_{0}}^{T_{w}}$ $\frac{\left( p_{s}-p_{0}\right) \left( \frac{1}{\rho _{v}}-\frac{1}{\rho _{w}}\right)}{l_{w}} = \ln T_{w} - \ln T_{0}$

Finally, we evaluate the boiling point Tw

$\ln T_{w}=\ln T_{0} + \frac{\left( p_{s}-p_{0}\right) \left( \frac{1}{\rho _{v}}-\frac{1}{\rho _{w}}\right)}{l_{w}},$ $T_{w} = T_{0}e^{\frac{\left( p_{s}-p_{0}\right) \left( \frac{1}{\rho _{v}}-\frac{1}{\rho _{w}}\right)}{l_{w}}}.$
• #### Numerical insertion

$T_{w} = T_{0}e^{\frac{\left( p_{s}-p_{0}\right) \left( \frac{1}{\rho _{v}}-\frac{1}{\rho _{w}}\right)}{l_{w}}}$ $T_{w}= 373.15\, \cdot \, e^{\frac{\left( 4\cdot{ 10^{5}}-101325\right) \left( \frac{1}{0.8}-\frac{1}{1000}\right)}{2.256\cdot{ 10^{6}}}}\, \mathrm{K}$ $T_{w} \dot{=} 440\, \mathrm{K}$