Energy comparison, evaporate water vs compress air, same mass

I would like to compare the energy requirements of the 2 following tasks:

1. Total energy to evaporate 1 litre of water at normal atmospheric pressure at normal temperature?

2. Total energy to compress air into a 1 litre container so that the compressed air had a mass of 1kg? What air pressure would atain this mass?

If its too hard to calculate could you give an estimate?

• I think you would need to specify the air's temperature as well. Regarding the water: you won't need to exert any energy. The water will evaporate by itself. Commented Jul 23, 2014 at 10:26

For (1): the energy required to vaporise a liquid is known as the enthalpy of vaporisation, or alternatively as the latent heat of vaporisation. For water at room temperature this is about 44kJ per mole, and a mole of water is 18g.

For (2): the density of air at room temperature and pressure is about 1.2 kg per cubic metre, so 1kg of air is 1/1.2 or about 0.83 m$^3$. Assuming the air behaves as an ideal gas the pressure can be calculated using Boyle's Law:

$$P_1V_1 = P_2V_2$$

so the pressure when you compress you 1kg of air down to one litre is:

$$P_2 = P_1 \frac{V_1}{V_2}$$

where $P_1$ is one atmosphere, $V_1$ is 0.83m$^3$ and $V_2$ is 1 litre.

The amount of work done in compressing the gas is a bit harder, but I go through the calculation in my answer to How much work is needed to compress a certain volume of gas?. The work is given by:

$$W = nRT ln \left( \frac{V_2}{V_1} \right)$$

where $n$ is the number of moles (1 mole of air $\approx$ 28.8g) and $R$ is the ideal gas constant.

• As you point out in your other calculation, the work of compression is very dependent on the actual process. Isothermal compression will give a lower bound, but the compression of such a small volume of air is likely to be adiabatic. Commented Jul 23, 2014 at 12:59