how much heat is needed to vaporize 2.72 moles of nitrogen? Cal? J?

Answer 1

The latent heat of vaporization is the amount of

"heat required to convert a unit mass of a liquid into vapor without a change in temperature".

For Nitrogen this value is 199 kJ/kg = 199 J/g

Assume we have liquid N2, convert 2.72 moles to grams;

m = n*Mr

m = 2.72 moles * 28.0134 g/mol

m = 76.2 g

199 J of energy is required per g of N2, but we have 76.2g so there are:

76.2*199J = 1.516x10⁴ J required.

1 Cal = 4.184 Joules

So 1.516x10⁴ / 4.184 = 3.62x10³ Cal or 3.62 kCal.

Try the calculation for 76.2g of water and report back on the energy required ( Water = 2256 kJ/kg = 2256 J/g). You will see that much more energy is required!