Question

If watervapour is assumed to be a perfect gas, molar enthalpy change for vaporization of 1 mol of water at 1 bar and 100°C is 41kJ/mol. Calculate the internal energy change when 1 mol of water is vapourised at 1 bar pressure and 100°C . What is the value of Delta ng

Answer 1

Answer : The internal energy change is, 37.89 J/mol

Explanation :

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $41kJ/mol=41000J/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 1

R = gas constant = 8.314 J/mol.K

T = temperature = $100^oC=273+100=373K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(41000J/mol)-[1mol\times 8.314J/mol.K\times 373K$

$\Delta U=41000J/mol-3101.122J/mol$

$\Delta U=37898.878J/mol=37.89J/mol$

Therefore, the internal energy change is 37.89 J/mol