Question

9gm H2o is vaporized at 100C and 1atm pressure.If latent heat of vaporization of water is x J/gm ,then ∆S is given as

Answer 1

Answer : The value of $\Delta S$ is (0.0241x) J/K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{vap}}{T_b}$

where,

$\Delta S$ = change in entropy

$\Delta H_{vap}$ = change in enthalpy of vaporization = (x) J/g = (9x) J

$T_b$ = boiling point temperature = $100^oC=273+100=373K$

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

$\Delta S=\frac{\Delta H_{Vap}}{T_b}$

$\Delta S=\frac{(9x)g}{373K}$

$\Delta S=(0.0241x)J/K$

Therefore, the value of $\Delta S$ is (0.0241x) J/K