Question

One mole of water at \[{\rm{100^\circ C}}\] is converted into steam at \[{\rm{100^\circ C}}\] at a constant pressure of 1 atm. The change in entropy is [heat of vaporization of water at \[{\rm{100^\circ C}}\]= 540cal/gm]
A. 8.74
B. 18.74
C. 24.06
D. 26.06

Answer 1

Hint: Vaporization is defined as the change of phase from the liquid phase to the vapor phase.
The enthalpy of vaporization \[\left( {{\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}} \right)\] or the latent heat of vaporization is the enthalpy needed to convert a liquid into a gas.

Formula Used:
\[{\rm{\Delta }}{{\rm{S}}_{{\rm{vap}}}}{\rm{ = }}\dfrac{{{\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}}}{{{{\rm{T}}_{\rm{b}}}}}\]
where
\[{\rm{\Delta }}{{\rm{S}}_{{\rm{vap}}}}\]=entropy change accompanying vapourization process
\[{\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}\]=enthalpy of vaporisation
\[{{\rm{T}}_{\rm{b}}}\]=boiling point

Complete step-by-step answer:
In the given reaction, one mole of water at \[{\rm{100^\circ C}}\] is converted into steam at a constant pressure of 1atm.
In this reaction, liquid water is getting converted to steam i.e., vapour state.
This process is called vapourisation.
The reaction will be as follows: -
\[{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right)\]
We have to find the entropy change for this reaction.
To find the entropy change we have to find out the enthalpy of vapourisation first.
It is given that heat of vapourisation or
\[{{\rm{q}}_{{\rm{vap}}}}\]=540cal/g
Mass of 1 mole of water=18g/mol
So, the enthalpy of vaporisation or
\[{\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}\]=\[{{\rm{q}}_{{\rm{vap}}}}\]×(mass of 1 mole of water)
=520 cal/g (18g/mol)
=9720 cal/mol
So, \[{\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}\] is 9720cal/mol
We are given the boiling point,
\[{{\rm{T}}_{\rm{b}}}\]=(100+273)K
=373K.
\[{\rm{\Delta }}{{\rm{S}}_{{\rm{vap}}}}{\rm{ = }}\dfrac{{{\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}}}{{{{\rm{T}}_{\rm{b}}}}}\]
\[{\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}\]=enthalpy of vaporisation=9720cal/mol
\[{{\rm{T}}_{\rm{b}}}\]=boiling point=373K
\[{\rm{\Delta }}{{\rm{S}}_{{\rm{vap}}}}{\rm{ = }}\dfrac{{{\rm{9720cal/mol}}}}{{{\rm{373K}}}}\]
=\[{\rm{26}}{\rm{.058calmo}}{{\rm{l}}^{{\rm{ - 1}}}}{{\rm{K}}^{{\rm{ - 1}}}}\]
=\[{\rm{26}}{\rm{.06calmo}}{{\rm{l}}^{{\rm{ - 1}}}}{{\rm{K}}^{{\rm{ - 1}}}}\]

So, option D is correct.

Note: While attempting the units of temperature, the heat of vaporisation, enthalpy of vaporisation, and entropy must be mentioned. Unit of entropy the final answer must be mentioned. The temperature which is expressed in celsius must be converted to Kelvin before the calculation of entropy change.