Question

what is the free energy change $\left( {\Delta G} \right)$ when 1.0 mole of water at $100^\circ C$ and 1 atm pressure is converted into steam at $100^\circ C$ and 2 atm pressure.
(A) zero cal
(B) 540 cal
(C) 517 cal
(D) None of the above .

Answer 1

In thermodynamics, free energy, also called Gibbs free energy, is the thermodynamic potential that can be used to calculate the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.
$\Delta G = 2.303RT\log \left( {\dfrac{{{P_2}}}{{{P_1}}}} \right)$
Where,
$\Delta G$ = change in free energy
$R$ =gas constant
$T$ =given temperature
${P_2}$ and ${P_1}$ =pressures.

Complete answer:
We know that Gibbs free energy is maximum free energy that can be converted into useful work.
Given:
${\text{ }}R = 2cal{\text{ }}.....{\text{(1)}}$
(because the value of gas constant R is calories is 2cal)
$T = 100^\circ C$
We will convert it into Kelvin.
$T = 100 + 273K$
$T = 373K........(2)$
${P_2} = 2atm.........(3)$
${P_1} = 2atm.........(4)$
The change in Gibbs free energy when 1.0 mole of water at $100^\circ C$ and 1 atm pressure is converted into steam at $100^\circ C$ and 1 atm pressure is 0 cal as the system at that point is in equilibrium.
Hence the change in Gibbs free energy when pressure of the steam is increased from 1 atm to 2 atm will be:
$\Delta G = 2.303RT\log \left( {\dfrac{{{P_2}}}{{{P_1}}}} \right)$
We will substitute the given values 1,2,3 and 4 into this equation:
$\Delta G = 2.303 \times 2 \times 373 \times \log \left( {\dfrac{2}{1}} \right)$
$\Delta G = 1718.038 \times \log 2$
$\Delta G = 517.180cal$
Or, $\Delta G = 517cal$
Hence the free energy change $\left( {\Delta G} \right)$ when 1.0 mole of water at $100^\circ C$ and 1 atm pressure is converted into steam at $100^\circ C$ and 2 atm pressure is 517 cal.
Hence the correct answer to this question is option C.

Note:
The answer here remained 517cal because initially the change in Gibbs free energy when 1.0 mole of water at $100^\circ C$ and 1 atm pressure is converted into steam at $100^\circ C$ and 1 atm pressure is 0 cal. Hence the total change is $\Delta G = 0 + 517cal$ , that is $\Delta G = 517cal$ .