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

Hint: 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 $ .