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Question: What is the free energy \(\left( {\Delta {\text{G}}} \right)\) when 1.0 moles of water at \({100^ \c...

What is the free energy (ΔG)\left( {\Delta {\text{G}}} \right) when 1.0 moles of water at 100C{100^ \circ }{\text{C}} and 1 atm pressure is converted into steam at 100C{100^ \circ }{\text{C}} and at 1 atm pressure?
A. 80 cal
B. 540 cal
C. 620 cal
D. zero

Explanation

Solution

Free energy or Gibbs free energy (ΔG)\left( {\Delta {\text{G}}} \right)gives us the maximum amount of work that can be done by a closed system, having constant temperature and pressure.

Formula used:
ΔG = nRT×lnpfpi\Delta {\text{G = nRT}} \times {\text{ln}}\dfrac{{{{\text{p}}_{\text{f}}}}}{{{{\text{p}}_{\text{i}}}}}
Where n = number of moles
R = gas constant (1.987calmol - 1K - 11.987{\text{calmo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}})
T = temperature in Kelvin (K)
pf{{\text{p}}_{\text{f}}}= initial pressure
pi{{\text{p}}_{\text{i}}}= final pressure

Complete step by step solution:
According to the question the temperature and pressure of the system remain constant, i.e., temperature = 100C{100^ \circ }{\text{C}} and pressure = 1 atm. Before and after the conversion.
Now, if we put up the given values in the mentioned formula we can easily calculate the free energy.
But before we continue to our formula, see that the given temperature is in Celsius, Therefore we have to convert it into kelvin by using the knowledge, 1C = 273K{1^ \circ }{\text{C = 273K}}
\Rightarrow100C^ \circ {\text{C}}= 373 K
Therefore, ΔG = nR×373×lnpfpi\Delta {\text{G = nR}} \times {\text{373}} \times {\text{ln}}\dfrac{{{{\text{p}}_{\text{f}}}}}{{{{\text{p}}_{\text{i}}}}}
\Rightarrow ΔG\Delta {\text{G}} = nR×373×2.303logpfpi{\text{ nR}} \times 373 \times 2.303\log \dfrac{{{{\text{p}}_{\text{f}}}}}{{{{\text{p}}_{\text{i}}}}} we have used the concept, ln = 2.303log, to convert lnpfpi{\text{ln}}\dfrac{{{{\text{p}}_{\text{f}}}}}{{{{\text{p}}_{\text{i}}}}} into logpfpi\dfrac{{{{\text{p}}_{\text{f}}}}}{{{{\text{p}}_{\text{i}}}}}
\Rightarrow ΔG\Delta {\text{G}} = nR ×373 \times 373 ×\times 2.303log112.303{\text{log}}\dfrac{1}{1} ,
\Rightarrow ΔG\Delta {\text{G}}= 0 (since the value of log101=0{\log _{10}}1 = 0 )

Hence, the correct answer to the question is option (D) i.e., zero

Note: The value of gas constant(R) varies with different S.I units. For this question, we have used the value with cal/molK units, sinceoption mentions the S.I unit of free energy in cal (calorie). Pay attention to the S.I units that can alter the correct answer. In the exam, these values would be given on the first page, if it is not given in the question itself.