Question:

Two moles of an ideal gas are allowed to expand reversibly and isothermally at $300\, K$ from a pressure of $1\, atm$ to a pressure of $0.1\, atm$. The change in Gibbs free energy is

Updated On: May 12, 2022
  • -11.488 J
  • +11.488 J
  • -11.488 kJ
  • zero
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The Correct Option is C

Solution and Explanation

Use the following equation,
$\Delta G=2303\, n R T \log \left(\frac{p_{2}}{p_{1}}\right)$
$\therefore \Delta G=2.303 \times 2 \times\left(8.314 J K^{-1} mol ^{-1}\right)$
$\times(300\, K) \times \log \left(\frac{0.1 atm }{1 atm }\right)$
$=-11488\, J =-11.488\, kJ$
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Concepts Used:

Gibbs Free Energy

The energy associated with a chemical reaction that can be used to do work.It is the sum of its enthalpy plus the product of the temperature and the entropy (S) of the system.

The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamically closed system. In completely reversible process maximum enthalpy can be obtained.

ΔG=ΔH−TΔS

The Conditions of Equilibrium

If both it’s intensive properties and extensive properties are constant then thermodynamic system is in equilibrium. Extensive properties imply the U, G, A.