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An ideal gas undergoes an isothermal process from state A (P = 1 atm, V = 20 L) to state B (P = 2 atm, V = 10 L), as shown on a V-P graph. Find the change in Gibbs free energy delta-G (in L atm) for this process. [Use ln 2 = 0.7]

1. 5
2. 2
3. 1
4. 4

Correct answer: 1

Solution

For an ideal gas at constant temperature, dG = V dP. So delta-G = integral from P1 to P2 of V dP = integral of (nRT/P) dP = nRT * ln(P2/P1). Here nRT = P1*V1 = 1 atm * 20 L = 20 L atm. And P2/P1 = 2. So delta-G = 20 * ln(2) = 20 * 0.7 = 14 L atm. None of the listed options (5, 2, 1, 4) match this calculation. The question may contain an error in the options or in the values given.

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