Question

One mole of an ideal gas is subjected to an isothermal increase in pressure from 100 kPa to 1000 kPa at 300 K. The change in Gibbs free energy of the system is ................ kJ mol\(^{-1}\). (Round off to one decimal place) 
[Given: Gas constant (R) = 8.3 J K\(^{-1}\) mol\(^{-1}\)]
 

Hint:

For isothermal processes, the change in Gibbs free energy can be calculated using the formula \(\Delta G = -nRT \ln \left(\frac{P_2}{P_1}\right)\).

Answer 1

Correct Answer: 5.6

Step 1: Gibbs free energy and pressure relationship.
The change in Gibbs free energy for an ideal gas under isothermal conditions is given by the formula: \[ \Delta G = -nR T \ln \left( \frac{P_2}{P_1} \right) \] where \(n\) is the number of moles, \(R\) is the gas constant, \(T\) is the temperature, \(P_2\) is the final pressure, and \(P_1\) is the initial pressure.
Step 2: Substituting the given values.
For 1 mole of gas: \[ \Delta G = -1 \times 8.3 \times 300 \times \ln \left( \frac{1000}{100} \right) \] \[ \Delta G = -2490 \times \ln(10) \] \[ \Delta G = -2490 \times 2.3026 \] \[ \Delta G = -5721.74 \, \text{J} \] Step 3: Conversion to kJ.
\[ \Delta G = -5.7 \, \text{kJ/mol} \] Step 4: Conclusion.
Thus, the change in Gibbs free energy is 0.0 kJ/mol.