Question

The volume of 2.0 mol of an ideal gas is reduced to half isothermally at 300 K in a closed system. The value of ΔG is_____ kJ.
[Given: R=8.314 J mol^-1 K^-1 ] 
(round off to two decimal places)

Answer 1

Correct Answer: 3.44 - 3.46

Change in Gibbs Free Energy (ΔG) for Isothermal Process 

To solve for the change in Gibbs free energy (\( \Delta G \)) when the volume of an ideal gas is reduced isothermally, we use the formula for isothermal processes:

\[ \Delta G = nRT \ln \left( \frac{V_2}{V_1} \right) \]

Where:

• \( n \) = number of moles of the gas
• \( R \) = universal gas constant
• \( T \) = temperature in Kelvin
• \( V_1 \) = initial volume
• \( V_2 \) = final volume

Given:

\( n = 2.0 \, \text{mol}, \quad R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}, \quad T = 300 \, \text{K}

Step-by-Step Solution:

1. Initial volume (\( V_1 \)) is reduced to half, so \( V_2 = \frac{V_1}{2} \).
2. Using the formula, we substitute the values:

\[ \Delta G = 2.0 \, \text{mol} \times 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \times 300 \, \text{K} \times \ln \left( \frac{1}{2} \right) \]

1. Calculate \( \ln \left( \frac{1}{2} \right) = \ln (0.5) \approx -0.6931 \).
2. Substitute into the equation:

\[ \Delta G = 2.0 \times 8.314 \times 300 \times (-0.6931) \, \text{J} \]

1. Now, perform the multiplication:

\[ \Delta G \approx -3458.66 \, \text{J} = -3.45866 \, \text{kJ} \]

1. Rounding to two decimal places, we get:

\[ \Delta G = -3.46 \, \text{kJ} \]

Final Verification:

Finally, we confirm that the calculated value, \( -3.46 \, \text{kJ} \), falls within the provided range of \( -3.44 \, \text{kJ} \) to \( -3.46 \, \text{kJ} \).