An ideal gas undergoes isothermal and reversible expansion from its in...
Answer:
Given:
- Isothermal and reversible expansion of an ideal gas
- Initial temperature (T) = 300 K
- Heat absorbed (q) = 90 kJ
To find:
- Change in Gibbs free energy (ΔG)
Solution:
We know that for isothermal and reversible processes,
ΔG = -TΔS + ΔH
Where,
- T is the temperature in Kelvin
- ΔS is the change in entropy
- ΔH is the change in enthalpy
Since the process is isothermal, the temperature (T) is constant. Therefore, ΔS = q/T, where q is the heat absorbed.
ΔS = q/T = 90 kJ / 300 K = 0.3 kJ/K
As the process is reversible, the change in enthalpy (ΔH) is equal to the heat absorbed (q).
ΔH = q = 90 kJ
Substituting these values in the equation for ΔG, we get:
ΔG = -TΔS + ΔH
ΔG = -(300 K)(0.3 kJ/K) + (90 kJ)
ΔG = -90 J + 90 kJ
ΔG = 89.91 kJ
Therefore, the change in Gibbs free energy is 89.91 kJ, which is equivalent to -300 J. Hence, the answer is option (d) -300 J.