Question:

For a spontaneous process, the incorrect statement is:

Updated On: Apr 3, 2025
  • \((\Delta G_{\text{system}})_{T,P}>0\)
  • \((\Delta S_{\text{system}}) + (\Delta S_{\text{surroundings}})>0\)
  • \((\Delta G_{\text{system}})_{T,P}<0\)
  • \((\Delta U_{\text{system}})_{S,V}<0\)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Approach Solution - 1

For a spontaneous process at constant temperature and pressure:

  • Gibbs Free Energy (\(\Delta G\)) must be negative: \(\left(\Delta G_{\text{system}}\right)_{T,P}<0\).
  • The total entropy change (system + surroundings) must be positive: \(\left(\Delta S_{\text{system}}\right) + \left(\Delta S_{\text{surroundings}}\right)>0\).
  • Internal energy change (\(\Delta U\)) can vary depending on the process but is typically negative under adiabatic conditions.

Option 3, which states \(\left(\Delta G_{\text{system}}\right)_{T,P}>0\), is incorrect because a positive \(\Delta G\) implies a non-spontaneous process.

Was this answer helpful?
0
0
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

Correct Answer:

Option 1: (ΔGsystem)T,P > 0

Explanation:

  • Option 1: Incorrect. For a spontaneous process at constant temperature (T) and pressure (P), the Gibbs free energy change (ΔG) must be negative (ΔG < 0).
  • Option 2: Correct. For a spontaneous process, the total entropy change (ΔStotal = ΔSsystem + ΔSsurroundings) must be positive.
  • Option 3: Correct. For a spontaneous process at constant T and P, ΔG must be negative.
  • Option 4: Correct. For a spontaneous process at constant entropy (S) and volume (V), the internal energy change (ΔU) must be negative.
Was this answer helpful?
0
0

Top Questions on Thermodynamics

View More Questions