In a totally irreversible isothermal expansion process for an ideal gas, ΔE = 0, ΔH = 0. Then ΔQ and ΔS will be_______________?
Correct answer: B. ΔQ = 0, ΔS = +ve
- A. ΔQ = 0, ΔS=0
- B. ΔQ = 0, ΔS = +ve
- C. ΔQ = 0, ΔS = -ve
- D. ΔQ = +ve, ΔS= +ve
Explanation
For the intended free, totally irreversible expansion, no work is done and an ideal gas has unchanged internal energy, so Q is zero; its entropy nevertheless increases. If “irreversible expansion” is not intended to mean free expansion, the heat transfer cannot be determined without more process details.
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