Entropy for a Reversible Process
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Q. Calculate the entropy change when 2 mol of an ideal gas expand isothermally and reversibly from an initial volume of 10 cm3 to 100 dm3 at 300 K.
28.294 J K-1
38.294 J K-1
45 J K-1
50 J K-1
Q. 1 mole of an ideal gas is allowed to expand isothermally at 27oC until its volume is tripled. Calculate △Ssys and △Suniv under the reversible expansion.
- △Ssys=9.1 J K−1mol−1
- △Suniv=△Ssys=9.1 J K−1mol−1
- △Ssurr=−9.1 J K−1mol−1
- △Suniv=0
Q. Calculate the change in entropy when 1 mole nitrogen gas expands isothermally and reversibly from an initial volume of 1 litre to a final volume of 10 litre at 27oC.
Q. When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its initial temperature, the change in entropy of gas (ΔS) is:
- Cp, mln2
- Cv, mln2
- (Cv, m−R)ln2
- Rln2
Q. The equilibrium constant for the reaction given below is 2.0×10−7 at 300 K. Calculate the standard entropy change if △H∘=28.40 kJmol−1 for the reaction:
PCl5(g)⇌PCl3(g)+Cl2(g)
PCl5(g)⇌PCl3(g)+Cl2(g)
- −33.6 Jmol−1K−1
- 33.6 Jmol−1K−1
- −43.6 Jmol−1K−1
- 43.6 Jmol−1K−1
Q. When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its initial temperature, the change in entropy of gas (ΔS) is
- Cp.m ln 2
- Cv.m ln 2
- R In 2
- (Cv, m−R) ln 2