(D) Unsolved Examples:
1. A quantity of an ideal gas in an isolated system is expanded
isothermally and reversibly at 127°C from a volume V, to V,
During expansion the gas absorbs 840.8 J of heat from a
reservoir in contact with it. Find the entropy change of the
air and total entropy of the system.
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1
Answer:
Work done in an isothermal reversible process is given as-
W=−2.303nRTlog
V
1
V
2
.....(1)
Given:-
n=1 mole
T=273K
V
1
=1dm
3
V
2
=50dm
3
R=8.314J
Substituting these values in eq
n
(1), we have
W=−2.303×1×8.314×273×log
1
50
⇒W=−5227.17×log50=8.88kJ
As the process is isothermal, i.e., temperature is constant.
∴ΔE=0
Now from first law of thermodynamics,
ΔE=q+W
∵ΔE=0
→q=−W=−(−8.88)=8.88kJ
Hence the value of W,ΔE and q are −8.88kJ,0 and 8.88kJ respectively.
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