Question

(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.​

Answer 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.