In an isothermal process temperature remains constant.
Consider pressure and volume of ideal gas changes from (P1, V1) to (P2, V2).
At any intermediate stage with pressure P and volume change from V to V+ΔV (ΔV small)
then from first law of thermodynamics ΔW=PΔV
Now taking ΔV approaching zero i.e. ΔV→0 and summing ΔW over entire process we get total work done by gas so we have W=∫V2V1PdV
where limits of integration goes from V1 to V2
as PV=nRT we have P=nRTV
Therefore
W=∫V2V1(nRTV)dV
On integrating we get, W=nRTln(V2V1)
Where n is number of Moles
in sample of gas taken.
and ln=loge
This can be also written as W=2.303nRTlog(V2V1)
This can also be expresses in terms of Initial Pressure and Final Pressure also
W=2.303nRTlog(P1P2)
We can easily conclude from Work done equation ,for V2>V1, W > 0 and for V2<V1, W < 0. That is, in an isothermal expansion, the gas absorbs
heat and does work while in an isothermal compression, work is done on the gas by the environment and heat is released.
Solved Example for Work done in a Isothermal Process
Question 1
2 moles of an ideal is isothermally expanded to 3 times its original volume at 300K . Calculate the Work done and heat absorbed by the gas?
Given ( R=8.31 J/mol-K and loge3=.4771) solution
Here n= 3, V2=3V1, T=300 K
Now using the above equation W=nRTln(V2V1)
Substituting the above values, we get W=5.48×103 J
Now heat absorbed = Work done= 5.48×103 J = 1.31×103 C
Question 2
An ideal gas undergoes isothermal process from some initial state i to final state f. Choose the correct alternatives.
(a) dU = 0
(b) dQ= 0
(c) dQ = dU
(d) dQ = dW solution
(a) and (d)
