NEET-XII-Physics
26: Laws of Thermodynamics
- #8Consider two processes on a system as shown in figure.
Figure
The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let â–µW1 and â–µW2 be the work done by the system in the processes A and B respectively.
(a) â–µW1 > â–µW2.
(b) â–µW1 = â–µW2.
(c) â–µW1 < â–µW2.
(d) Nothing can be said about the relation between â–µW1 and â–µW2.digAnsr: cAns :
(c) ∆W1 < ∆W2
Work done by the system, `` ∆W=P∆V``
Here,
P = Pressure in the process
`` ∆V`` = Change in volume during the process
Let Vi and Vf​ be the volumes in the initial states and final states for processes A and B, respectively. Then,
`` \Delta {W}_{1}={P}_{1}\Delta {V}_{1}``
`` \Delta {W}_{2}={P}_{2}\Delta {V}_{2}``
`` \,\mathrm{\,But\,}\Delta {V}_{2}=\Delta {V}_{1},\left[\left({V}_{{f}_{1}}-{V}_{{i}_{1}}\right)=\left({V}_{{f}_{2}}-{V}_{{i}_{2}}\right)\right]``
`` \Rightarrow \frac{\Delta {W}_{1}}{\Delta {W}_{2}}=\frac{{P}_{1}}{{P}_{2}}``
`` ``
`` \Rightarrow \Delta {W}_{1}<\Delta {W}_{2}\left[\because {P}_{2}>{P}_{1}\right]``
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