In case of a reversible process (following pvn = constant), work obtained for trebling the volume (v1 = 1 m3 and v2 = 3 m3) is maximum, when the value of 'n' is
A. 0
B. 1
C. 1.44
D. 1.66
Answer: Option A
Solution (By Examveda Team)
The work done for a process described by equation $$P{V^n} = {\text{constant}}{\text{.}}$$Is given by $$w = \frac{{CV_2^{1 - n} - CV_1^{1 - n}}}{{1 - n}}$$
So, the work obtained will be maximum when n is minimum so, $$n = 0.$$
This Question Belongs to Chemical Engineering >> Chemical Engineering Thermodynamics
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