In the previous problem, the temperature of the isothemal process is how much lower the maximum temperature :

Options

A. `(P_(0) V_(0))/(nR)`
B. `(P_(0) V_(0))/(2nR)`
C. `(3P_(0) V_(0))/(2nR)`
D. `(P_(0) V_(0))/(4nR)`

Answer

D

Explanation

d. From the last problem, from Eq. (ii), at point `A(V = V_(0))`
`T = (2 P_(0) V_(0))/(nR)`
This is the temperature which is constant along isothermal process
`T_(max)` is for` V = (3)/(2) V_(0)`
`T_(max) = (9 P_(0) V_(0))/(4 nR)` [from Eq. (iii)]
`T_(max) - T = (P_(0) V_(0))/(4nR)`