In the previous problem, the temperature of the isothemal process is how much lower the maximum temperature :A. `(P_(0) V_(0))/(nR)`B. `(P_(0) V_(0))/(2nR)`C. `(3P_(0) V_(0))/(2nR)`D. `(P_(0) V_(0))/(4nR)`

In the previous problem, the temperature of the isothemal process is h

1.	In the previous problem, the temperature of the isothemal process is how much lower the maximum temperature :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» Correct Answer - D 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)`