An ideal gas undergoes a circular cycle as shown in the figure. Find the ratio of maximum temperature of cycle to minimum temperature of cycle.

Options

A. `((1+sqrt(2))/(sqrt(2)-1))^(2)`
B. `((2+sqrt(2))/(2-sqrt(2)))^(2)`
C. `((3+sqrt(2))/(3-sqrt(2)))^(2)`
D. `((4+sqrt(2))/(4-sqrt(2)))^(2)`

Explanation

The maximum temperature will occur at point `A` and minimum temperature will accurate point `B` of the cycle, so
At point `A`
`(P_(A))/(P_(B))=(V_(A))/(V_(B))=2+cos45^(@)=(2sqrt(2)+1)/(sqrt(2))=(4+sqrt(2))/(2)`
`nRT_(A)=P_(A)V_(A)=((4+sqrt(2))/(2))^(2)P_(0)V_(0)`
Similarly at point `B`
`(P_(B))/(P_(0))=(V_(B))/(V_(0))=2-cos45^(@)=(4-sqrt(2))/(2)`
`nRT_(B)=P_(B)V_(B)=((4-sqrt(2))/(2))^(2)P_(0)V_(0)`
`rArr (T_(A))/(T_(B))=((4+sqrt(2))/(4-sqrt(2)))^(2)`