An ideal gas undergoes the cyclic process shown in a graph below :

Options

A. `T_(1) = T_(2)`
B. `T_(1) gt T_(2)`
C. `V_(a) V_(c ) = V_(b) V_(d)`
D. `V_(a) V_(b) = V_(c) V_(d)`

Answer

B, C

Explanation

For adiabatic process `'bc'`
`T_(1)V_(B)^(gamma-1)=T_(2)V_(c)^(gamma-1) ....(i)`
For adiabatic process `'da'`
`T_(2)V_(d)^(gamma-1)=T_(1)V_(a)^(gamma-1) ....(ii)`
Multiplying Eqs. `(i)` and `(ii)`, we get
`impliesT_(1)T_(2)(V_(b)V_(d))^(gamma-1)=T_(1)T_(2)(V_(a)V_(c))^(gamma-1)`
` implies V_(b)V_(d)=V_(a)V_(c)`
Since adiabatic expansion leads to cooling,
so `T_(1)gtT_(2)`.