The indicator diagram for two processes 1 and 2 carrying on an ideal gas is shown in Fig. If `m_(1)` and `m_(2)` be the slopes `(dP//dV)` for process 1 and process 2, respectively, then

Options

A. `m_(1) = m_(2)`
B. `m_(1) gt m_(2)`
C. `m_(1)ltm_(2)`
D. `m_(2)C_(V) = m_(1)C_(p)`

Answer

C, D

Explanation

Slope of isothermal process
`m_(1) = (DeltaP)/(DeltaV) = -(P)/(V)`
Slope of adiabatic process
`m_(2) = (DeltaP)/(DeltaV) = -(gammaP)/(V) = -gammam_(1) = (C_(p))/(C_(V)) m_(1)`