An ideal gas with constant heat capacity `C_(V)=(3)/(2)nR` is made to carry out a cycle that is depicted by a triangle in the figure given below.

The following statement is true about the cycle-

Options

A. The efficiency is given `1- (P_1V_1)/(P_2V_2)`
B. The efficiency is given `1- 1/2(P_1V_1)/(P_2V_2)`
C. Net heat absorbed in the cycle is `(P_2-P_1)(V_2-V_1)`
D. Heal absorbed in part AC is given by `2(P_2V_2 - P_1V_1) +1/2 (P_1V_2 - P_2V_1)`

Answer

B

Explanation

`C_V=3/2 R, C_P=C_V+R=(5R)/(2) implies f=3`
`W=1/2 (V_2-V_1)(P_2-P_1)`
For BA, `Q=n C_P Delta T=n(5/2 R)xxDelta T=5/2(P_1V_1-P_2V_2)`
For AC, `Q_(AC)=1/2(P_1+P_2)(V_2-V_1)+nC_VDeltaT`
Now, `nC_V Delta T=3/2(P_2V_2-P_1V_1)`
Now, `eta=W/Q=(1/2 (V_2-V_1)(P_2-P_1))/(1/2xx(P_1+P_2)(V_2-V_1)+3/2(P_2V_2-P_1V_1))`
Using formula for heat we can calculate ~ heat absorbed in AC.