An ideal gas undergoes a four step cycle as shown in the P − V diagram below. During this cycle, heat is absorbed by the gas in

($\alpha$) steps 1 and 2           ($\beta$) steps 1 and 3
($\gamma$) steps 1 and 4           ($\delta$) steps 2 and 4

Solution

Step 1 (A to B): Isobaric

$V \propto T$

$T_B > T_A $

$Q \propto C_p \Delta T > 0 $

Step 2 (B to C): Isochoric

$P \propto T$

$T_C < T_B$

$Q \propto C_v \Delta T < 0$

Step 3 (C to D): Isobaric

$V \propto T$

$T_D < T_C$

$Q \propto C_p \Delta T < 0$

Step 4 (D to A): Isochoric

$P \propto T$

$T_A > T_D$

$Q \propto C_v \Delta T > 0$

Answer: ($\gamma$)