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
(A) steps 1 and 2 (B) steps 1 and 3
(C) steps 1 and 4 (D) steps 2 and 4
Solution
$V \propto T$
$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: (C)