Q: An Ideal gas undergoes a four step cycle as shown in 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
Click to See Solution :
Sol:Process 1:
P = constant , Volume increases and temperature also increases
Q = W + ΔU
W = +ve , ΔU = positive
Heat is positive and supplied to the gas
Process 2 :
V = constant , Pressure decreases
T ∝ P , (V = constant )
Temperature decreases , W = 0
ΔT is negative and ΔU = f/2 nRΔT
ΔU is also negative
Q = ΔU + W
Heat is negative and rejected by gas
Process 3 :
P = constant , Volume decreases
Temperature also decreases
W = PΔV = negative
ΔU = f/2 (nRΔT) = negative
Heat is negative and rejected by gas
Process 4 :
V = constant , Pressure increases
W = 0
PV = n RT => temperature increases
ΔU = f/2 (nRΔT) is positive
ΔQ = ΔU + W = Positive