Question 133: Thermodynamics & KTG

Question

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

Accepted Answer

Process 1
p = constant, Volume increases and temperature also increases
Q = W + $$\Delta$$U
$$	herefore$$ W = positive, $$\Delta$$U = positive
$$\Rightarrow$$ Heat is positive and supplied to the gas.
Process 2
V = constant, Pressure decreases
T $$\propto$$ pV [as V = constant]
$$\Rightarrow$$ Temperature decreases
W = 0
$$\Delta$$T is negative and $$\Delta$$U = $${f \over 2}$$nR$$\Delta$$T
$$	herefore$$ $$\Delta$$U is also negative
Q = $$\Delta$$U + W
$$	herefore$$ Heat is negative and rejected by gas.
Process 3
p = constant, Volume decreases
$$\Rightarrow$$ Temperature also decreases
W = p$$\Delta$$V = negative
$$\Delta$$U = $${f \over 2}$$nR$$\Delta$$T = negative
$$	herefore$$ Heat is negative and rejected by gas.
Process 4
V = constant, Pressure increases
W = 0 (as V = constant)
pV = nRT $$\Rightarrow$$ Temperature increase
$$\Rightarrow$$ $$\Delta$$U = $${f \over 2}$$nR$$\Delta$$T is positive
$$\Delta$$Q = $$\Delta$$U + W = positive

Answer

steps 1 and 2

Answer

steps 1 and 3

Answer

steps 1 and 4

Answer

steps 2 and 4

Thermodynamics & KTG

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

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