JEE Advance - Physics (2021 - Paper 1 Online - No. 2)

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

JEE Advanced 2021 Paper 1 Online Physics - Heat and Thermodynamics Question 33 English
steps 1 and 2
steps 1 and 3
steps 1 and 4
steps 2 and 4

Explanation

Process 1

p = constant, Volume increases and temperature also increases

Q = W + $$\Delta$$U

$$\therefore$$ 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

$$\therefore$$ $$\Delta$$U is also negative

Q = $$\Delta$$U + W

$$\therefore$$ 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

$$\therefore$$ 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

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