QJEEAdv21Ph1977: JEE Advanced 2021 Physics - Class 11 JEEAdv21Ph1977 Solution: An ideal gas undergoes a four step cycle as shown in the P-V…

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