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Q.

An ideal gas is made to go through a cyclic thermodynamical process in four steps. The amount of heat involved are Q₁ = 600 J, Q₂ = -400 J, Q₃ = -300 J and Q₄ = 200 J, respectively. The corresponding work involved are W₁ = 300 J, W₂ = -200 J, W₃ = -150 J and W₄. The value of W₄is ……joule

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Detailed Solution

From the first law of thermodynamics,

$ΔQ = ΔU + ΔW$ ……………(i)

For a cyclic process, $ΔU = 0$

$∴ ΔQ = ΔW$

Now, $ΔQ = Q_{1} + Q_{2} + Q_{3} + Q_{4}$

$= 600 J - 400 J - 300 J + 200 J = 100 J$

and $ΔW = W_{1} + W_{2} + W_{3} + W_{4}$

$\begin{matrix} \Rightarrow ΔW = 300 J - 200 J - 150 J + W_{4} \\ = - 50 J + W_{4} \end{matrix}$

Substitute the value of $ΔQ$ and $ΔW$ in Eq. (i), we get

$\begin{matrix} 100 J = - 50 J + W_{4} \\ W_{4} = 150 J \end{matrix}$

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