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Standard XII

Physics

Thermodynamic Processes

An ideal gas ...

Question

An ideal gas is taken through a cyclic thermodynamic process through four steps. The amount of heat exchanged in the steps are $Q_{1} = + 5960 J, Q_{2} = - 5600 J, Q_{3} = - 3000 J, Q_{4} = + 3600 J$ respectively. The corresponding quantities of internal energy changes are $Δ U_{1} = + 3760 J, Δ U_{2} = - 4800 J, Δ U_{3} = - 1800 J$ and $Δ U_{4}$ . Find the value of $Δ U_{4}$ and work done during the cyclic process.

+ 2840 J, + 960 J

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+ 2830 J, + 900 J

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- 2930 J, - 960 J

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- 2930 J, + 960 J

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Solution

The correct option is A $+ 2840 J, + 960 J$
In a cyclic process, total change in internal energy is zero.
$Δ U = [Δ U_{1} + Δ U_{2} + Δ U_{3} + Δ U_{4}] = 0$
From the data given in the question,
$+ 3760 - 4800 - 1800 + Δ U_{4} = 0$
$\Rightarrow Δ U_{4} = + 2840 J$
Using first law of thermodynamics for cyclic process, we can say that
$Δ W = Δ Q$
$= + 5960 - 5600 - 3000 + 3600 = + 960 J$ .
Hence, option (a) is the correct answer.

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An ideal gas is taken through a cyclic thermodynamic process through four steps. The measures of heat involved in these steps are: $Q_{1} = 5960 J, Q_{2} = - 5585 J, Q_{3} = - 2980 J$ and $Q_{4} = 3645 J,$ respectively. The corresponding quantities of work involved are $W_{1} = 2200 J, = W_{2} = - 825 J$ and $W_{3} = 1100 J$ respectively.