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Entropy for a Reversible Process

A reversible ...

Question

A reversible heat engine carries $1 m o l$ of an ideal monatomic gas around the cycle $A B C A$ , as shown in the diagram. The process $B C$ is adiabatic. Call the processes $A B$ , $B C$ and $C A$ as 1, 2 and 3 and the heat ${(Δ Q)}_{r}$ , change in internal energy ${(Δ U)}_{r}$ and the work done ${(Δ W)}_{r}$ , $r = 1, 2, 3,$ respectively. The temperature at $A$ , $B$ , $C$ are $T_{1} = 300 K$ , $T_{2} = 600 K$ and $T_{3} = 455 K$ . Indicate the pressure and volume at $A$ , $B$ and $C$ by $P_{r}$ and $V_{r}$ , $r = 1, 2, 3,$ respectively. Assume that initially pressure $P_{1} = 1.00 a t m$ . Suppose the gas had been taken along an isothermal process from $B$ instead of along the adiabatic process $B C$ shown, so as to reach the same volume $V_{3}$ , and the total heat $Q_{i}$ is taken during the complete cycle involving the isothermal, while $Q_{a}$ is the total heat taken in the cycle $A B C A$ shown in the diagram. Then

Q_{i} < Q_{a}

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Q_{i} = Q_{a}

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Q_{i} > Q_{a}

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Q_{i}

will be greater or less than

Q_{a}

depending upon the value of

T_{3}

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Solution

The correct option is A $Q_{i} < Q_{a}$
Given:
1 mol of ideal monatomic gas.
Processes AB=1, BC=2, CA=3
Heats, $Δ Q_{1}, Δ Q_{2}, Δ Q_{3}$
Change in internal energy, $Δ U_{1}, Δ U_{2}, Δ U_{3}$
Work done, $Δ W_{1}, Δ W_{2}, Δ W_{3}$
Temperature, $Δ T_{1} = 300 K, Δ T_{2} = 600 K, Δ T_{3} = 455 K$
Pressure, $P_{1} = 1.00 a t m, P_{2}, P_{3}$
Volume, $V_{1}, V_{2}, V_{3}$
Total heat during isothermal process, $Q_{i}$
Total heat during adiabatic process, $Q_{a}$
To find:
Relation between $Q_{i}, Q_{a}$
Case(i): When BC is adiabatic process,
In the cyclic process : $Δ = 0$
$⟹ Δ Q_{a} = Δ W_{a}$
And, $Δ W_{a} = Δ W_{1} + Δ W_{2} + Δ W_{3} . . . . (i)$
As BC is adiabatic in this case hence,
$Δ Q_{2} = 0 ⟹ Δ W_{2} = - Δ U_{2} . . . . . . (i i)$
Hence from equation(i) and (ii) we get,
$Δ Q_{a} = Δ W_{a} = Δ W_{1} - Δ U_{2} + W_{3} . . . . . (i i i)$

Case (ii): When BC is isothermal process,
In the cyclic process : $Δ = 0$
$⟹ Δ Q_{i} = Δ W_{a}$
And, $Δ W_{i} = Δ W_{1} + Δ W_{2} + Δ W_{3} . . . . . (a)$
As BC is isothermal in this case hence,
Hence temperature is constant.
$⟹ Δ T = 0 ⟹ Δ U_{2} = 0$
$Δ W_{2} = P Δ V = n R Δ T = 0......... (b)$
Hence $Δ Q_{a} = Δ W_{a} = W_{1} + 0 + W_{3} = W_{1} + W_{3} . . . . . (c)$

From equation(iii) and (c),
$Q_{a} > Q_{i}$

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