Question

An ideal gas is taken through a cycle $1231$ (see figure) and the efficiency of the cycle was found to be $25\%$. When the same gas goes through the cycle $1341$, the efficiency is $10\%$. Find the efficiency of the cycle $12341$.

• A. $18.6\%$
• B. $23.2\%$
• C. $32.5\%$
• D. $44\%$

Answer 1

The correct option is C $32.5\%$

Cycle 1-2-3-1
From $1\to 2$, work done is +ve ($\because$ volume increases) and $\Delta U$ is also positive ($\because$ temperature increases).
Similarly, for $2\to 3$, $W$ as well as $\Delta U$ are positive.
Thus, in both the processes $1\to 2$ and $2\to 3$, the gas absorbs heat.
Let the total heat absorbed in the process $1\to 2\to 3$ be $Q_1$. Similarly, one can argue that the gas rejects heat (say $Q_0$) to the surrounding in the process $3\to 1$. Also, let work done in the cycle be $W_1$.


Using first law of thermodynamics for complete cycle -
$\Delta Q=\Delta U+W$ $[\Delta U_{\text{cycle}}=0]$
$Q_1-Q_0=W_1$ $\dots \dots \left(i\right)$
And efficiency $\eta =\frac{W_1}{Q_1}$
$\Rightarrow \frac{1}{4}=\frac{W_1}{Q_1}$ $\dots \dots \left(ii\right)$

Cycle 1-3-4-1


$Q_0=$ heat gained by the gas in process $1\to 3$.
[due to symmetry, it will be the same as heat rejected in previous]
Let $Q_2=$ heat rejected by the gas in process $3\to 4\to 1$
Let $W_2=$ work done in cycle.
$\eta =\frac{W_2}{Q_0}$ $\Rightarrow \frac{1}{10}=\frac{W_2}{Q_1-W_1}$ [using (i)]
$\Rightarrow$ $Q_1-W_1=10W_2$
$\Rightarrow$ $Q_1=W_1+10W_2$ $\dots \dots \left(iii\right)$
Using (ii) and (iii)
$4W_1=W_1+10W_2$
$3W_1=10W_2$ $\dots \dots \left(iv\right)$

Efficiency of cycle 1-2-3-4-1 is given by:


$\eta =\frac{W_1+W_2}{Q_1}=\frac{W_1+W_2}{W_1+10W_2}$
$=\frac{1+\frac{W_2}{W_1}}{1+10\frac{W_2}{W_1}}=\frac{1+\frac{3}{10}}{1+3}$ [using (iv)]
$=\frac{13}{40}$
i.e $\eta =\frac{13}{40}\times 100=32.5\%$