An ideal gas is taken through 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%. The efficiency of the cycle 12341 is E%. Then $2\times E$ will be?

Cycle 1-2-3-1
From$1\to 2$, work done is $+ve(\because$ volume increases) and ΔU is also positive ($\because$temperature increased)
Similarly, for $2\to 3,W$ as well as ΔU are positive.
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 $3\to 1$. Let work done in the cycle be $W_1$ .

Using first law of thermodynamics for complete cycle-

$\begin{array}{rr} & \mathrm{\Delta }Q=\mathrm{\Delta }U+W\\ & Q_1-Q_0=W_1\end{array}$

$\begin{array}{rr} & \left[\mathrm{\Delta }{U}_{\text{cycle }}=0\right]\end{array}$ 

And efficiency $\eta =\frac{W_1}{Q_1}$

$\frac{1}{4}=\frac{W_1}{Q_1}$

Cycle 1-3-4-1
$Q_0=$ heat gained by the gas in process $1\to 3$ .
$Q_2=$heat rejected by the gas in process $3\to 4\to 1$
$W_2=$ work done in cycle.

$\eta =\frac{W_2}{Q_0}$

$\frac{1}{10}=\frac{W_2}{Q_1-W_1}\mathrm{ }$[using (i)]

$\Rightarrow Q_1-W_1=10W_2\Rightarrow Q_1=W_1+10W_2$

Using (i) and (iii)

$\begin{array}{rr} & 4W_1=W_1+10W_2\\ & 3W_1=10W_2\end{array}$

Efficiency of cycle 1-2-3-4-1

$\begin{array}{rr} & \eta =\frac{W_1+W_2}{Q_1}=\frac{W_1+W_2}{W_1+10W_2}\\ & =\frac{1+\frac{W_2}{W_1}}{1+10\frac{W_2}{W_1}}=\frac{1+\frac{3}{10}}{1+3}\mathrm{ }\text{ [using 4] }\end{array}$

$=\frac{13}{40}$

In percentage $\mathrm{ }\eta =\frac{13}{40}\times 100=32.5\mathrm{\%}$