Question

An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are $Q_1=5960\text{J},Q=-5585\text{J},Q_3=-2980\text{J}$ and $Q_4=3645\text{J}$ respectively. The corresponding quantities of work done by gas for each steps are $W_1=2200\text{J},W_2=-825\text{J},W_3=-1100\text{J}$ and $W_4=765\text{J}$ respectively. The efficiency of the cycle is

• A. $10.82\%$
• B. $20.85\%$
• C. $60\%$
• D. $79.18\%$

Answer 1

The correct option is A $10.82\%$

For a cyclic process, the efficiency of cycle is given by,
$\eta =\frac{\text{Net work done}}{\text{Total heat input}}\times 100$
Heat input for the cycle is $Q_1\&Q_4$ due to $+ve$sign of $Q$
$\eta =\left(\frac{W_1+W_2+W_3+W_4}{Q_1+Q_4}\right)\times 100$
$\Rightarrow \eta =\left\{\frac{2200-825-1100+765}{5960+3645}\right\}\times 100$
$\therefore \eta =10.82\%$