Question

The area enclosed by a $PV$ graph for one thermodynamic cycle of a heat engine is $610\text{J}$. The total thermal energy absorbed by the gas in one cycle is $1300\text{J}$. Determine the efficiency of the cycle.

• A. $47\%$
• B. $23\%$
• C. $9.4\%$
• D. $36.8\%$

Answer 1

The correct option is A $47\%$

Given that, the area enclosed by $PV$ diagram for one thermodynamic cycle of a heat engine is $610\text{J}$.
It means, work done $W=610\text{J}$.
and, thermal energy absorbed by the gas in one cycle is $Q=1300\text{J}$
Efficiency of heat engine in one cycle
$\eta =\frac{\text{Work done}}{\text{Heat absorbed}}=\frac{W}{Q}=\frac{610}{1300}$
$=0.469\approx 47\%$
Thus, option (a) is the correct answer.