Question

The thermodynamic cycle of an engine operating with an ideal monatomic gas is represented by the $P-V$ diagram as shown in the above figure. Find the efficiency and the amount of heat extracted from the source in a single cycle.

• A. $P_0{V}_0,9.1\%$
• B. $\left(\frac{13}{2}\right)P_0{V}_0,15.4\%$
• C. $\left(\frac{11}{4}\right)P_0{V}_0,10.5\%$
• D. $4P_0{V}_0,12.5\%$

Answer 1

Answer: (B)

Let $Q$ be the heat extracted per cycle,
$C_v=$ specific heat at constant volume,
$C_p=$ specific heat at constant pressure
$Q=n{C}_v\left(dT\right)+n{C}_p\left(dT\right)\Rightarrow Q=n{C}_v\left(2T_0-T_0\right)+n{C}_p\left(4T_0-2T_0\right)$
For monoatomic gas,
$C_v=$ $\frac{fR}{2}$
$C_p=\frac{(f+2)R}{2}$
$C_v=\frac{3R}{2},C_p=\frac{5R}{2}$
$Q=\frac{n(3R)T_0}{2}+\frac{n5R}{2}\left(2T_0\right)=\frac{13}{2}nR{T}_0$
$Q=\frac{13}{2}{P}_0{V}_0$
$W=$ area bounded by curve $=P_0{V}_0$
Let $\eta =$ efficiency during the cycle,
$\eta =\frac{W}{Q}\times 100\Rightarrow \eta =\frac{P_0{V}_0}{\frac{13P_0{V}_0}{2}}\times 100\Rightarrow \eta =\frac{2}{13}\times 100\Rightarrow \eta =15.4\%$