Question

An ideal monoatomic gas is carried around the cycle ABCDA as shown in the fig. The efficiency of the gas cycle is

• A. $\frac{4}{21}$
• B. $\frac{2}{21}$
• C. $\frac{4}{31}$
• D. $\frac{2}{31}$

Answer 1

The correct option is A $\frac{4}{21}$

Heat is absorbed only during processes AB and BC
Gas is monoatomic.Hence $C_v=\frac{3}{2}R,C_p=\frac{5}{2}R$
Heat absorbed during AB, $\Delta Q_{AB}=n{C}_v\Delta T=\frac{3}{2}nR\Delta T=\frac{3}{2}V\Delta P=3P_o{V}_o$

Heat absorbed during BC, $\Delta Q_{BC}=n{C}_p\Delta T=\frac{5}{2}nR\Delta T=\frac{5}{2}P\Delta V=\frac{15}{2}{P}_o{V}_o$

Hence, net heat absorbed=$\Delta Q=\Delta Q_{AB}+\Delta Q_{BC}=\frac{21}{2}{P}_o{V}_o$

Also, net work done=$\Delta W$=area under P-V graph=$2P_o{V}_o$
Hence efficiency=$\frac{\Delta W}{\Delta Q}=\frac{4}{21}$