Answer:
The thermal efficiency of an engine is [tex]0.15[/tex] that is 15%
Explanation:
Specific heat of a gas at constant volume is
[tex]C_{v}=\frac{fR}{2}[/tex]
here, [tex]f=[/tex]degree of freedom
[tex]R=[/tex]universal gas constant
Thermal efficiency of a cycle is
[tex]\frac{total workdone}{gross heat absorbed}[/tex]
Gross heat absorbed is amount of heat that absorbed.
ω[tex]=P_{o} V_{o}[/tex]
[tex]Q_{AB}=[/tex]Δ[tex]V_{AB}+[/tex]ω[tex]_{AB}[/tex][tex]=nC_{v}(T_{f}-T_{i} )[/tex]
[tex]=n\frac{f}{2}(RT_{f}-RT_{i} )[/tex][tex]=\frac{f}{2}(P_{o}V_{o} )[/tex]
[tex]Q_{BC}=n(\frac{f}{2}+1 )R(T_{f}-T_{i} )[/tex]
[tex]=(\frac{f}{2} +1)(4P_{o}V_{o}-2P_{o}V_{o} )[/tex]=[tex]=(\frac{f}{2}+1)2P_{o}V_{o}[/tex]
[tex]Q_{CD}=-ve and Q_{DA}=-ve[/tex] these 2 are not the part of gross heat
η[tex]=\frac{P_{o}V_{o} }{(\frac{3f}{2}+2 )(P_{o}V_{o})}[/tex]
[tex]=\frac{1}{(\frac{3f}{2}+2 )}[/tex]
For monoatomic gas [tex]f=3[/tex]
η[tex]=\frac{2}{13}[/tex]
[tex]=0.15[/tex]
