Question 495: Thermodynamics & KTG

Question

An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is :
(Take Cv = 1.5 R, where R is gas constant)

Accepted Answer

Work done by engine = area under closed curve = P
0
V
0
According to the principle of calorimetry,
Heat given to the system,
Q = Q
AB
+ Q
BC
= nC
v
$$\Delta $$T
AB
+ nC
p
$$\Delta $$T
BC
= $${3 \over 2}$$ (nR T
B
$$-$$ nRT
A
) + $${5 \over 2}$$ (nRT
c
+ nRT
B
)
= $${3 \over 2}$$ (2P
0
V
0
$$-$$ P
0
V
0
) + $${5 \over 2}$$ (4P
0
V
0
$$-$$ 2P
0
V
0
)
= $${13 \over 2}$$ P
0
V
0
$$	herefore\,\,\,$$ Thermal efficiency
= $${W \over Q}$$ = $${{{P_0}\,{V_0}} \over {{{13} \over 2}{P_0}{V_0}}}$$ = $${2 \over {13}}$$ = 0.15

Answer

0.24

Answer

0.15

Answer

0.32

Answer

0.08

Thermodynamics & KTG

An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is : (Take Cv = 1.5 R, where R is gas constant)

Choose the correct answer: