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 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)

  • Option 1)


  • Option 2)


  • Option 3)



  • Option 4)




Answers (1)


As we learnt in

Specific heat of gas at constant volume -

C_{v}= \frac{fR}{2}

- wherein

f = degree of freedom

R= Universal gas constant


 Thermal Efficiency of Cycle =\frac{Total\ work\ done}{Gorss\ Heat\ absorbed}

Gross Heat absorved is amount of heat that is absorbed only.

\omega = P_{0}V_{0}

Q_{AB}=\Delta V_{AB}+\omega_{AB}=nC_{v}\left(T_{f}-T_{i} \right )=n\frac{f}{2}\left(RT_{f}-RT_{f} \right )=\frac{f}{2} \left(P_{0}V_{0} \right )

Q_{BC}=n\left(\frac{f}{2}+1 \right )R\left(T_{f}-T_{i} \right )=\left(\frac{f}{2}+1 \right ).\left(4P_{0}V_{0}-2P_{0}V_{0} \right )=\left(\frac{f}{2}+1 \right )\left(2P_{0}V_{0} \right )

Q_{CD}=-Ve, \ \;Q_{DA} =-ve

These two are not the part of gross heat.

\therefore\ \; \eta=\frac{P_{0}V_{0}}{\left(\frac{3f}{2}+2 \right )(P_{0}V_{0})}=\frac{1}{\frac{3f}{2}+2}

For monoatomic gas f = 3


Correct option 2

Option 1)


This is an incorrect option.

Option 2)


This is the correct option.

Option 3)



This is an incorrect option.

Option 4)



This is an incorrect option.

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