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)

     0.24

  • Option 2)

    0.15

  • Option 3)

     0.32

     

  • Option 4)

     0.08

     

 

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

\eta=\frac{2}{13}=0.15

Correct option 2


Option 1)

 0.24

This is an incorrect option.

Option 2)

0.15

This is the correct option.

Option 3)

 0.32

 

This is an incorrect option.

Option 4)

 0.08

 

This is an incorrect option.

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