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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Starting from the same initial conditions , an ideal gas expands from volume $\;V_{1}\;$ to $\;V_{2}\;$ in three different ways . The work done by the gas is $\;W_{1}\;$ if the process is purely isothermal , $\;W_{2}\;$ if purely isobaric and $\;W_{3}\;$ if purely adiabatic . Then :

$(a)\;W_{2} >W_{1} >W_{3}\qquad(b)\;W_{2} >W_{3} >W_{1}\qquad(c)\;W_{1} >W_{2} >W_{3}\qquad(d)\;W_{1} >W_{3} >W_{2}$

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1 Answer

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Answer : $\;W_{2} >W_{1} >W_{3}$
Explanation :
Curve 1 - isobaric process
Curve 2 - isothermal process
Curve 3 - idiabatic process
Since work done Area under PV graph
$= W_{2} >W_{1} >W_{3}$
answered Apr 9, 2014 by yamini.v
 

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