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  4. A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is - 180 J, The gas absorbs 250 J of heat along the path ab and 60 J along the path be. The work done (in joule) by the gas along the path abc is: <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/d125ab474e7b6e27463dcb7c24ff8049-.png />

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Q. A sample of an ideal gas is taken through the cyclic process $abca$ as shown in the figure. The change in the internal energy of the gas along the path ca is $- 180 J$, The gas absorbs $250 J$ of heat along the path ab and $60 J$ along the path be. The work done (in joule) by the gas along the path $abc$ is :
image

Thermodynamics

Solution:

$\Delta U_{ac} = -\left(\Delta U_{ca}\right) = - \left(-180\right) =180J$
$Q = 250 +60 = 310J$
Now $Q=\Delta U + W$
or $310 =180 +W$
or $W = 130 J$