Q. 18

Figure shows the variation in the internal energy U with the volume V of 2.0 mol of an ideal gas in a cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K respectively. Calculate the heat absorbed by the gas during the process.



Answer :

Given


Number of moles n=2


Temperature at b Tb=500K


Temperature at c Tc=300K


From the graph it is clear that


Tb=Ta and Td=Tc


Thus, path ab and cd are isothermal paths.


We know that work done by the gas is given as


ΔW=PΔV


Where ΔV =change in volume


P =pressure


Again, from the graph we can see that ΔV=0 for path bc and da.


Therefore, work done along path bc and da are zero.


So, total work done ΔW=Wab+Wcd


We know that work done in an isothermal process is given as



Where n=number of moles


R=gas constant =8.31J/Kmol


T=temperature


Vf=final volume


Vi=initial volume




Wab = 8310×ln2


Similarly,





Wcd = -4986 ×ln2


So, total work done ΔW=Wab+Wcd


=8310×ln2 - 4986×ln2


=3324×ln2


ΔW =3324×0.693=2304.02J


From first law of thermodynamics, we know that,


ΔQ=ΔU+ΔW


Where ΔQ=heat supplied to the system


ΔU=change in internal energy


ΔW=work done by the system


Process ABCA is a cyclic process. The system is brought back to its initial state. Since internal energy is a state function, change in internal energy will be zero.


So, ΔU=0.


So, first law becomes


ΔQ=ΔW=2304.02J


Thus, heat absorbed by the system is 2304.02J.


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