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arpon
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Homework Statement
##dz=Mdx+Ndy## is an exact differential if ##(\frac{\partial M}{\partial y})_x=(\frac{\partial N}{\partial x})_y##.
By invoking the condition for an exact differential, demonstrate that the
reversible heat ##Q_R## is not a thermodynamic property.
Homework Equations
##dQ=dU+PdV##
##dU=C_VdT##
The Attempt at a Solution
##dQ=C_VdT+PdV##
So, we have to show that, ##(\frac{\partial C_V}{\partial V})_T \neq (\frac{\partial P}{\partial T})_V##
Now,
##LHS=(\frac{\partial C_V}{\partial V})_T##
##=(\frac{\partial}{\partial V}((\frac{\partial U}{\partial T})_V))_T~~~~##[##C_V=(\frac{\partial U}{\partial T})_V##]
##=(\frac{\partial}{\partial T}((\frac{\partial U}{\partial V})_T))_V~~~~## [Symmetry of second derivatives]
So, we need to show that
##(\frac{\partial U}{\partial V})_T \neq P+F(V)~~~## [where, ##F(V)## is an arbitary function of ##V##]
##(\frac{\partial U}{\partial V})_T - P \neq F(V)##
If the equation of state is given, this can easily be proved. How can I prove this in general? I also do not understand what would be the difference for irreversible heat.
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