Enthalpy change and Entropy change; given the equation for the substance

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Homework Statement



Equation for the substance given for a certain range of pressure and temperature:

v=(RT/p) - C/T3

Where C=Constant

Show that
(i) The change of enthalpy for this substance in an isothermal process is

(h2-h1)T=4C/T3(p1-p2)T

(ii) The change of entropy for this substance in an isothermal process is

(S2-S1)T=Rln(p1/p2)+3C/T4(p1-p2)T


The Attempt at a Solution



taking h=U+pv

dh=du+d(pv) {du=0 for isothermal process}

dh=d(pv)=-Cdp/T3

but result which i need to show has factor 4 also; which i have no idea how to get.
 

Answers and Replies

  • #2
physicsworks
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Welcome to PF, sac.lalit! :smile:
Your mistake is to treat [tex]dU=0[/tex] in isothermal process. If it were ideal gas with the usual gas law
[tex]p(V,T)=\frac{RT}{V}[/tex]
you would be right, but this is not the case.
In fact, you have
[tex]V(p,T) = \frac{RT}{p} - \frac{C}{T^3}~~ \text{or}[/tex]
[tex]p(V,T) = \frac{RT^4}{(VT^3+C)} \left(*\right)[/tex]
and that's why the usual expression
[tex]dU = C_V dT[/tex]
need to be changed (Note also that if [tex]C=0[/tex] in the above equation (*), you get the usual form of equation of state).
To change U in appropriate way, you must rederive it from the basical first law of thermodynamics in its form:
[tex]dS = \frac{dU}{T} + \frac{p}{T}dV \left( ** \right)[/tex]
Now, with [tex]U=U(T,V)[/tex] (caloric equation) you have:
[tex]dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV[/tex]
and after substituting this into (**):
[tex]dS = \frac{1}{T} \left(\frac{\partial U}{\partial T}\right)_V dT + \frac{1}{T} \left[ p + \left(\frac{\partial U}{\partial V}\right)_T \right] dV[/tex]
From this equation:
[tex]\left(\frac{\partial S}{\partial T}\right)_V = \frac{1}{T} \left(\frac{\partial U}{\partial T}\right)_V[/tex]
[tex]\left(\frac{\partial S}{\partial V}\right)_T = \frac{1}{T} \left[ p + \left(\frac{\partial U}{\partial V}\right)_T \right] dV[/tex]
and remembering that
[tex]\left(\frac{\partial^2 S}{\partial V \partial T}\right) = \left(\frac{\partial^2 S}{\partial T \partial V}\right)[/tex]
we finally get:
[tex]T \left(\frac{\partial p}{\partial T}\right)_V = \left(\frac{\partial U}{\partial V}\right)_T + p \left(***\right)[/tex]


From this equation we can derive U (caloric equation) if we know thermal equation of state [tex]p=p(V,T)[/tex]
In fact, from (*):
[tex]\left(\frac{\partial p}{\partial T}\right)_V = \frac{RT^3 (VT^4 + C)}{(VT^3+C)^2}[/tex]
and substituting this into (***)
[tex]\left(\frac{\partial U}{\partial V}\right)_T = \frac{3RCT^4}{(VT^3+C)^2}[/tex]
Then
[tex]dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV = C_V dT + \frac{3RCT^4}{(VT^3+C)^2}dV[/tex]
or, using (*) for [tex]V(p,T)[/tex] from which
[tex]\frac{dV}{dp} = -\frac{RT}{p^2}[/tex]
and substituting the expression for [tex]p=p(V,T)[/tex] we can rewrite
[tex]dU = C_V dT - \frac{3C}{T^3}dp[/tex]
________________________________________________________________

Now, as you wrote, but using the new expression for [tex]dU[/tex]:
[tex]dH = dU + pdV + Vdp = C_V dT - \frac{3C}{T^3}dp -\frac{RT}{p}dp + \frac{RT}{p}dp - \frac{C}{T^3}dp = C_V dT - \frac{4C}{T^3} dp [/tex]
For isothermal process
[tex]dH = - \frac{4C}{T^3}dp[/tex]
Now the factor 4 pops up :smile:
AND:
[tex]\Delta H_{12} = - \frac{4C}{T^3} \Delta p_{12}[/tex]
 
Last edited:

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