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Unformatted text preview: Relating thermodynamic quantities A,G,H,U,S to measurable quantities (equation of state and heat capacities) If A(V,T) or G(P,T) is known for some material , all of its other thermodynamic properties can be determined dA = dU − TdS − SdT = (− PdV + TdS ) − TdS − SdT = − PdV − SdT similarly dG = dA + PdV + VdP = VdP − SdT On the other hand, viewing A as a function of V and T or G as a function of P,T, we can write the formal identities: dA = ⎜ ⎛ ∂A ⎞ ⎛ ∂A ⎞ ⎟ dV + ⎜ ⎟ dT ⎝ ∂V ⎠T ⎝ ∂T ⎠V ⎛ ∂G ⎞ ⎛ ∂G ⎞ dG = ⎜ ⎟ dP + ⎜ ⎟ dT ⎝ ∂P ⎠T ⎝ ∂T ⎠ P Comparing the two expressions for dA, we find ⎛ ∂A ⎞ P = −⎜ ⎟ ⎝ ∂V ⎠T ⎛ ∂A ⎞ S = − ⎜ ⎟ ⎝ ∂T ⎠V
To calculate the internal energy, we now use the definition of A = U
TS and the above expression for the entropy: ⎛ ∂A ⎞ U = A + TS = A − T ⎜ ⎟ ⎝ ∂T ⎠V
If we wish now to compute the heat capacity we use: ∂⎛ ⎛ ∂U ⎞ ⎛ ∂A ⎞ ⎞ CV = ⎜ ⎜ A−T ⎜ ⎟= ⎟ ⎟ ⎝ ∂T ⎠V ∂T ⎝ ⎝ ∂T ⎠V ⎠V
So the equation of state and the heat capacities are not independent: They all can be derived from the same fundamental quantity A(V,T) (or G(P,T)). Any thermodynamic quantity can be calculated if we know either A or G. In principle, one can calculate A or G if the interaction energies among the molecules are known – this is the main goal of statistical mechanics. Unfortunately, in practice this is not easy except for very simple, idealized systems. In some (rare) cases, such as an ideal gas, A and G can be calculated analytically from first principles. Thermodynamics takes a different approach and starts with the quantities that can be measured and from those derives A(V,T), G(V,T) and other thermodynamic functions. A useful relationship from calculus (aka Maxwell’s relationship). consider a function of 2 variables, F(x,y). x and y can be V, T, or P, and F can be G, S, U, H, A etc. We have: dF = F(x+dx, y+dy) – F(x,y) = A(x,y) dx + B(x,y) dy = (∂F/∂x) dx + (∂F/∂y) dy A(x,y) = (∂F/∂x) is the partial derivative of F(x,y) with respect to x, that is when it is calculated, all other variables (y in this case) are kept constant. One can also use the notation A(x,y) = (∂F/∂x)y to emphasize that y is kept constant. We will be using this kind of notation in thermodynamics because we need to specify which variable we are working with. Both A(x,y) and B(x,y) are functions of x and y themselves. Consider the partial derivative (∂A/∂y)x = (∂/∂y) ∂F/∂x = ∂2F/∂x∂y = (∂/∂x) ∂F/∂y = (∂B/∂x)y This is a very important relationship that will allow us to derive many useful formulas. It can also be used to test whether A(x,y) dx + B(x,y) dy represents an exact differential of some function. Exercise: verify the validity of the above identity for F ( x, y) = x 2 y 7 How to calculate the entropy as a function of (V,T). Below we will be considering only equilibrium processes so instead of inequalities we will have equalities. We have previously learned two relationships: dG = V dP – S dT dA =
PdV – S dT (1) (2) Consider the second one. We have also established that ⎛ ∂A ⎞ P = −⎜ ⎟ ⎝ ∂V ⎠T
and ⎛ ∂A ⎞ S = − ⎜ ⎟ ⎝ ∂T ⎠V
Applying Maxwell’s trick: ⎛ ∂P ⎞ ⎛ ∂S ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂V ⎠T
(3) Let us calculate the change of entropy in a process where both V and T are changed. We have dS = ⎜ We do not have any convenient way of calculating or measuring ⎜ ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂P ⎞ ⎛ ∂S ⎞ ⎟ dV + ⎜ ⎟ dT = ⎜ ⎟ dV + ⎜ ⎟ dT ⎝ ∂V ⎠T ⎝ ∂T ⎠V ⎝ ∂T ⎠V ⎝ ∂T ⎠V (4) ⎛ ∂S ⎞ ⎟ yet. However this ⎝ ∂T ⎠V derivative can be easily expressed in terms of the heat capacity, which can be measured. Consider a process at a constant V. Then we have δq = n CV dT = T dS so that ⎜ Eq. 4 finally becomes: ⎛ ∂S ⎞ ⎟ = CV / T = nCV / T ⎝ ∂T ⎠V C ⎛ ∂P ⎞ dS = ⎜ ⎟ dV + V dT T ⎝ ∂T ⎠V
(5) Exercise: show that Eq. 5 reduces to S = Rn ln V + CV ln T for an ideal gas whose CV is temperature independent. How to express U in terms of measurable quantities. We have dU = TdS – P dV. For dS we obtained: dS = n C v dT/T + ⎜ two equations, we have: dU = C v n dT + [T ⎜ For ideal gases PV = nRT and T ⎜ ⎛ ∂P ⎞ ⎟ dV . Combining these ⎝ ∂T ⎠V ⎛ ∂P ⎞ ⎟
P] dV ⎝ ∂T ⎠V ⎛ ∂P ⎞ ⎟ = P so we get ⎝ ∂T ⎠V dU = C v n dT, an expression we used (without proof) before. Calculating S and H as a function of P,T (Not discussed in the class) Start with G now: dG = V dP – S dT We also have the identity: dG(P,T) = ⎜ Comparing the two eqs. , V = ⎜ Now use: ⎛ ∂G ⎞ ⎛ ∂G ⎞ ⎟ dP + ⎜ ⎟ dT . ⎝ ∂P ⎠T ⎝ ∂T ⎠ P ⎛ ∂G ⎞ ⎛ ∂G ⎞ ⎟ and
S = ⎜ ⎟ ⎝ ∂P ⎠T ⎝ ∂T ⎠ P ⎛ ∂ ⎛ ∂G ⎞ ⎞ ⎛ ∂ ⎛ ∂G ⎞ ⎞ ∂ 2G =⎜ ⎜ = ⎜⎜ ⎟⎟ ⎟⎟ ⎝ ∂T ⎝ ∂P ⎠T ⎠ P ⎝ ∂P ⎝ ∂T ⎠ P ⎠T ∂P∂T
or ⎛ ∂V ⎞ ⎛ ∂S ⎞ ⎜ ⎟ = − ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂P ⎠T
Now we are ready to calculate the change of entropy in a process where both P and T are changed. We have dS = ⎜ We do not have any convenient way of calculating or measuring ⎜ ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎛ ∂S ⎞ ⎟ dP + ⎜ ⎟ dT = − ⎜ ⎟ dP + ⎜ ⎟ dT ⎝ ∂P ⎠T ⎝ ∂T ⎠ P ⎝ ∂T ⎠ P ⎝ ∂T ⎠ P ⎛ ∂S ⎞ ⎟ yet. However this ⎝ ∂T ⎠ P derivative can be easily expressed in terms of the heat capacity cp, which can be measured. Consider a process at a constant P. Then we have δq = n C p dT = T dS so that ⎜ Finally, we obtain, for the change of entropy ⎛ ∂S ⎞ ⎟ = n C p /T ⎝ ∂T ⎠ P Cp ⎛ ∂V ⎞ dS = − ⎜ dP + n dT ⎟ T ⎝ ∂T ⎠ P
or S(2)
S(1) = − Note that this integral can be performed along any path we like and the result will always be the same Exercise: Using the above result, show that ∫
1 2 Cp ⎛ ∂V ⎞ dT ⎜ ⎟ dP + ∫ n T ⎝ ∂T ⎠ P 1 2 S = − Rn ln P + nCP ln T for an ideal gas whose CP is temperature independent. We have dH = d(U+PV) = dU + P dV + V dP =
P dV + T dS + P dV + V dP = TdS + V dP and dS = C p n dT/T
⎜ which gives us: ⎛ ∂V ⎞ ⎟ dP , ⎝ ∂T ⎠ P dH = C p n dT + (V
T ⎜ ⎛ ∂V ⎞ ⎟ )dP ⎝ ∂T ⎠ P For an ideal gas, the second term is identically equal to zero. ...
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 Spring '11
 Makarov

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