lecture 13 - An example of using our general...

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Unformatted text preview: An example of using our general relationships for U(V,T), S(V,T) etc. : Entropy of a non ­ideal gas. Consider irreversible expansion of one mole of a gas, in which it occupies the volume 2V where V is its initial volume; Assume no heat transfer between the gas and its surroundings. We have previously found that for an ideal gas the change of the gas entropy is R ln 2. Now we’d like to see how our result changes if the gas is non ­ideal and obeys the equation of state ( P + a / V 2 )V = RT Since entropy is a function of state, we can calculate it if we know the final temperature of the gas after expansion. To do that, we can use the fact that the internal energy of the gas is conserved (no heat exchange, no work done on the gas), so we use dU = C v n dT + [T ⎜ ⎛ ∂P ⎞  ­P] dV ⎟ ⎝ ∂T ⎠V P = RT / V − a / V 2 and R ⎛ ∂P ⎞ ⎜ ⎟ = ⎝ ∂T ⎠V V dV ⎛ RT RT a ⎞ dU = CV dT + ⎜ − + 2 ⎟ dV = CV dT + a 2 V V⎠ V ⎝V To simplify calculations, let’s assume that CV is constant. But can we? Generally speaking, CV is a function of both V and T. Here we view CV(V,T) as something that we know from measurements. However whatever this dependence is, it must be consistent with the equation of state. We know that both CV(V,T) and the equation of state P=P(V,T) can be derived from the same function A(P,V). How do we know that CV(V,T) and P(V,T) can be independent of one another? They are not. To see this, consider the expression for dU: dU = CV n dT + [T ⎜ ⎛ ∂P ⎞  ­P] dV ⎟ ⎝ ∂T ⎠V Since this is an exact differential, we can write the Maxwell relationship: ∂ ⎛ ∂CV ⎞ ⎜ ⎟= ⎝ ∂V ⎠T ∂T ⎡ ⎛ ∂P ⎞ ⎤ ⎛ ∂2P ⎞ T ⎜ ⎟ − P ⎥ = T ⎜ 2 ⎟ ⎢ ∂T ⎝ ∂T ⎠V ⎣ ⎝ ⎠V ⎦ The rhs of this expression can be calculated from the equation of state. This expression therefore tells us how CV depends on the volume if the equation of state is known. The assumption that CV is a function of T only cannot be generally made! ⎛ ∂2P ⎞ Luckily, for the particular problem at hand, you can check that T ⎜ = 0 and so the assumption 2⎟ ⎝ ∂T ⎠V that CV does not explicitly depend on V is correct. This is not generally true! Thermodynamics does not impose any constraints on the temperature dependence of CV (except that it should be positive for a uniform material – I will leave this statement without a proof but it sounds reasonable: materials get hotter when given heat). Thus our assumption that CV is constant does not contradict anything we know. Back to our problem. For a constant CV, we have ⎛1 1⎞ ΔU = CV (T (2) − T (1)) − a ⎜ − ⎟ = 0 ⎝ V (2) V (1) ⎠ or, using V(2)=2V(1)=2V, T(2)=T(1) ­ a 2CVV For positive a, the gas became colder after expansion. This makes sense: positive a means attractive interaction among particles. Therefore gas expansion results in an increase of the interaction energy, which must be compensated for by a decrease in the kinetic energy of the molecules. To calculate the change of entropy, use dS = CV dT ⎛ ∂P ⎞ dT R + ⎜ ⎟ dV = CV + dV , T ⎝ ∂T ⎠V TV which gives, after integration: S(2) ­S(1)= CV ln T − a /(2CVV ) + R ln 2 T where T is the initial temperature. Note that the first term is negative. That means if a/V is large enough S(2) ­S(1) would be negative, which is prohibited by the 2nd law! This tells us that this equation of state may be inconsistent with the 2nd law for certain values of a and/or V and T. The relationship between Cv and Cp (NOTE: this has not been discussed in the class) We have derived two different expressions for the entropy: dS = − ⎜ Let us view the volume as a function of P and T: V = V(P,T). Then dV = ⎜ Substituting this into dS: Cp Cv ⎛ ∂P ⎞ ⎛ ∂V ⎞ dT = ⎜ ⎟ dP + n ⎟ dV + n dT T T ⎝ ∂T ⎠ P ⎝ ∂T ⎠V ⎛ ∂V ⎞ ⎛ ∂V ⎞ ⎟ dT + ⎜ ⎟ dP ⎝ ∂T ⎠ P ⎝ ∂P ⎠T Cp ⎛ ∂V ⎞ ⎛ ∂P ⎞ −⎜ dT = ⎜ ⎟ dP + n ⎟ T ⎝ ∂T ⎠ P ⎝ ∂T ⎠V ⎡⎛ ∂V ⎞ ⎤ Cv ⎛ ∂V ⎞ ⎢⎜ ∂T ⎟ dT + ⎜ ∂P ⎟ dP ⎥ + n T dT ⎠P ⎝ ⎠T ⎦ ⎣⎝ By collecting the terms proportional to dT, we get n( C p  ­ C v )/T= ⎜ ⎛ ∂P ⎞ ⎛ ∂V ⎞ ⎟⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠ P or n( C p  ­ C v ) = T ⎜ Therefore we can express cp ­cv in terms of the equation of state. For the ideal gas, we obtain cp ­cv = nR. It is often a reasonable approximation to assume that the volume of a solid or a liquid does not change. This results in n( C p  ­ C v ) ~ 0. Physically, this is a consequence of the fact that the work associated with expansion of a solid or a liquid is small (unlike the case of a gas). ⎛ ∂P ⎞ ⎛ ∂V ⎞ ⎟⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠ P ...
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