The given equation of state is P = k T v b e a /

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The given equation of state is

P=kTvbea/kTv(1)(1)P=kTvbea/kTv

It follows that

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At the critical point, both these derivatives vanish - with the result that

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whence vc=2bvc=2b and kTc=a/4bkTc=a/4b. Equation (1) then gives: Extra \left or missing \rightExtra \left or missing \right and hence kTc/Pcvc=e2/23.695kTc/Pcvc=e2/23.695.

(a) For large v, the given equation of state may be approximated as

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Comparing this with eqns. (10.3.7-10), we see that the coefficient B2B2 in the present case is formally the same as the one for the van der Waals gas, viz. b(a/kT)b(a/kT).

(b) We note that the derivative (P/v)T(P/v)T for the Dietrici gas can be written as

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Clearly, if T>TcT>Tc, then (P/v)T(P/v)T is definitely negative; the same is true at T=TcT=Tc - except for the special case v=a/2kTc=2bv=a/2kTc=2b when (P/v)T(P/v)T is zero. In any case, for all TTc,PTTc,P is a monotonically decreasing function of vv with the result that, for any given TT and PP, we have a unique vv.

(c) For T<Tc,PT<Tc,P is a non-monotonic function of vv generally decreasing with vv but increasing between the values

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For any given TT, we now have (for a certain range of PP ) three possible values of vv such that

v1>vmax>v2>vmin>v3;v1>vmax>v2>vmin>v3;

see Figs. 12.2 and 12.3. We further note that

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Clearly, vmin<vc<vmaxvmin<vc<vmax and, hence, v3<vc<v1v3<vc<v1.

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