A gas of interacting atoms has an equation of state and heat capacity at constant volume given by the expressions p(T,V )= aT^1/2 + bT^3 + c V ^- 2, Cv(T, V)= dT^1/2V + eT^2V + fT^1/2, where a through f are constants which are independent of T and V . (a) Find the differential of the internal

Question

A gas of interacting atoms has an equation of state and heat capacity at constant volume given by the expressions

[latex]p(T, V)=a T^{1 / 2}+b T^{3}+c V^{-2}[/latex]

[latex]C_{v}(T, V)=d T^{1 / 2} V+e T^{2} V+f T^{1 / 2}[/latex],

where a through f are constants which are independent of T and V.

(a) Find the differential of the internal energy dU(T,V )in terms of dT and dV.

(b) Find the relationships among a through f due to the fact that U(T,V) is a state variable.

(c) Find U(T,V) as a function of T and V.

(d) Use kinetic arguments to derive a simple relation between p and U for an ideal monatomic gas (a gas with no interactions between the atoms, but whose velocity distribution is arbitrary). If the gas discussed in the previous parts were to be made ideal, what would be the restrictions on the constants a through f?

Accepted Answer

[latex] ext { (a) We have } d U(T, V)=C_{v} d T+\left(\frac{\partial U}{\partial V} ight)_{T} d V ext {, }[/latex]

[latex]p=-\left(\frac{\partial F}{\partial V} ight)_{T}=-\left(\frac{\partial U}{\partial V} ight)_{T}+T\left(\frac{\partial p}{\partial T} ight)_{V}[/latex].

[latex] ext { Hence } d U=\left(d T^{1 / 2} V+e T^{2} V+f T^{1 / 2} ight) d T-\left(\frac{a}{2} T^{1 / 2}-2 b T^{3}+c V^{-2} ight) d V[/latex].

(b) Since U(T,V) is a state variable dU(T,V) is a total differential, which requires

[latex]\frac{\partial}{\partial V}\left(\frac{\partial U}{\partial T} ight)_{V}=\frac{\partial}{\partial T}\left(\frac{\partial U}{\partial V} ight)_{T}[/latex].

that is,

[latex]d T^{1 / 2}+e T^{2}=-\left(\frac{1}{4} a T^{-1 / 2}-6 b T^{2} ight)[/latex].

Hence a = 0, d = 0, e = 6b.

(c) Using the result in (b) we can write

[latex]d U(T, V)=d\left(2 b T^{3} V ight)+f T^{1 / 2} d T-c V^{-2} d V[/latex].

Hence

[latex]U(T, V)=2 b T^{3} V+2 f T^{3 / 2} / 3+c V^{-1}+ ext { const }[/latex].

(d) Imagine that an ideal reflecting plane surface is placed in the gas.
The pressure exterted on it by atoms of velocity v is

[latex]p_{v}=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2 \pi} \frac{N}{V} \frac{\sin heta d heta d \varphi}{4 \pi} v \cos heta \cdot 2 m v \cos heta=\frac{1}{3} \frac{N}{V} m v^{2}[/latex].

The mean internal energy density of an ideal gas is just its mean kinetic energy density, i.e.,

[latex]u=\frac{1}{2} \overline{m v^{2}} \cdot \frac{N}{V}[/latex].

[latex] ext { The average pressure is } p=\bar{p}_{v}=2 u / 3, ext { giving } p V=\frac{2}{3} U[/latex]. For the gas discussed above to be made ideal, we require the last equation to be satisfied:

[latex]\left(b T^{3}+\frac{c}{V^{2}} ight) V=\frac{2}{3}\left(2 b T^{3} V+\frac{2}{3} f T^{3 / 2}+\frac{c}{V}+ ext { const. } ight)[/latex].

[latex] ext { i.e., } \quad 3 b T^{3} V+4 f T^{\frac{3}{2}}-\frac{3 c}{V}= ext { const. }[/latex]

It follows that b and f cannot be zero at the same time. The expression for p means that b and c cannot be zero at the same time.

Question 2.166: A gas of interacting atoms has an equation of state and heat...

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