Modified equation of state of an ideal gas

[gdumont]

I have this problem, it doesn't seem very complicated but I can't figure out how to do it.

A gas obeys to the equation of state
P(V-b) = Nk\zeta
where k is the Boltzman's constant. The internal energy of the gas is a function of
\zeta alone. Show that \zeta=T using a Carnot cycle.

Any help greatly appreciated

Guillaume

 

[gdumont]

Any ideas?

 

[Physics Monkey]

Here is a hint to get you started.

The thermodynamic temperature T is defined in terms of the efficiency of a Carnot engine as
<br /> \eta = 1-\frac{T_C}{T_H},<br />
so what you want to do is calculate the efficiency of your Carnot cycle using your equation of state, the first law, and whatever other information may be available to you. Upon comparing the two, you will find that \zeta, which is an empirical temperature for the gas as defined by the zeroth law, is in fact equal (with a suitable choice of scale) to the thermodynamic temperature i.e. you want to find that \zeta is proportional to T.

 

[gdumont]

\zeta [...] is an empirical temperature for the gas as defined by the zeroth law.

I don't really get why is that... any chance you can explain it to me

Thanks

 

[Physics Monkey]

I can certainly try, the zeroth law is usually taken as saying that it is possible to find a quantity, called empirical temperature, which is the same for two systems in equilibrium. This is probably fairly obvious, and it can be more or less rigorously shown using the zeroth law. The experimental observation that PV is constant in the limit of a dilute gas allows us to identify PV as being proportional to an empirical temperature for dilute or ideal gases. By defining the temperature of the triple point of water to be 273.16 K exactly, one defines the Kelvin scale in terms of an ideal gas temperature. The coeffecient then turns out to be N k so that we have the ideal gas law PV = N k \zeta. This is one way to define a temperature scale, another way is in terms of the Carnot engine where we define the thermodynamic temperature T in terms of the efficiency as \eta = 1 - \frac{T_C}{T_H}. One very basic question one can ask is, "do these temperature scales coincide?" This is the subject of your problem.

While that is perhaps all very interesting, the point of the problem is just to show that \zeta can be taken equal to T, which is defined in terms of the Carnot efficiency. To do this, you need to calculate the efficiency of your gas based Carnot engine and then compare it to the definition of T.