Section 1

Thermodynamic state In classical mechanics the state of a body or a particle in motion is fully determined by its position in space r = zyxr ),,( and its velocity r = wvuv ),,( . For an consisting of a large number of molecules (1023), one would need a very large number of parameters (6x1023) to define its state , something that at first sight seems impossible.

Concept of statistical equilibrium. Although the position and velocities of the molecules of a gas change with time, the time mean of quantities like the kinetic of the particles remain unchanged when the system has reached a statistical equilibrium. managed to connect such quantities with macroscopic quantities like and that remain unchanged as well in statistical equilibrium. So we are able to determine the state of a by the knowledge of a sufficient (but small) number of macroscopic variables, known as the state variables. In our case these are pressure p, temperature T, V and number of molecules n (or mass).

Thermodynamic diagrams

A single point represents the state in these diagrams. But one could think that a point gives information about only two quantities (pressure and volume for the p-V diagram). How are the rest of them defined?

Equation of state The mathematical expression relating the state variables is called the . In our case it is the perfect gas law = *TnRpV . Of course there are more realistic models for gases that have a different equation of state (for example the Van der

⎛ an 2 ⎞ Waals equation of state ⎜ p + ⎟ )( =− *TnRnbV ). ⎜ 2 ⎟ ⎝ V ⎠ So in a p-V diagram, if you know p, V , then you can find the temperature T of the state by using the perfect gas law.

Quasi-static processes In order to go from one state to the other, the change is carried out in small steps and very slowly, so that the system is effectively in a state of equilibrium throughout (quasi-static process). The advantage of such a change is that it is reversible (doing the change in reverse order and very slowly again leads us to our initial state)

Fist law of = + dudwdq

Heat dq : flow of energy between two bodies/gasses of different temperature

v dw : From classical mechanics : r ===⋅= pdVpAdxFdxsdFdw Important note : In calculus, if you have a differential dG ,you can integrate

B between the values A, B to get : −= GGdG Similarly, you could think that you ∫ B A A

B can integrate −= wwdw . But this is not the case here, since there is no physical ∫ AB A meaning assigning a certain amount of work to a specific state (saying for example that “state A has 10 J of work”). Mathematically, the problem is that dw is not an exact differential (the same holds for dq ). Usually in thermodynamic books the notation “ d ′ “ is used to denote that fact. Since Wallace & Hobbs do not follow this notation, we will also use the simple notation, but have in mind that by “d” we denote a small quantity (for work and heat) and not an exact differential.

Another version of the first law is that : ∫ dwdq =− 0 . In other words dq is not an exact differential, dw is not an exact differential, but their difference ( du ) is. In fact it can be shown, using statistical mechanics that is a function of temperature only.

Various processes • Isothermal (constant temperature)

VB VB RT VB 1 W pdV === RTdV = VVRTdV )/ln( >− BA V ∫∫∫ V AB VA VA VA

• Isochoric (constant volume)

=+= dupdVdwdq

dq du cV =⇒== V dTcdu dT V dT V

• Isobaric (constant pressure)

−+=+= = + pVudVdppVddwpdVdwdq )()( 142 43 h

dq dh c p == dT p dT p Note that the h, plays the role of internal energy here. dq du dV c p +== p dT p dT p dT p dV R RdTVdppdVRTpV =⇒=+⇒= dT p =cdu dT du dV V c p += p ⎯⎯→c p cV +=⎯⎯⎯ R dT p dT p

Why is c p larger than cV ? We can clearly see, just by looking at the processes, that in an , all the heat that we pump into the system, goes into changing the internal energy and thus the temperature of the gas. On the other hand in the , some of the heat goes into work (moving the piston) and one needs a larger amount of heat to achieve the same temperature difference.

• Adiabatic (dq=0)

−=⇒+= dudwdudwdq

Problem : There is a gas in a chamber that has volume V A . It expands isothermally to a state with volume VB and the work done during the expansion is W . It is then heated under constant volume and then compressed isobarically to reach the initial state (as shown in the p-V diagram below). Calculate all the states, as well as the work and heat during all of the processes.

Calculation of the states :

We’ve already calculated the work done during an :

W >− BA = ABA )/ln( TVVRTW A =⇒ VVR AB )/ln(

Now from the perfect gas law for state A :

RTA TW A AA A pRTVp A ==⇒= VA VVV ABA )/ln( From the perfect gas law for state B :

RTA TW A BB A pRTVp B ==⇒= VB VVV ABB )/ln(

The volume in state C is the same as the volume in state B ( =VV BC ) and the pressure is the same as the pressure in state A ( = pp AC ). So from the perfect gas law for state C :

Vp BA VTW BA CC C BA TRTVpRTVp CC ==⇒=⇒= R VVRV ABA )/ln(

Calculation for the work done

For A->B : It is given by the problem : >− BA =WW

For B->C : pdVdw 0 ⇒== W >− CB = 0

VA VA For C->A : WpdVdw −===⇒= VVpdVppdV )( >− AC ∫∫ A BAA VB VB

Calculation for the heat :

For A->B : =+= V + = ⇒ >− =WQdwdwdTcdwdudq >− BABA

TC For B->C : −==⇒=+=+= TTcdTcQdTcpdVdTcdwdudq )( V V >− CB ∫ V BCV TB

For C->A :

TA VA −+−=+=⇒+=+= VVpTTcpdVdTcQpdVdTcdwdudq )()( V >− AC ∫∫V CAV CAA TC VC