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 ideal gas 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 energy of the particles remain unchanged when the system has reached a statistical equilibrium. Statistical mechanics managed to connect such quantities with macroscopic quantities like pressure and temperature that remain unchanged as well in statistical equilibrium. So we are able to determine the state of a thermodynamic system 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, volume 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 equation of state. 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 thermodynamics = + dudwdq
Heat dq : flow of energy between two bodies/gasses of different temperature
v Work 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 heat 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 internal energy 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 enthalpy 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 isochoric process, 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 isobaric process, 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 isothermal process :
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