Register to reply 
Change in Temperatureby tan90
Tags: temperature change 
Share this thread: 
#1
Apr409, 07:47 PM

P: 22

1. The problem statement, all variables and given/known data
This is an experiment done on solid at high pressure. If the pressure is increased by an amount [tex]\Delta[/tex]p, this being done under condition where the sample is thermally insulated and at a sufficiently slow rate that the process can be regarded as quasistatic, what is the resulting change of temperature [tex]\Delta[/tex]T of the sample ? If[tex]\Delta[/tex]p is fairly small, derive an expression for [tex]\Delta[/tex]T in terms of [tex]\Delta[/tex]p, the absolute temperature T of the sample, its specific heat at constant pressure c_{p} (in ergs g^{1} deg ^{1}), its density rho (in g/cm^{3}), and its volume coefficient of thermal expansion [tex]\alpha[/tex](in deg ^{1}) 2. Relevant equations dv = [tex]\alpha[/tex]V dt Maxwell's equations 3. The attempt at a solution I started from an expression for coefficient of thermal expansion and tried to relate it with maxwell's equation. The constant terms in maxwell's equations are very confusing. 


#2
Apr509, 06:58 AM

Sci Advisor
HW Helper
PF Gold
P: 2,532

Well, first of all, what variable are you going to hold constant?
(P.S. They're usually referred to as Maxwell relations to distinguish them from all that Bfield and Efield stuff.) 


#3
Apr509, 10:58 AM

P: 22

Since the questions says "If [tex]\Delta[/tex] p is fairly small, " is it ok to consider p as a constant? Other variables [tex]\Delta[/tex]T and [tex]\Delta[/tex]V do change. I am rather confused.
Thank you 


#4
Apr509, 11:10 AM

Sci Advisor
HW Helper
PF Gold
P: 2,532

Change in Temperature
Unfortunately, we can't assume [itex]\Delta P[/itex] is zero because we have to divide [itex]\Delta T[/itex] by [itex]\Delta P[/itex] (that is, we want to find [itex](\partial T/\partial P)_X[/itex], where X is the variable we hold constant). Since we've eliminated pressure, temperature, and volume, what's left? We're blocking heat transfer; besides energy, what flows during heat transfer?



#5
Apr509, 11:16 AM

P: 22

P,V,T all change. It means that the only entity left in Maxwell's relations is S. So are you suggesting[tex]\Delta[/tex]S is a constant?



#6
Apr509, 11:18 AM

Sci Advisor
HW Helper
PF Gold
P: 2,532

Yes, S is constant, or [itex]\Delta S[/itex] is zero.



#7
Apr509, 11:26 AM

P: 22

Thank you ! Can you please explain a bit further about what is actually going on in the problem? Its rather hard to picture the problem for me.



#8
Apr509, 11:35 AM

Sci Advisor
HW Helper
PF Gold
P: 2,532

The experiment involves pressurizing a solid. Will the solid heat up or cool down, and by how much for a given increase in pressure? The effect is somewhat related to thermal expansion, in which we observe how the volume of a solid changes for a given temperature increase at constant pressure. You should be able to write the differential term (e.g., [itex](\partial Y/\partial Z)_X[/itex]) for each effect, and manipulate them via Maxwell relations.
It turns out that even the sign of the answer can vary between materialssome materials heat up under pressure, some cool downso I wouldn't try to intuit the answer through visualization. You have to go through the (admittedly abstract) calculations. 


#9
Apr509, 12:00 PM

P: 2,157

To find the partial derivative at constant S, you can write:
[tex]dS =\left(\frac{\partial S}{\partial P}\right)_{T}dP + \left(\frac{\partial S}{\partial T}\right)_{P}dT[/tex] If you equate dS to zero and solve for the ratio dT/dP you got your partial derivative at constant S. Now, you can express the partial derivate of S w.r.t. T at constant P in terms of the heat capacity at constant pressure. But what to make of the partial derivative of S w.r.t. P at constant T? Well, we have the findamental thermodynamic relation: dE = T dS  P dV We can write: T dS = d(TS)  S dT and P dV = d(PV)  V dP This means that: dE = d[TS  PV] SdT + V dP > d[E + PV  TS] = S dT + V dP Define G = E + PV  TS. The above relatiomn implies that minus S is the partial derivative of G w.r.t. T and constant P and that V is the partial derivative of G w.r.t. P at constant T. Then the fact that te second derivative of G w.r.t. P and T does not depend on the order of differentiation implies that: [tex]\left(\frac{\partial S}{\partial P}\right)_{T}=\left(\frac{\partial V}{\partial T}\right)_{P} [/tex] So, the partial derivative of S can be expressed in terms of the thermal expansion coefficient. 


#10
Apr509, 04:25 PM

P: 22

Brilliant !!!



Register to reply 
Related Discussions  
Kelvin Temperature  Celsius Temperature Change  General Physics  12  
Chemistry: Finding a temperature change using a temperature/gram ratio  Biology, Chemistry & Other Homework  3  
Temperature Change  Introductory Physics Homework  9  
Temperature change  Introductory Physics Homework  3 