# Change in Temperature

by tan90
Tags: temperature change
 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 $$\Delta$$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 quasi-static, what is the resulting change of temperature $$\Delta$$T of the sample ? If$$\Delta$$p is fairly small, derive an expression for $$\Delta$$T in terms of $$\Delta$$p, the absolute temperature T of the sample, its specific heat at constant pressure cp (in ergs g-1 deg -1), its density rho (in g/cm3), and its volume coefficient of thermal expansion $$\alpha$$(in deg -1) 2. Relevant equations dv = $$\alpha$$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.
 P: 22 Since the questions says "If $$\Delta$$ p is fairly small, " is it ok to consider p as a constant? Other variables $$\Delta$$T and $$\Delta$$V do change. I am rather confused. Thank you
 Sci Advisor HW Helper PF Gold P: 2,532 Change in Temperature Unfortunately, we can't assume $\Delta P$ is zero because we have to divide $\Delta T$ by $\Delta P$ (that is, we want to find $(\partial T/\partial P)_X$, 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?
 P: 22 P,V,T all change. It means that the only entity left in Maxwell's relations is S. So are you suggesting$$\Delta$$S is a constant?
 Sci Advisor HW Helper PF Gold P: 2,532 Yes, S is constant, or $\Delta S$ is zero.
 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., $(\partial Y/\partial Z)_X$) for each effect, and manipulate them via Maxwell relations. It turns out that even the sign of the answer can vary between materials--some materials heat up under pressure, some cool down--so I wouldn't try to intuit the answer through visualization. You have to go through the (admittedly abstract) calculations.
 P: 2,157 To find the partial derivative at constant S, you can write: $$dS =\left(\frac{\partial S}{\partial P}\right)_{T}dP + \left(\frac{\partial S}{\partial T}\right)_{P}dT$$ 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: $$\left(\frac{\partial S}{\partial P}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{P}$$ So, the partial derivative of S can be expressed in terms of the thermal expansion coefficient.