Math 1321
Week 7 Lab Worksheet
Due Thursday 02/28
1. Isothermals The gas law for a fixed mass m of an ideal gas at absolute temperature T ,
pressure P , and volume V is P V = mRT , where R is the gas constant.
(a) The level curves of T are called isothermals because at all points on such a curve
the temperature is the same. Sketch some of the isothermals given by the ideal gas
law for a gas constant R = 0.25 and a mass m = 1.25. Hint: The temperature can
be viewed as a function of pressure and volume.
(b) Show that
∂P ∂V ∂T
= −1
∂V ∂T ∂P
(c) Show that, for an ideal gas,
T
∂P ∂V
= mR
∂T ∂T
Solution:
(a) The plot below is a contour plot of the Temperature T for varying values of
volume V and pressure P . The dark lines represent contours which are
constant in T, thus they represent the isothermals of the ideal gas law for the
chosen parameters R = 0.25 and m = 1.25.
Isothermals for Ideal Gas Law
10
8
6
4
Pressure
2
0
−2
−4
−6
−8
−10
0
20
40
60
80
100
Volume
mRT
∂P
so ∂V
= −mRT
; V = mRT
V
V2
P
∂P ∂V ∂T
−mRT mR V
−mRT
=
=
2
∂V ∂T ∂P
V
P mR
PV
(b) P =
PV
∂T
V
so ∂V
= mR
; T = mR
, so ∂P
= mR
. Thus
∂T
P
= −1, since P V = mRT . Notice this is
exactly negative of the quantity we would arrive at if we treated all of the
terms ∂P , ∂V , and ∂T as fractions and canceled “like” terms. Thus we see
derivatives are “almost” like fractions but still something a little different.
, so ∂P
= mR
. Also,
(c) By part (a), P V = mRT =⇒ P = mRT
V
∂T
V
∂V
mR
PV
mRT
, we have
P V = mRT =⇒ V = P and ∂T = P . Since T = mR
∂P ∂V
P V mR mR
T ∂T ∂T = mR V P = mR.
2. Frost Penetration In a study of frost penetration it was found that the temperature
T at time t (measured in days) at a depth x (measured in feet) can be modeled by the
function
T (x, t) = T0 + T1 e−λx sin(ωt − λx)
where ω = 2π/365 and λ is a positive constant.
(a) Find
(b) Find
∂T
.
∂x
∂T
.
∂t
What is its physical significance?
What is its physical significance?
(c) Show that T satisfies the heat equation Tt = kTxx for a certain constant k.
(d) If λ = 0.2, T0 = 0, and T1 = 10, graph T (x, t).
(e) What is the physical significance of the term −λx in the expression sin(ωt − λx)?
Solution:
(a)
∂T
= T1 e−λx [cos(ωt − λx)(−λ)] + T1 (−λe−λx sin(ωt − λx)
∂x
= −λT1 e−λx [sin(ωt − λx) + cos(ωt − λx)]
This quantity represents the rate of change of temperature with respect to the
depth below the surface at a given time t.
(b)
∂T
= T1 e−λx [cos(ωt − λx)(ω)] = ωT1 e−λx cos(ωt − λx)
∂t
This quantity represents the rate of change of temperature with respect to time
at a fixed depth x.
(c)
Txx
∂ ∂T
=
∂x ∂x
= −λT1 e−λx [cos(ωt − λx)(−λ) − sin(ωt − λx)(−λ)] + · · ·
+e−λx (−λ) [sin(ωt − λx) + cos(ωt − λx)]
= 2λ2 T1 e−λx cos(ωt − λx)
From part(b) we saw, Tt = ωT1 e−λx cos(ωt − λx) =
the function T satisfies the heat equation.
ω
T .
2λ2 xx
So with k =
ω
,
2λ2
(d)
(e) The term −λx is a phase shift: it represents the fact that since heat diffuses
slowly through soil, it takes time for changes in the surface temperature to affect
the temperature at deeper points. As x increases, the phase shift also increases.
For example, when λ = 0.2, the highest temperature at the surface is reached
when t ≈ 91, whereas at a depth of 5 feet, the peak temperature is attained at
t ≈ 149, and at a depth of 10 feet, at t ≈ 207.