March 10, 2006
Name:________________________________
PHY3513/Kumar
Midterm Exam II
You are allowed a formula sheet, a calculator, pencil and papers. The formula sheet
must not contain anything other than the formulae.
The exam has to be finished within the allotted time of 50 minutes.
1. The equation of sublimation and the vaporization curves of a particular
material are given by
ln P = 0.04 – 6/T (sublimation)
ln P = 0.03 – 4/T (vaporization)
where P is in atmospheres.
(a) Find the temperature of the triple point.
The triple point is the intersection of the two curves. The temperature of the triple
point is given by .01 = 2/T or T = 200K.
(b) Show that the specific latent heats of vaporization and sublimation are 4R
and 6R respectively.
Recall that dP/dT = Δs/Δv. If we approximate Δv ≅ vv = RT/P and Δs = l/T, we can
read of the latent heats for vaporization and sublimation as 4R and 6R respectively
(c) Find the latent heat of fusion.
lfusion + lvaporization = lsublimation Hence lfusion = 2R.
2. Show that the difference between the isothermal and adiabatic
compressibilities is
κ - κs = Tvβ2/cp
In Chapter 7 (Eq. 7.21), there is a proof of the statement that κs = κ/γ where γ = cp/cv.
It follows therefore that κ - κs = κ(1-1/γ) = κ [(Tvβ2/κ]/cp. This last result also makes
use of the fact that cp-cv = Tvβ2/κ. The result then follows.
What is the value of this difference for a monatomic ideal gas?
For a monatomic ideal gas, γ = 5/3 and κ = 1/P, thus κ - κs = 0.4/P
3. Consider the Joule cycle heat engine consisting of two isobars and 2 adiabats.
Assuming that the working substance is an ideal gas with the constant
specific heat capacities cp and cv, show that the efficiency of this engine can
be written as
η = 1 – (P1/P2)(1-1/γ)
Recall that the efficiency of a heat
engine is η = W/QH = 1 – Qc/QH
2
2
3
P
Qc = cp(T4 – T1) and QH = cp(T3 – T2)
4
1
and η = 1 –[P1 (v4 –v1)/P2 (v3 – v2)]
But Pvγ = constant , leads to the final
answer.
1
V