BME 250 Thermodynamics Homework #1 Solution
1.)
F = x3y2
dF = 3 x 2 y 2 dx + 2 x 3 ydy
Criteria of Exactness :
(a)
2F 2F
=
xy yx
6x 2y = 6x 2y
F = x cos xy
dF (cos xy xy sin xy ) dx x 2 sin xydy
=
Criteria of Exactness :
(b)
2F 2F
=
xy yx
2 x sin xy x
BME 250 Homework 2 Solutions
1.) Problem 1s solution is a separate document!
2.) The PV diagram is
In the first step the gas is compressed isentropically to a volume of V=0.004958m3. The
gas temperature at the end of this step is
" 0.02479 m 3 % 2 / 5
" V
BME 250 Homework 3 Solutions
1.)
(a.) False.
The second law of thermodynamics states that the entropy change for a closed, adiabatic
system is zero for a reversible process and greater than zero for an irrersible process. A
spontaneous process is an irrev
BME 250 Homework 4 Solutions
1) The Enthalpy is expressed in its canonical variables as:
dH = TdS + VdP + dN
# H &
So, the derivative % ( is written as
$ T 'V ,N
# H &
# S &
# P &
% ( = T% ( + V % (
$ T 'V ,N
$ T 'V ,N
$ T 'V ,N
# S &
Cv = T% (
$ T 'V ,N
BME 250
Thermodynamics
Midterm Exam
05/14/2009
Question 1: (20 points)
a) Derive the following expression starting from the differential of U in terms of its
canonical variables:
U
P
= T
P
V T , N
T V , N
b) Evaluate (U / V ) T , N for: i) an ideal