February 7, 2007
Name:_______ ____
PHY 3513 Midterm Exam I
Solutions
1. The condition for dz = Mdx + Ndy to be an exact differential is ∂M
= ∂N
.
∂y
∂x
It is possible to make an inexact differential in to an exact differential by using an
integrating factor. Consider dz = 2 ydx + 3xdy , show that dz is inexact. Find an
integrating factor μ, such that dw = μdz is an exact differential.
If we follow the instructions then μ(x, y) must satisfy
∂ (2 yμ ) ∂ (3xμ )
∂μ
∂μ
or 2 μ + 2 y
= 3μ + 3x
=
∂y
∂y
∂x
∂x
∂μ
∂μ
− 2y
+ μ = 0 . This can be solved by separation of variables, Assume that μ =
∂x
∂y
f(x)g(y). Then the equation can be turned into
3x
3x ∂f 2 y ∂g
−
+ 1 = 0 Since all terms are constants, and depend on different
f ∂x g ∂y
independent variables, they must each be a constant. Solving the two equations, we get
μ = x m/3y(m+1)/2. There are two special cases, m = 0, which is an x independent solution,
μ = y0.5 and m = -1 which leads to a y independent answer, μ = x-1/3
2. A Carnot engine is operated between two heat reservoirs at temperatures of 300K
and 200K.
a. If the engine receives 1000 kilocalories from the reservoir at 300K, how much
heat does it reject to the reservoir at 200K?
W = QH/3 = 333.33 kcal and Qc = 1000 – 333.33 = 666.67 kcal
b. When the engine is operated as a refrigerator, it takes 1000 kilocalories from the
reservoir at 200K. How much heat does it deliver to the reservoir at 300K?
W = Qc/2 = 500 kcal = QH-Qc, therefore QH = 1500 kcal.
c. What is the efficiency of the engine in (a) and the coefficient of performance in
(b).
η =1-200/300 = 1/3 = W/QH
C = Qc/W = Tc/(TH-Tc) = 2
3. The temperature of an ideal gas at an initial pressure P1 and volume V1 is
increased at constant volume until the pressure is doubled. The gas is then
expanded isothermally until the pressure drops to its original value where it is
compressed at constant pressure until the volume returns to its initial value.
a. Sketch these processes in the P-V plane and in the P-T plane.
P
P
V
T
b. Compute the work in each step and the net work done in a cycle if n = 2 kilomoles
and P1 = 2 atm and V1 = 4m3.
No work is done in the isochoric heating. In the isothermal expansion, the work done is
V
W = nRT ln( 3 ) = 2×8314×97.47×ln 2 = 1.123×106 J
V1
The isobaric compression consists of work = PΔV = 2×1.013×105 (4-8) = -8.10×105 J.
Net work = 11.23 – 8.1 = 3.13×105 J
4. A volume of 10 m3 contains 8kg of oxygen at a temperature of 300K. Find the
work necessary to decrease the volume to 5 m3.
The molecular weight of O2 is 32 kg/kmole. We have 0.25 kmoles.
P = 0.25×8314×300/10 = 0.624×105 Pa
a. At constant pressure
Work = P(5-10) = -3.12 × 105 J
b. At constant temperature
Work = nRT ln(V2/V1) = -4.32×105 J
c. What is the temperature at the end of the process in (a)
Volume decreases by a factor of 2 at constant pressure. For an ideal gas Tf = 150K
d. What is the pressure at the end of the process in (b)
Final pressure at constant temperature Pf = 1.25×105 Pa
e. Sketch both processes in the P-V plane.
P
isothermal
isobar
V