HOMEWORK 6 (Due on Oct. 28 at beginning of class)
1. Show that for a refrigerator cycle:
Q1 Q2
+
0
T1
T2
2. (a) A solid of mass m has specic heat c. Show that the change in the entropy of the mass
whe
PHYS 3141 Midterm Exam
Spring 2010
Problem 1: The box on the left has adiabatic walls. A partition separates
a region lled with gas from a region of vacuum. At a certain moment, the
partition disinteg
HOMEWORK 3 (Due on Sep. 21 at beginning of class)
1. A cylinder A, having VA = 30 litres of an ideal monoatomic gas, is connected to a second
cylinder, B, of circular cross-section, as shown in Fig. 3
Changed with the DEMO VERSION of CAD-KAS PDF-Editor (http:/www.cadkas.com).
Changed with the DEMO VERSION of CAD-KAS PDF-Editor (http:/www.cadkas.com).
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1
PHYS 3141 MIDTERM EXAM 1
September 26, 2011
NAME:
1. One mole of an ideal gas undergoes an adiabatic, stirring process. As a result, the
pressure of the gas changes from 3175 Pa to 104 Pa while the
1
HOMEWORK 9 (Due on Dec. 2 at beginning of class)
1. We have seen that the equation of state and the temperature-dependence of the heat
capacity allows the determination of the fundamental relation.
1
HOMEWORK 8 (Due on Nov. 23 at beginning of class)
1. The molar heat capacity of solids exhibits a temperature dependence at suciently low
temperatures, as shown in the gure below, where we plot cV a
HOMEWORK 1 (Due on August 31, at beginning of class)
1. The temperature of a boiling water bath is measured with a constant volume gas thermometer
using the triple point of water as the standard refer
HOMEWORK 5 (Due on Oct. 14 at beginning of class)
1. Consider the free expansion (Q = 0 = W ) of a compressible uid. You can show this process
is irreversible by bringing the uid back to its original
HOMEWORK 4 (Due on Sep. 30 at beginning of class)
Where you require data on the thermodynamic properties of water, refer to Fig. B3 in Appendix
B of the book (attached graph at the end of the assignme
HOMEWORK 2 (Due on Sep. 9 at beginning of class)
1. Let x = x(y, z ). Then:
x
y
dx =
x
z
z
y
y
z
dy +
x
dz
Alternatively, we can also say: y = y (x, z ). Then:
y
x
dy =
dx +
z
dz
(a) By eliminating dy