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Mid Term Exam – General Physics II
Apr 21, 2000
1.
One mole of an ideal gas of diatomic molecules, whose molecules are rotating but
not oscillating, undergoes a thermodynamic cycle that consists of an isothermal,
an adiabatic and a constant volume processes as shown in the following P-V
diagram. In the case, p1=1 atm=1×105 Pa, T1 =600K, (P,T,V at state 1) and
R=8.314 J/mol K:
A. What is the temperature of state 3?
B. Calculate the heat emitted by the gas during the constant volume process 23.
(If you don’t have a calculator, just write down the expression.)
C. Calculate the work done by the gas and the heat that the gas absorbs during
the isothermal process 12.
D. Calculate the entropy difference between the state 1 and 2. What is the
entropy difference between the state 2 and 3?
Volume
2. The ideal gas law PV=nRT assumes that the molecules do not interact at all. If we
consider the effects of intermolecular forces, the equation of state becomes the
van der Waals equation:

an 2 

P  V2 
V  nb   n R T


Here a, b are two small constants. Calculate the work done by the gas when it
undergoes an isothermal expansion process from volume Vi to Vf . (Hint: Just
repeat what you did in 1B, using the van der Waals equation this time instead of
the ideal gas law.)
3.
Put an electric dipole, consisting of a point charge q and a point charge –q
separated by a small distance d, in a uniform electric field E.
A. Calculate the total electric force by the field on the electric dipole.
B. Calculate the torque on the electric dipole around the axis which is
perpendicular to E and goes through the center of the electric dipole.
C. Find the frequency for a small oscillation of this dipole around its equilibrium
position. Assume that the rotational inertia of the dipole is I.
4.
Place a point charge q at the point with coordinates (0,0, a) above an infinitely
large plane (平面) conductor sheet located at the x-y plane, that is, z=0, as shown
in the diagram below.
A. Draw the electric field lines and equipotential lines on the plane x=0.
B. Now remove the conductor sheet and place a negative charge –q at (0,0, -a).
The electric field in the upper half space z>0 is actually the same as in the
case when the conductor sheet exists at z=0. This is called the method of
image charge. Show that the directions of the electric field on the line x=0,
z=0 is indeed the same in the two cases.
C. Use the method of image charge described in B to calculate the electric field
at the origin x=y=z=0.
D. Could you use the result of C and the Gauss’s law to calculate the surface
charge density of the conductor surface at the origin? (Hint: Draw a tiny
cylindrical Gauss surface around the origin. If this cylinder is tiny enough, the
electric field is a constant a bit above the surface.) What is the total charge on
the whole conductor sheet?
5.
An aluminum wire having a cross-sectional area of 4.00×10-6 m2 carries a current
of 5.00 A.
A. Calculate the current density.
B. Find the drift speed of the electrons in the wire. The density of aluminum is
2.70 g/cm3. The molar mass of aluminum is 27 g/mol. Assume that each atom
supplies one conduction electron.
C. The resistivity of aluminum is 2.75×10-8  m. What is the electric field in the
wire?