南京大学 大学物理 期末真卷

dQ = 0.
Then,
dU = − pdV
or
CV dT +
or
a ⎞ a ⎛ RT dV = −⎜ − 2 ⎟dV 2 V ⎝V − b V ⎠
CV dT dV . =− R T V −b
Integrating both sides of above equation, after some algebra, we obtain the adiabatic equation where
U = U 0 + CV T −
a , V
where CV is assumed to be constant. (a) Determine the adiabatic equation of the one mole of Van der Waals gas. (b) If the gas expands freely into the vacuum from the volume V1 to the volume V2 , what is the temperature change in this process? Solution: (a) In an adiabatic process
W = nRT0 ln
And the entropy of the substance is given by
V . V0
V ⎛T ⎞ ⎟ S = nR 0 ⎜ ⎟ V ⎜ ⎝ T0 ⎠
x
where x is a fixed constant. (a) Calculate the Helmholtz free energy of the substance. (b) Find the equation of state of the substance. Solution: (a) From dF = − SdT − pdV we have
o o
1 atm. , what is the change of the entropy? (The specific heat of the ice is c pi = 2.1 J/g ⋅ K and
the specific heat for the water is c pw = 4.2 J/g ⋅ K . The latent heat absorbed during the ice-water phase transition is L = 333 J/g .) Solution: The change of the entropy is
Therefore,
⎞ ⎟ ⎟ ⎠
⎤ − 1⎥. ⎥ ⎦
x +1
V 1 ⎡⎛ T ⎞ V ⎢⎜ ⎟ F (T ,V ) = F (T0 ,V0 ) − nRT0 ln − nRT0 0 V0 V x + 1 ⎢⎜ T ⎟ ⎣⎝ 0 ⎠
(b) The equation of state is
⎤ − 1⎥. ⎥ ⎦
B ′(T ) + f (T ) V where f (T ) is some function of T . Thus, the heat absorbed in the expansion is B ′(T ) ΔQ = TΔS = T ( S1 − S 0 ) = T ln 2 + T . 2V0 S = ln V −
γB
2. (20 pts. ) A gas obeys the equation of state
p=
T B(T ) + 2 V V
where B (T ) is a function of temperature only. The gas is expanded isothermally at temperature T from the volume V0 to the volume V1 = 2V0 . The process is reversible. (a) Find the work done in the expansion. (b) Find the heat absorbed in the expansion. Solution: (a) The work done in the expansion is
W =
2V0
V0
+ ∫ pdV = ∫ ⎜ ⎝V
V0
2V0
⎛T
B(T ) ⎞ B(T ) dV = T ln 2 + . 2 ⎟ 2V0 V ⎠
1
(b) By using the Maxwell relation, we have
1 B ′(T ) ⎛ ∂S ⎞ ⎛ ∂p ⎞ ⎜ ⎟ =⎜ ⎟ = + V2 ⎝ ∂V ⎠ T ⎝ ∂T ⎠V V where B ′(T ) represents the derivative of B (T ) with respect to T . Integrating the both side of
June 6, 2008 Solution to the Practice Final Examination of University Physics Select five out of following six problems.
Friday
1. (20 pts. ) A person on Earth observes two rocket ships moving directly toward each other and colliding. At time t = 0 in the Earth frame, the Earth observer determines that rocket A, traveling to the right at v A = 0.8c , is at point a , and rocket B is at point b , traveling to the left at v B = 0.6c .They are separated by a distance l = 4.2 × 10 8 m . (a) In the Earth frame, how much time will pass before the rockets collide? (b) How fast is rocket B approaching in A’s frame? How fast is rocket A approaching in B’s frame? (c) How much time will elapse in A’s frame from the time rocket A passes point a until collision? How much time will elapse in B’s frame from the time rocket B passes point b until collision? Solution: (a) In the Earth frame
3
⎛ ∂U ⎞ ⎛ ∂f ⎞ ⎜ ⎟ = f −T⎜ ⎟ . ⎝ ∂l ⎠ T ⎝ ∂T ⎠ l (b) Show that the small temperature rise ΔT that takes place in a slow adiabatic stretching of the strip from the natural length l 0 is given by l 1 ⎛ ∂f ⎞ ΔT =∫ ⎜ ⎟ dl l0 C T l ⎝ ∂T ⎠ l where C l is the heat capacity of the strip at constant length l .
⎛ a a⎞ ΔU = CV (T2 − T1 ) − ⎜ ⎟ = 0. ⎜V − V ⎟ 1 ⎠ ⎝ 2
Therefore
(T2 − T1 ) =
a CV
⎛ 1 1⎞ ⎜ − ⎜V V ⎟ ⎟. 1 ⎠ ⎝ 2
2
4. (20 pts. ) If 20 grams of ice at temperature of − 10 C is changed into water of 10 C at
nRT0 nRT0V0 1 ⎡⎛ T ⎛ ∂F ⎞ ⎢⎜ p = −⎜ = + ⎟ V T V 2 x + 1 ⎢⎜ ⎝ ∂V ⎠ T ⎣⎝ 0
⎞ ⎟ ⎟ ⎠
x +1
⎤ − 1⎥ . ⎥ ⎦
6. (20 pts. ) A strip of rubber has length l when subject to a tensile force f . If the volume change duo to the extension of the strip can be neglected, the differential of the internal energy is then dU = TdS + fdl . (a) Show that
b−a l = = 1s . v A +vB v A +vB A B (b) In the Earth frame, v x = v A and v x = −v B .Then in the A’s frame, u = v A , B vx −u −v B −v A B ′ vx = = = −0.95c . v Bv A B u 1−v x 1+ 2 c2 c In the B’s frame, u = −v B , v xA − u v +vB A ′ = A = 0.95c . vx = u v v 1 − v xA 2 1 + A 2 B c c t=
3. (20 pts. ) The equation of state for one mole of Van der Waals gas can be written as
Baidu Nhomakorabea
above equation, we obtain
a ⎞ ⎛ ⎜ p + 2 ⎟(V − b) = RT V ⎠ ⎝
and the internal energy for the Van der Waals gas is given by
γ = (CV + R ) CV . This equation can also be rewritten as
a ⎛ ⎜p+ 2 V ⎝ ⎞ γ ⎟(V − b) = Const. ⎠
T (V − b) γ −1 = Const.
In the case of a free adiabatic expansion, the internal energy is unchanged so that
F (T0 ,V ) − F (T0 , V0 ) = −W = −nRT0 ln
and
V V0
⎛T ⎜ ⎜T ⎝ 0
x +1
V F (T ,V ) − F (T0 ,V ) = − ∫ SdT = − ∫ nR 0 T0 T0 V
T T
⎞ ⎟ ⎟ dT ⎠
x
V 1 ⎡⎛ T ⎢⎜ = −nRT0 0 V x + 1 ⎢⎜ T ⎣⎝ 0
ΔS =
T0
T0 −10
∫
mc pi dT T
+
T0 +10
T0
∫
mc pw dT T
+
mL T0
= mc pi ln
273 283 mL + mc pw ln + = 29 J K . 263 273 T0
5. (20 pts. ) There is a substance of n moles . When the substance expands from a volume V0 to a volume V at constant temperature T0 , the work done by it is
(c) In each of the two rocket frames, the time to collision is dilated with respect to that in the Earth frame. So
tA = tB =
t
γA
t
= t 1− = t 1−
2 vA = 0.6 s , c2 2 vB = 0.8 s . c2