Homework #2 Solutions
Question 1) Adiabatic Process for an ideal gas. We are given:
∆U = Q + U
f
N kB T
2
Adiabatic Process means: Q = 0. Hence, from the lst Law,
U=
∆U = W
And compressive work can be written as W = −P ∆V . With the above information:
f
N kB ∆T = −P ∆V
2
Now use the ideal gas Law P V = N kB T to eliminate Pressure P,
N kB T
f
N kB ∆T = −
∆V
2
V
Divide both sides by T ,
∆V
f ∆T
=−
2 T
V
Integrate both sides from the initial state (Ti , Vi ) to the final state (Tf , Vf ):
Z
Z Vf
f Tf dT
dV
=−
2 Ti T
V
Vi
f
Tf
Vf
ln
= − ln
2
Ti
Vi
After exponentiating both sides of the equation,
f
Tf 2
Vi
( ) =
Ti
Vf
f
Multiply both sides by Vf Ti 2 , then
f
f
Vf Tf 2 = Vi Ti 2
So, in the other words,
f
VT 2 =C
where C is a constant. Now, we can use the ideal gas law to eliminate T in favor of P:
f
V(
PV 2
) =C
N kB
1
Both N and kB are constant, so we can write
f
P 2V
f
2
+1
= C0
Where C 0 is another constant. Now raise both sides to the power 2/f to get
PV
2
f
(1+ f2 )
= C 00
and C 00 is also a constant.
Define γ ≡
f +2
f ,
then we have
P V γ = constant
Discussion: Alternative Method:
dW = −P dV
f
f
f
dU = d( N KB T ) = d( P V ) = (V dP + P dV )
2
2
2
For an adiabatic process, Q = 0, then
dU = dW ⇒ −P dV =
f
(V dP + P dV )
2
This can be written as
dV
dP
=−
V
P
Integrate both sides from the initial state (Vi , Pi ) to the final state (Vf , Pf ) to get:
γ
P V γ = constant
Question 2a) We know that (from the above question), Pi Vi γ = Pf Vf γ . Hence
Vf = (
where γ ≡
f +2
f
Pi γ1
) Vi
Pf
= 75 . Hence,
Vf = (1Liter)(
1atm 5
) 7 = 0.25L
7atm
b) The word done in compression is
Z
Vf
W =−
P dV
Vi
From the equation
P V γ = constant
we can get
P =
2
C
Vγ
Hence
Z
Vf
W = −C
Vi
Where C =
Pf Vfγ
=
Pi Viγ ,
C
dV
=
[V 1−γ − Vi1−γ ]
γ
V
γ−1 f
so that
W =
1
1
[Pf Vfγ Vf1−γ − Pi Viγ Vi1−γ ] =
[Pf Vf − Pi Vi ]
γ−1
γ−1
Before the calculation, we should change the units to SI,
1atm = 1.013 × 105 N/m2
1Liter = 10−3 m3
Then we can get the result
W = 189.9N · m = 189.9J
Discussion Alternative method:
W =
f
f
N KB (Tf − Ti ) = (Pf Vf − Pi Vi )
2
2
f
c) We know from the last problem that V T 2 = constant. Hence,
f /2
Vf Tf
f /2
= Vi Ti
So
Tf = (
Vi f2
1L 2/5
) Ti = 300K(
)
= 522K = 249o C
Vf
0.25L
which is pretty hot!
Discussion Alternative methods: Method 1) From the ideal gas law,
Tf
Pf Vf
7atm × 0.25L
7
=
≈
=
Ti
Pi Vi
1atm × 1L
4
7
Ti = 525K
4
Method 2) From the result of question 2b,(adiabatic process Q = 0)
Tf ≈
dU = W
you get the change of internal energy, then
dU =
f
N kB (Tf − Ti )
2
3
where N kB =
Pi Vi
Ti .
Question 3) Note that this probability is nothing but g(Nh , Nt )/2N where g(Nh , Nt ) = NhN!N! t ! is the
multiplicity function for spins that you calculated in class. I use Nh to denote the number of heads and
Nt for the number of tails.
3a) P (5, 5) =
10! 1
5!5! 210
= 0.246 (i.e. 24.6%)
3b) P (4, 6) =
10! 1
6!4! 210
= 0.205 (i.e. 20.5%)
3c) The probabilities and the corresponding average is given in Table 1. Taking the square root of the
Nh
0
1
2
3
4
5
6
7
8
9
10
Table 1:
Probability
( N2 − Nh )2
0.001
25
0.010
16
0.044
9
0.117
4
0.205
1
0.246
0
0.205
1
0.117
4
0.044
9
0.010
16
0.001
25
sum=1
average=2.508
average of ( N2 − Nh )2 we get δNrms = 1.58 and δNrms /N = 0.158.
Discussion How to get the root-mean squared deviation? The formula should take into consideration the
probability of each deviation from the mean. And ”hi” represents the average.
NX
h =10
1
1
1
δNrms = h( N − Nh )2 i1/2 = [
( N − Nh )2 P ( N − Nh )]1/2
2
2
2
Nh =0
Where
P ( 12 N
− Nh ) is the probability of each Nh as shown in the table above.
Let us compare this to the gaussian approximation. Remember
that the spin excess 2s is equal to Nh − Nt
q
2
2 − 2s
and the probability in terms of s is given by P (s) = πN e N and < ( N2 − Nh )2 >=< s2 >, valid for
large N and |s| << N .
Z ∞r
2 − 2s2 2
N
N
2
2
e N s ds =
< ( − Nh ) >=< s >=
2
πN
4
−∞
√
√
Hence δNrms = N /2 and δNrms /N = 1/(2 N ). Putting N=10 we√get 0.158, which is very close to
the exact solution. Remember for this problem N is pretty small so 1/ N is not very small. In the next
problem the approximation will work much better.
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Question 4) Again we will use the probability distribution function g(Nh , Nt )/2N as in question 3.
However now the numbers are so big that we have to use the gaussian approximation (Try typing 5000!
in your calculator!)
q
q
2 0
2
e = π10000
= 8 × 10−3
4a) P (5000, 5000) = P (s = 0) = πN
4b) P (6000, 4000) = P (s = 1000) =
q
2
2
−2 1000
10000
π10000 e
= P (s = 0) × e−200
4c) We did the hard work for this problem in question 3c. Set N=10000 there to get δNrms = 50 and
δNrms /N = 5 × 10−3 .
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