Assignment 3. Tyler Shendruk February 26, 2010

Transcription

1 Assignment 3 Tyler Shendruk February 6, 00 Kadar Ch. 4 Problem 7 N diatomic molecules are stuck on metal. The particles have three states:. in plane and aligned with the ˆx axis. in plane but aligned with the ŷ axis 3. out of plane aligned with the ẑ axis. There is an energy cost associated with standing up but no energy cost associated with being in plane.. Part a. How many microstates have the smallest value of energy? The lowest energy state is when none of the molecules are paying the cost of standing up i.e. when all molecules line in plane: min 0 There are two orientations which each of the N molecules can take when it s in place so the number of states is Ω min N. What is the largest microstate energy? is maximized when every particle is out of plane: max N 3 and only a single state i.e. when every single particle is perpendicular to the planecan achieve that so Ω max 4

2 . Part b. For a microcanonical ensemble of energy haw many states are there? We have to ask how many ways can we get that energy? That requires N z 5 molecules to be aligned perpendicular to the plane. So then the number of ways for this to occur is Ω, N ΩN z, N N z!n N z! That is the number of ways to have N z molecules out of the plane and N N z molecules in the plane but the N N z molecules can be arranged in N Nz ways as we showed in the last part. So then the total number of states to get is where N z /.. What s the entropy? The entropy is Ω, N N z!n N z! N Nz 6 S k B ln Ω k B ln N z!n N z! N Nz k B ln + N N z ln N z!n N z! Break up the fraction and apply Stirling s formula S k B ln ln N z! ln N N z! + N N z ln k B N ln N N N z ln N z + N z N N z ln N N z + N N z + N N z ln k B N ln N N z ln N z N N z ln N N z + N N z ln k B N ln N N z ln N z + N z ln N N z ln N N N z ln N N z + N N z ln }{{} 0 k B N N z ln N + N z ln N ln N z N N z ln N N z + N N z ln Nz N Nz k B N z ln N N z ln + N N z ln N N Nz Nk B N ln Nz + N N z N Nz ln N N z ln. N N N N If we write N z / we can say that the entropy is S Nk B N ln + ln N N N ln. N 7

3 .3 Part c Calculate the heat capacity. We want to use C d/dt but we don t know energy as a function of temperature yet so first we must use T S N Nk B N ln N Nk B N ln + N N N ln N k B ln N ln + N ln k B ln + ln N + ln ln N N N Now we rearrange to find T ln + ln N N exp + ln N exp ln exp N exp + N exp We can now find the heat capacity C d dt N exp C N k B + T exp N N. 8 + exp exp N N ln.4 Part d What s the probability that a specific molecule is standing up at the energy? 3

4 We are looking for the probability that some specific molecule has an energy and all the other molecules take care of the rest of the energy. This is the probability of the configuration of N particles leading to energy divided by the probability of all the other N particles in a configuration resulting in an energy of which was left over. p, N pn p, N p, N p, N Ω, N Ω, N Ω, N Ω, N N! N z! N N z! N z! N N z! N Nz N! N z! N N z! N! N z! N z! N z N N z! N N z! N Nz N Nz N Nz which of course if we had thought about the question before trying something difficult, we knew was the answer all along. If we want this as a function of temperature, we can just substitute the values of N z and in to find pn N z N N N N exp + pn exp Part e What s the largest possible value of energy at any temperature? Since T > 0, the term exp is minimized when T. Therefore, 4

5 the maximum internal energy is max lim T T lim T N + N exp + max N 3 Sethna Ch. 3 Problem 7 Monatomic He gas mixed with diatomic H gas. Both are ideal-ish so they are only weakly coupled. The question states as a fact: monatomic Ω 3N/ diatomic Ω 5N/.. Part a. Calculate the probability of the monatomic gas being at an energy. The probability is the probability that the monatomic gas is at and the diatomic gas is at when the total energy is. So p Ω Ω. Ω The key is determining what Ω is. Ω d Ω Ω d 3N/ 5N/ We know the integral 0 0 d 3N/ 5N/ d N 3N/ 0 x m x n dx ΓmΓN Γm + n a numerical value not dependant on x. 3 5

6 So then the probability is p Ω Ω Ω 3N/ 5N/ Ω 3N/ 5N/ p N 3N/ 4. The maximum occurs when dω Ω d dω Ω d 5 where p is the maximum and. We find 3N/ 3N 5N 5N/ 3N 5N 3N/ N/ That s a little theoretical for me. We could have found the exact same solution from the more physically transparent strategy of saying the state with the maximum probability is the equilibrium state and entropy must be maximized at equalibrium. Therefore ds d ds d d k B ln Ω d d k B Ω d d k B ln 3N/ d d k B ln 5N/ d k B 3N d ln d 3N k B k B 3N k B 5N k B 5N k B 5N d ln d 6

7 which is rearranged to find 3 8 in accordance with our previous result. The third useful way to find it is to just take the derivative of probability q. 4 with respect to and set it equal to zero. That energy is an extrema.. Part b Take the logarithm of the probability, do a Taylor expansion and then reexponentiate. What is the approximation to the probability distribution? ln p ln p + ln p + ln p valuate the derivatives as ln p 3N ln + 5N ln 3N + 5N N 3 5. Therefore when we remember q. 6, we have ln p N 3 5 N N On the other hand, the second derivative is ln p { } N 3 5 N

8 which we again evaluate at and find ln p N N N N N N 5 N. Now we can put all of this back into our logarithm and the take the exponent of both sides. ln p ln p N 5 exp ln p exp ln p 56 N 5 Comparing this to a gaussian p p exp px we identify the mean to be exp πσ 5 56 N x µ σ 8 < >, 9 meaning the distribution is symmetric i.e. has approximately zero skew and that the standard deviation is 5 σ. 0 6 N We know the definition of the standard deviation is σ < > which is the energy fluctuation. So we see that the energy fluctuation is N 8

9 3 Harden Problem 3 A polymer is made up of N segments which are either in a relaxed state α of length a contracted state β of length a/ β s energy is larger than α s we can say the α state energy is zero. 3. Part a Calculate the number of states for a polymer of energy 0 < < N i.e. not all compressed. The energy is fixed so this is microcanoncial again the length is N α a + N β a where N α is the number of segments in the α state, etc which add up to the total number of segments N N α + N β. 3 The number of segments in the β state determine the energy to be H N β. 4 We ll start by finding the number of states Ω as a function of N β which means we want to rearrange qs and 3 in terms of N β. a N α a + N β a N α + N β a N N β + N β a N N β nn 5 N a + N β N β N a Using q. 4, we find the number of states ΩN, N β ΩN, N β! N N β! 6 7 8! N! 9

10 We re going to need it later so we also calculated the number of states as a function of length: ΩN, N β N β! N N β! ΩN, N a! N N a! ΩN, N a! a N! 9 3. Part b Calculate the entropy as a function of energy and as a function of length. S k B ln Ω SN, k B ln SN, k B ln! N! N a! a N!

11 3.3 Part c We use the Stirling approximation and the very important formula T S N k B ln! N! k B ln ln! ln N! k B ln! + ln N! k B ln + N ln N k B ln + N ln N N k B ln + N ln N k B ln + / + ln N k B ln + ln N k B ln ln N k B k B ln ln N N. We can now rearrange for as a function of T : ln N N exp Ne /k BT e /k BT N + e /k BT T e /k BT N + e /k BT N + N N / 3 The length should be directly related to. From q. 7 and 4, we find the

12 length to be N β N a N a a N Na a Na a N Na Na Na + e /k BT + e /k BT + e / + e /k BT + e /k B T + e /k BT 33