Chem 175 HW 3 JasperSK

import numpy as np import matplotlib.pyplot as plt import scipy from scipy import constants

1d)

The wording of this question confused me a little. In particular that we want the fraction of unfolded at different values of external force. I interpret this just as the probability of being in the unfolded state because you are either in the unfolded state or the folded state. Thus the probability of the unfolded state gives you the fraction in the folded state compared to the number in either state so the fraction in the folded state. This is the first graph below. That being said I went to office hours and asked about this and was told that you wanted the ratio of the probability of the unfolded state over the probability of the folded state. Thus I have provided this as the second graph.

KBT = 4.114 B = 1/KBT E0 = 0 E1 = 312.4 deltaL = 22 fext=np.linspace(13.3,15.5, 200) pUnfolded = (np.e**(-B * E1))/((np.e**(-B * deltaL * fext)) + (np.e**(-B * E1))) plt.plot(fext,pUnfolded) plt.title('Fraction Unfolded With Respect to External Force') plt.ylabel('Fraction Unfolded') plt.xlabel('External Force (pN)') plt.show()

KBT = 4.114 B = 1/KBT deltaE = 312.4 deltaL = 22 fext=np.linspace(13.3,15.5, 200) pUnfoldedOverpFolded = ((np.e**(B * deltaL * fext)) * (np.e**(-B * deltaE))) plt.plot(fext,pUnfoldedOverpFolded) plt.title('Fraction Unfolded over Fraction Folded With Respect to External Force') plt.ylabel('Fraction Unfolded over Fraction Folded') plt.xlabel('External Force (pN)') plt.show()

2a)

T = np.linspace(0,10,100000) mass = 1.6735575*(10**(-27)) length = 1 * (10**(-9)) h = scipy.constants.Planck k = scipy.constants.Boltzmann SoverKB = 0.5*np.log(2 * np.pi * mass * k * T )+np.log(length)-np.log(h) plt.plot(T,SoverKB) plt.title('Entropy over KB with respect to temerature') plt.ylabel('Entropy over KB') plt.xlabel('Temperature (K)') plt.show()

below K=1 we start to get negative entropy.

b) We get a non physical result because we let q_tran be less than 1. This doesn't make any sense because this says that there are no accessible states for the system to be in however it has to be in some state so q_tran should never be lower than 1. One reason this error came up is because we assumed that the temperature was high enough so we could approximate the summation in the partition function by an integral. Here however we let the temperature get arbitrarily low and thus this approximation makes no sense.

T = np.linspace(0,10,100000) mass = 1.6735575*(10**(-27)) length = 1 * (10**(-9)) h = scipy.constants.Planck k = scipy.constants.Boltzmann a = (h**2)/(8*mass*(length**2)*k) SoverKB = np.log(1+np.e**((-a)/T)) plt.plot(T,SoverKB) plt.title('Entropy over KB with respect to temerature') plt.ylabel('Entropy over KB') plt.xlabel('Temperature (K)') plt.show()