Assignment6.ipynb

{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "3815a7b2",
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "import sympy as sp\n",
    "from IPython.display import display"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9c6f7b65",
   "metadata": {},
   "source": [
    "8.2 Consider a polymer chain represented by N = 20 beads connected by springs \n",
    "of root-mean-square size b = 5 A diffusing in a melt of similar chains with \n",
    "bead friction coefficient < = 3 x loP1\" g sP1 a t  22 \"C. \n",
    "(i) What is the root-mean-square end-to-end distance R of this chain? \n",
    "(ii) What is best model describing the dynamics of the chain in a melt? What is friction co-eff for chain?\n",
    "(iii) What is the diffusion coefficient D of the chain? \n",
    "(iv) Estimate the longest relaxation time r of the chain. \n",
    "(v) Estimate the viscosity of the melt. \n",
    "\n",
    "the friction coefficient of the chain? "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e957acaf",
   "metadata": {},
   "source": [
    " # The best model is the Rouse Model"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 75,
   "id": "938af1b8",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$\\displaystyle RMS = \\left(b^{2} n\\right)^{0.5}$"
      ],
      "text/plain": [
       "Eq(RMS, (b**2*n)**0.5)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "in angstrom\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle RMS = 22.3606797749979$"
      ],
      "text/plain": [
       "Eq(RMS, 22.3606797749979)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle friction_{coeff} = n \\xi$"
      ],
      "text/plain": [
       "Eq(friction_coeff, n*xi)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "in g/s\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle friction_{coeff} = 6.0 \\cdot 10^{-9}$"
      ],
      "text/plain": [
       "Eq(friction_coeff, 6.0e-9)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle D = \\frac{T k_{B}}{n \\xi}$"
      ],
      "text/plain": [
       "Eq(D, T*k_{B}/(n*xi))"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "in m^2/s\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle D = 6.785 \\cdot 10^{-10}$"
      ],
      "text/plain": [
       "Eq(D, 6.785e-10)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\tau_{R} = \\frac{RMS^{2} n \\xi}{T k_{B}}$"
      ],
      "text/plain": [
       "Eq(\\tau_{R}, RMS**2*n*xi/(T*k_{B}))"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "in seconds\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\tau_{R} = 7.36874871039057 \\cdot 10^{-9}$"
      ],
      "text/plain": [
       "Eq(\\tau_{R}, 7.36874871039057e-9)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "I arrive at this expression by equating the two extressions for Khun monomer relaxation time. Here, I assume that estimate means use the equations given in the problem\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\eta_{s} = \\frac{\\xi}{b}$"
      ],
      "text/plain": [
       "Eq(\\eta_{s}, xi/b)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "in units of g/ang*s\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\eta_{s} = 6.0 \\cdot 10^{-11}$"
      ],
      "text/plain": [
       "Eq(\\eta_{s}, 6.0e-11)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "RMS = sp.Symbol(\"RMS\")\n",
    "n = sp.Symbol(\"n\")\n",
    "b = sp.Symbol(\"b\")\n",
    "D = sp.Symbol(\"D\")\n",
    "kb = sp.Symbol(\"k_{B}\")\n",
    "T = sp.Symbol(\"T\")\n",
    "f = sp.Symbol(r\"xi\")\n",
    "t = sp.Symbol(r\"\\tau_{R}\")\n",
    "t0 = sp.Symbol(r\"\\tau_{0}\")\n",
    "fric = sp.Symbol(\"friction_coeff\")\n",
    "eta = sp.Symbol(r\"\\eta_{s}\")\n",
    "RMS_eq = sp.Pow(n*sp.Pow(b,2), 1/2)\n",
    "display(sp.Eq(RMS,RMS_eq))\n",
    "print(\"in angstrom\")\n",
    "display(sp.Eq(RMS, RMS_eq.subs({n:20,b:5})))\n",
    "fric_eq = n*f\n",
    "display(sp.Eq(fric, fric_eq))\n",
    "print(\"in g/s\")\n",
    "display(sp.Eq(fric, fric_eq.subs({f:3e-10, n:20})))\n",
    "D_eq = (kb*T)/(n*f)\n",
    "display(sp.Eq(D,D_eq))\n",
    "print(\"in m^2/s\")\n",
    "display(sp.Eq(D,D_eq.subs({f:3e-10,kb:1.38e-23, T:22+273, n:20})*1e3))\n",
    "t_eq = (n*f*sp.Pow(RMS,2))/(kb*T)\n",
    "display(sp.Eq(t,t_eq))\n",
    "print(\"in seconds\")\n",
    "display(sp.Eq(t,t_eq.subs({f:3e-13,kb:1.38e-23, T:22+273, RMS:22.36e-10, n:20})))\n",
    "eta_eq = f/b\n",
    "print(\"I arrive at this expression by equating the two extressions for Khun monomer relaxation time\\\n",
    ". Here, I assume that estimate means use the equations given in the problem\")\n",
    "display(sp.Eq(eta,eta_eq))\n",
    "\n",
    "print(\"in units of g/ang*s\")\n",
    "display(sp.Eq(eta,eta_eq.subs({f:3e-10,b:5})))\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7dc94c3b",
   "metadata": {},
   "source": [
    "Consider a polymer chain repregented by N = 100 beads connected by springs \n",
    "of root-mean-square size b = 5 A in a dilute 0-solution at 22.5 \"C with solvent \n",
    "viscosityqS=0.6cP=6x 10P3g(cms)P'. \n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "814cbe67",
   "metadata": {},
   "source": [
    " # Best model is zimm because it is a dilute solution"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 72,
   "id": "ce50d7b7",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\xi_{Z} = R \\eta_{s}$"
      ],
      "text/plain": [
       "Eq(\\xi_{Z}, R*\\eta_{s})"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "in units of g/s\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\xi_{Z} = 3.0 \\cdot 10^{-9}$"
      ],
      "text/plain": [
       "Eq(\\xi_{Z}, 3.0e-9)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle D_{z} = \\frac{T k_{B}}{\\eta_{s} b n^{0.5}}$"
      ],
      "text/plain": [
       "Eq(D_{z}, T*k_{B}/(\\eta_{s}*b*n**0.5))"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "in units of m^2/s\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle D_{z} = 1.35792 \\cdot 10^{-9}$"
      ],
      "text/plain": [
       "Eq(D_{z}, 1.35792e-9)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\tau_{z} = \\frac{\\eta_{s} b^{3} n^{1.5}}{T k_{B}}$"
      ],
      "text/plain": [
       "Eq(\\tau_{z}, \\eta_{s}*b**3*n**1.5/(T*k_{B}))"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "in seconds\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\tau_{z} = 1.84105101920584 \\cdot 10^{-8}$"
      ],
      "text/plain": [
       "Eq(\\tau_{z}, 1.84105101920584e-8)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\eta_{sp} = \\frac{T \\phi \\tau_{z} k_{B}}{\\eta_{s} b^{3} n}$"
      ],
      "text/plain": [
       "Eq(\\eta_{sp}, T*\\phi*\\tau_{z}*k_{B}/(\\eta_{s}*b**3*n))"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\eta_{sp} = 0.99942912$"
      ],
      "text/plain": [
       "Eq(\\eta_{sp}, 0.99942912)"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "#R = 10*5\n",
    "etaZ = sp.Symbol(r\"\\xi_{Z}\")\n",
    "R = sp.Symbol(\"R\")\n",
    "Dz = sp.Symbol(\"D_{z}\")\n",
    "tz = sp.Symbol(r\"\\tau_{z}\")\n",
    "etaZ_eq = eta*R\n",
    "phi = sp.Symbol(r\"\\phi\")\n",
    "eta_sp = sp.Symbol(r\"\\eta_{sp}\")\n",
    "display(sp.Eq(etaZ,etaZ_eq))\n",
    "print(\"in units of g/s\")\n",
    "display(sp.Eq(etaZ,etaZ_eq.subs({R:50,eta:6e-11})))\n",
    "Dz_eq = (kb*T)/(eta*b*sp.Pow(n,1/2))\n",
    "display(sp.Eq(Dz,Dz_eq))\n",
    "print(\"in units of m^2/s\")\n",
    "display(sp.Eq(Dz,Dz_eq.subs({T:22.2+273,kb:1.38e-23,eta:6e-4, b:5e-10,n:100})))\n",
    "tz_eq = (eta*sp.Pow(b,3)*sp.Pow(n,3/2))/(kb*T)\n",
    "display(sp.Eq(tz,tz_eq))\n",
    "print(\"in seconds\")\n",
    "display(sp.Eq(tz,tz_eq.subs({eta:6e-4,n:100,b:5e-10,T:22.2+273,kb:1.38e-23})))\n",
    "eta_sp_eq = (kb*T*phi*tz)/(n*sp.Pow(b,3)*eta)\n",
    "display(sp.Eq(eta_sp,eta_sp_eq))\n",
    "display(sp.Eq(eta_sp,eta_sp_eq.subs({eta:6e-4,n:100,b:5e-10,kb:1.38e-23,T:22.2+273,tz:1.84e-8,phi:0.1})))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "030856d2",
   "metadata": {},
   "source": [
    "What is the physical reason that the 2/3 exponent in Eq. (8.70) for sub- \n",
    "diffusive Zimm motion is independent of solvent quality? \n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "dbc663f5",
   "metadata": {},
   "source": [
    " # Sub-diffusive motion of monomers is independent of solvent quality in the Zimm model because, in the zimm model, the entire polymer moves through solution like a stokes sphere with size proportional to the pervaded volume of the polymer which depends on the polymer size rather than the degree of polymerization. In the Zimm model, the relaxation time measures the time it takes for the polymer to diffuse a distance proportional to its size, mathematically, this means that, for a monomer, the scaling parameter will cancel out. Physically, the independence on the scaling parameter can be explained based on the fact that the friction depends on the size of the polymer rather than the number of monomers."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "7f757a52",
   "metadata": {},
   "outputs": [],
   "source": []
  }
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