University:
Amherst College
Course:
PHYS 230 | Statistical Mechanics and Thermodynamics
Academic year:

2021
Views:

167

Pages:

8
Author:

Kathryn Duarte

Boiling Point of Water The relevant data are (per mole at STP): H₂o (e) SH (KJ) (\Delta G(kJ) S(J/k)) -285.83 -237.13 69.91 [V(m^{3})] [18~cm^{3}=18~\pi10^{-6}m^{3}] H₂o (g) -241.82 -228.57 188.83 [25l=25\pi_{10}^{-3}m^{3}] I molar volume at room temp. So \[\Delta fge=(-24,82)-(-285.83)=44.81kT\] \[SG_{jL}=(-228.57)-(-237.13)=8.56~kJ^{-}\] \[ss_{ge}=18883-69.91=0.1189~kT/K\] \[sV_{gl}=25\kappa_{0}o^{-3}-i8\kappa_{1o}=25\kappa_{to}^{-3}\mu^{3}.\] For G=H-TS \[\Delta G=\Delta H-T\Delta S\] \[8.56~kJ\div\] we expect \[6~k5\cong44.01~k5-(298k)(0.1189^{kJ}/k)\] \(=8.58~kJ\) (pretty close)✓ 2 Water boils when (G_{e}=G_{g}) (no Aru energy difference bit two states), so [O=\Delta G_{ge}=\Delta H_{ge}-T\Delta S_{ge}] ③ Fran \[\rightarrow T_{bsillyy}=\frac{\Delta4}{\Delta~\delta}=\frac{44,si~kJ}{0.11g92~kJ}=370~k\sim97^{\circ}C.\] \(\Delta H=\Delta U+P\Delta V \Delta U_{ge}=\Delta H_{ge}-P~\Delta V_{g}e,\) so \[\Delta U_{ge}=444.01KJ-(i\circ\frac{5N}{\mu\epsilon})(2{\pi\epsilon o}^{-3}{m^{3}})\] \[=444.si~kJ-2.5~kJ=|4i.5~kJ\] [n(x-ze,t)] [n(x+le,t)] [n(x,t+t)] comes from [n(x-ze,t)] with probability p and from [n(x+le,t)] with probability q. [n(x,t+t)=p\cdot n(x-ze,t)+q\cdot n(x+le,t)] (1) Taylor expand each of these: [=p\left[n(x,t)+(-ze)\frac{\partial n}{\partial x}+\frac{1}{2}(-ze)^2\frac{\partial^2 n}{\partial x^2}\right]] [+q\left[n(x,t)+(le)\frac{\partial n}{\partial x}+\frac{1}{2}(le)^2\frac{\partial^2 n}{\partial x^2}\right]] [n(x,t+t)=(p+q)n(x,t)+(-2p+2q)\frac{\partial n}{\partial x}+\frac{1}{2}(4p+4q)\frac{\partial^2 n}{\partial x^2}] [n(x,t+t)-n(x,t)=(q-2p)\frac{\partial n}{\partial x}+\frac{1}{2}(4p+4q)\frac{\partial^2 n}{\partial x^2}] [\frac{\partial n}{\partial t}=\frac{1}{t}\left[(q-2p)\frac{\partial n}{\partial x}+\frac{1}{2}(4p+4q)\frac{\partial^2 n}{\partial x^2}\right]] (2) So [D=\frac{1}{2}(4p+4q)] [v=\frac{1}{t}(2p-2q)] or [D=\frac{1}{2}(8-2p)] [v=\frac{1}{t}(2p-8)] 5. If (v=0), (2p-8=0 \rightarrow p=\frac{8}{2}); since (p+q=1), [p=\frac{1}{2}, q=\frac{1}{2}] and [D=\frac{1}{2t}\left(2^2+\frac{4}{3}\right)=\frac{1}{2t}\left(\frac{10}{3}\right)] which is twice (\frac{1}{2t}). 6. This is the (+2/-1) walk with (e=

10\), \(t=1\) turn and \[p=p_{win}=\frac{12}{38}, q=prob. of losing=\frac{26}{38}\]. So, \[v=\$10 \cdot \left(2\cdot \frac{12}{38}-\frac{26}{38}\right)=0.53 \$/turn\] \[D=\frac{(10)^2}{2\cdot 1 turn}\left(\frac{26}{38}+\frac{4}{38}\right)=9.76 \/turn.]

Physics Midterm 1 Solutions

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