# Biophysics Textbook Questions - Chapter 11

#### Problem 11-10

\begin {align} \text{molar mass of} \; H_2O & = [(2\times 1)+ 16]\text{g/mol}\\ & = 18 \; \text{g/mol}\\ & = 0.018 \; \text{kg/mol} \end {align}\\ \text{mass of} \; 1\; \text m ^3 \text{of ice is}\; 0.92 \times 10^3 \;\text{kg}

$\therefore \;\text{# of moles of}\; H_2O\; \text{in}\; 1 \;\text m ^3 \text{of ice} \\ \quad = \frac{0.92 \times 10^3\; \text{kg}}{0.018 \; \text{kg/mol} } \\ \quad = 5.11 \times 10^4 \; \text{mol} \\ \therefore \text{# of molecules of} \; H_2O\; \text{in}\; 1\; \text m^3 \; \text{of ice} \\ \quad = 5.11\times 10^4\; \text{mol} \times 6.02 \times 10^{23}\; \text {mol}^{-1} \\ \quad = 3.08 \times 10^{28}\\ \therefore \text{volume per molecule} = \frac{1\; \text m^3}{3.08 \times 10^{28}}\\ = 3.2\times 10^{-32}\; \text m^3 \\ (\text{or} \; 3.3 \times 10^{-29}\; \text m^3, \; \text{depending on how many extra digits are carried in intermediate answers)}$

#### Problem 11-11

P_2 = P_3 \text{(same level in continuous fluid)} \\ \begin {align} \text {But} \; P_3 & = 0 \; \text{(gauge)} \\ \therefore \; P_2 & = 0 \; \text{(gauge)} \end {align} \\ \text{Use} \; \Delta P = P_2 - P_1 = \rho gh \quad [1] \\ \text{Convert}\; P_1 \; \text{to}\; \text{pascals}:
\begin{align} P_1 &= \rho_{Hg}\;gh_{Hg} \quad (Hg: mercury)\\ & = - (13.6 \times 10^3 \text{kg/m}^3)(9.80\; \text{m/s}^2) \times (80 \times 10^{-3} \text m) \\ & = -1.0_7 \times 10^4 \; Pa \end {align} \\ \therefore [1] \Rightarrow 0 - \underbrace {-1.0_7 \times 10^4}_{P_1} = (0.92 \times 10^3\; \text{kg/m}^3)\times (9.80\; \text{m/s}^2) \\ \Rightarrow h = 1.2\; \text m

#### Problem 11-12

First , consider the monometer connecting B and the atmosphere:
By inspection of the manomater levels, $P_B> P_{atm}$
$\Delta P = P_B - P_{atm}= \rho gh,$
Easier to use gauge pressures:
$​ \therefore \text{write (gauge)} ​P_B - 0 = \rho gh, \text{where 0 is the gauge pressure corresponding to}\; P_{abs} = P_{atm} \\ \text{(gauge)} P_B = \rho gh, \quad [1]$

Now cosider the monometer connecting $A$ and $B$
By inspection of the manometer levels, $P_A < P_B$
\therefore \; P_A = P_B - \rho gh_2 \\ \begin {align} \therefore \text{from} \; [1], \; P_A & = \rho gh,_1 - \rho gh_2 \\ & = \rho g (h_1 - h_2) \end {align}
\begin {align} \therefore \; P_A & - (1000\; \text{kg/m}^2)(9.80\; \text{m/s}^2)[(0.30- 0.10)\text m] \\ & = 2.0 \times 10^3 \; Pa \quad (1.9_6 \times 10^3 \; Pa) \; (\text{2.0 is the gauge pressure}) \end {align}

\begin {align} \text{Then, absolute}\; P_a &= P_{A, Gauge} + 1\; atm \\ & = (2.0 \times 10^3 + 1.01 \times 10^5) Pa \\ & = 1.03 \times 10^5\; Pa \end {align}

#### Problem 11-15

\begin {align} \frac{P}{P_o} & = e^{-mg (h_2-h_1)/k_BT} \\ \therefore \ell_n \bigg( \frac{P}{P_o} \bigg) & = -mg (h-2 - h_1)/k_bT \\ \therefore h_2 - h_1 & = \frac{\bigg[ - \ell _n (\frac{P}{P_o})\bigg]k_BT}{mg} , \\ \text{where m} & = \frac{29\; \text g}{\text{mol}} \times \frac{1\; \text{kg}}{1000\; \text{g}} \times \frac{1\; \text{mol}}{6.02 \times 10^{23}\; \text{molecules}} \\ & = 4.82 \times 10^{-26}\; \text{kg} \quad (\text{per molecule}) \\ \therefore h_2 - h_1 & = \frac{[-\ell_n(0.10)](1.38\times 10^{-23}\; \text{J/K})(300\; \text K)}{(4.82 \times 10^{-26\; \text{kg}})(9.8\; \text{m/s}^2)} \\ & = 2.0 \times 10^4 \; \text m \end {align}

#### Problem 11-16

Find gauge pressure inside bubble first, then abs. press.

(gauge) $P_1 = 0$ (open to atmosphere)

$P_2 - P_1 = \rho_m gh_m (\text {where m is manometer})$

$\therefore P_2 - 0 = (13.6\times 10^3\; \text{kg/m}^3)(9.80\; \text{m/s}^2)(0.200\; \text m) \\ \therefore P_2 = 3.99_8 \times 10^3 \; Pa$

\begin {align} \text{Then,}\; P_3 & = P_2 + \rho_w gh_w (\text{where w is water}) \\ &= (3.99_8 \times 10^3 \; Pa)+(1000\; \text{kg/m}^3)(9.80\; \text{m/s}^2)(0.200 \; \text m) \\ & = 5.95_8 \times 10^3 \; Pa \\ \text{Finally,} \; P_4 & = P_3 + \frac{2 \gamma}{r}\\ & = (5.95_8\times 10^3 Pa)+ \frac{2(72.8 \times 10^{-3}\; \text{N/m})}{0.0500 \times 10^{-3}\; \text m} \\ & = 8.87 \times 10^3 \; Pa \end {align}

$\therefore \text{gauge press. inside bubble is} \; 8.90 \times 10^3 \; Pa$

\begin {align} P_{abs} & = P_{gauge} + 1\; atm\\ & = (8.87 \times 10^3 + 1.013 \times 10^5) Pa\\ & = 1.10 \times 10^5 \; Pa \end {align}

#### Problem 11-18

$P_{gauge} = 0\; \text{at top of container} \\ \therefore \; \text{inside bubble}$

\begin {align} P_{gauge} & = 0 +(\rho gh)_{oil} + (\rho gh)_{water}+ \frac{2 \gamma}{r} \\ \therefore P_{gauge} & = (0.50\times 10^3 \; \text{kg/m}^3)(9.8\; \text{m/s}^2)(5.0\; \text{m}) \\ &+ (1000\; \text{kg/m}^3)(9.8\; \text{m/s}^2)(15\; \text{m}) \\ &+ \frac{2(72.8 \times 10^{-3}\; \text{N/m})}{8.0 \times 10^{-5}\; \text m} \\&= 1.7 \times 10^5\; Pa \end {align}

#### Problem 11-20

Subscripts: '$s$' - surfactant; '$w$' - water

\begin {align} y & = \frac{2 \gamma \cos \theta}{\rho gr} \\ \therefore \frac{y_s}{y_w} & = \frac{\bigg( \frac{2 \gamma_s \cos\theta}{\rho_s gr} \bigg)}{\bigg( \frac{2\gamma_w \cos \theta _w}{\rho_w gr} \bigg)} \\ \therefore \frac{y_s}{y_w} & = \frac{2 \gamma_s \cos \theta_s}{\rho_s gr} \times \frac{\rho_w gr}{2\gamma_w \cos \theta _w} \\ & = \frac{\gamma_s (\cos\theta _s)\rho_w}{\rho_s \gamma_w \cos \theta_w} \end {align}

\begin {align} \theta_s &= \theta_w \therefore \cos \theta_s = \cos \theta_w\\ \text{and} \; \rho_s &= 0.80\; \rho_w \\ \therefore \frac{y_s}{y_w} & = \frac{\gamma_s\rho_w}{0.80\; \rho_w \gamma_w} = \frac{\gamma_s}{0.80\; \gamma_w} \\ \therefore \gamma_s & = 0.80\; \gamma_w \frac{y_s}{y_w} \\ & = 0.80(72.8 \times 10^{-3} \; \text{N/m}) \frac{0.020 \; \text m}{0.10\; \text m}\\ & = 1.2 \times 10^{-2} \; \text{N/m} \end {align}

#### Problem 11-21

Contant angle = 0

$\therefore$ shape of surface is hemispherical, i.e., there already is a hemispherical "bubble."

$\therefore$ the pressure required to move the bubble to the bottom of the tube is just $\rho g(h_1+h_2).$

$\rho g (h_1+h_2) \\ = (1000\; \text{kg/m}^3)(9.80\; \text{m/s}^2)[(0.020+0.050)\text m] \\ = 6.9\times 10^2\; Pa$

#### Problem 11-22

(a) At C, $P_{gauge} = 0.$

But $P_B = P_C$ (same level in continuous fluid)

$\therefore P_B = 0 \; Pa$

(b)

\begin {align} P_B & = P_A + \rho gh_2 \\ \therefore P_A & = P_B - \rho gh_2 \\ &= (0 \; Pa)-(8.0\times 10^3 \; \text{kg/m}^3)(9.8\; \text{m/s}^2)(0.04 \times 10^{-2}\; \text m) \\ &= - 31\; Pa \end {align}

Note: since surface $S$ is curved [a portion of a single-surfaced bubble], then $P_A$ and $P_D$ are related as follows:

$P_A = P_D - \frac{2\gamma}{r_{surface}} + \rho gh_1,$

where $P_D = 0 \; Pa$

$r_{surface} =$ radius of curvature of surface $S$ (not given)

$\not =$ radius of capillary tube