# Biophysics Textbook Questions - Chapter 13

#### Exercise 13-13 $C_1 = 0.50 \frac{\text{mol}}{\text{dm}^3}$

$C_2 = 0.05 \frac{\text{mol}}{\text{dm}^3}$

\begin {align} \Delta x & = 10 \text{nm}\\ & = 10 \times 10^{-9}\text m \end{align}

\begin {align} J &= \frac{1}{A} \frac{dn}{dt}\\ \text{and}\; \frac{dn}{dt} & = -DA \frac{C_2-C_1}{\Delta x} \\ \therefore \underbrace{\frac{1}{A} \frac{dn}{dt}} & = -D \frac{C_2 - C_1}{\Delta x} \\ \therefore J & = -D \frac{C_2 - C_1}{\Delta x} \\ \therefore D & = \frac{J\Delta x}{-(C_2 - C_1)} \quad  \end {align}

Convert $C_2-C_1$ to $\frac{\text{mol}}{\text m^3}$ :

\begin {align} C_2-C_1 & = (0.05-0.50) \frac{\text {mol}}{\text{dm}^3} \\ & = - 0.45 \frac{\text{mol}}{\text{dm}^3} \times \bigg( \frac{10\; \text{dm}}{1\; \text m} \bigg)^3 \\ & = -4.5 \times 10^2 \frac{\text{mol}}{\text m}^3 \\ \therefore  \Rightarrow D & = \frac{(1.20\; \text{mol} \cdot \text m^{-2}\cdot \text s ^{-1})(10 \times 10^{-9}\; \text m)}{-(-4.5 \times 10^2 \text{mol/m}^3)} \\ & = 2.67 \times 10^{-11}\; \text{m}^2/\text s \end {align}

But the membrane is only $5.0\; \%$ pores.

\begin {align} \therefore \text {true} \; D & = \frac{100}{5.0} \times 2.67 \times 10^{-11} \text{m}^2/\text s \\ & = 5.3 \times 10^{-10} \text m^2/\text s \end {align}

#### Exercise 13-19

Osmotic pressure $\Pi = RTC \quad [\text{text equation 13-55}]$ where $C$ is the concentration in moles per cubic metre.

(OR: Start with equation 13-56, which is onthe formula sheet:$\Pi = RTC/M$ where $C$ is the concentration in kilograms per cubic metre and $M$ is the molar mass. Then write $C=m/V$, where $m$ is the mass of solute and $V$ is the volume of solution. Hence, $\Pi - RTm/MV= RTn/V = RTC$, where $n(= m/M)$ is the number of moles of solute, and $C(= n/V)$ is again the concentration in moles per cubic metre.)

For two types of solute, A and B, add the two osmotic pressures:

$\Pi = RT(C_A+C_B)$

When the two solutions are mixed, the total volume doubles and hence each concentration is halved.

$C_A = \frac{1}{2}\bigg( 0.50 \frac{\text{mol}}{\text{dm}^3} \times \frac{1000 \; \text{dm}^3}{1\; \text m^3} \bigg) = 250 \frac{\text{mol}}{\text m^3}$

The given concentration of solution B is $0.75 \; \text{mol/dm}^3$ of NaCl. However, in the equation $\Pi = RTC$, the concentrations must be expressed as total number of moles of solute entities per unit volume. Therefore, since NaCl ionizes, the concentration of solution B must first be doubled: $2(0.75\; \text{mol/dm}^3)= 1.5\; \text{mol/dm}^3$. Then, dividing the concentration by two because the volume doubles upon mixing, and converting to $\text{mol/m}^3$:

$C_B = \frac{1}{2}\bigg( 1.5 \frac{\text{mol}}{\text{dm}^3} \times \frac{1000\; \text{dm}^3}{1\; \text m^3}\bigg) = 750 \frac{\text{mol}}{\text m^3}$

The osmotic pressure is:

\begin {align} \Pi & = RT(C_A + C_B) \\ & = (8.315 J \cdot K^{-1} \cdot \text{mol}^{-1})(277 K) \bigg( 250 \frac{\text{mol}}{\text m^3} + 750 \frac{\text{mol}}{\text m^3}\bigg) \\ & = 2.3 \times 10^6\; Pa \end {align}

#### Exercise 13-20

(a) The osmolarity of a solution is the number of moles of solute entities per litre of solution. Since the glycerol does not ionize, the $0.500$ osmoles. Because the CaCl ionizes, the $0.500$ molar solution of NaCl prepared in water at $4^\circ \text C$ has osmolarity of $1.000$ osmoles.

It is normally assumed that a dilute solution of any solute in wateer at $4^\circ \text C$ has the same density as the water alone $(1000\; \text{kg/m}^3 = 1.000\; \text{kg/dm}^3 = 1.000 \; \text{kg/L})$, and hence $1 \text L$ of either the glycerol solution of the NaCl solution has a mass of $1\text{kg}$ at $4^\circ \text C$.

At $60^\circ \text C$ the water has undergone thermal expansion and has a density of only $983\; \text{kg/m}^3 = 0.983\; \text{kg/dm}^3= 0.983\; \text{kg/L}$. This meant that $1\text{kg}$ of solution, which had a volume of $1\text L$ at $4^\circ \text C$, now has a volume greater than $1\text L$. This volume is given by:

$\text{volume}= \frac{\text{mass}}{\text{density}} = \frac{1\; \text{kg}}{0.983\; \text{kg/L}} = \bigg(\frac{1}{0.983} \bigg) \text L = 1.017 \; \text L$

The osmolarity of the glycerol solution at $60^\circ \text C$ is then

$\frac{0.500\; \text{osmoles}}{1.017\; \text L} = 0.492\; \frac{\text{osmoles}}{\text L} = 0.493\; \text{osmolar}$

Similarly, the osmolarity of the NaCl solution at $60^\circ \text C$ is

$\frac{1.000\; \text{osmoles}}{1.017\; \text L} = 0.983\; \frac{\text{osmoles}}{\text L}= 0.983\; \text{osmolar}$

(b) The osmolarity of a solution is the number of moles of solute entities per kilogram of solvent. For a dilute solution in water at $4^\circ \text C$, the osmolality of the solution is the same as the osmalarity because $1\; \text L$ of solution contains $1\text{ kg}$ of solvent. Thus, the osmolality of the glycerol solution is $0.500$ osmolal, and the osmolality of the NaCl solution is $1.000$ osmolal.

As either the glycerol of NaCl solution expands in the volume as the temperature changes from $4^\circ \text C$ to $60^\circ \text C$, the number of kilorgrams of solvent remains the same. Hence, the osmolality does not change, and so the osmolality of either solution is the same at $60^\circ \text C$ as it was at $4^\circ \text C$

#### Exercise 13-25 \begin {align} F_{Buoyant}= \rho_wV_g, \text{(where w is water)}, \text{where}\; V &= \text{volume of bacterium} \\ & = \frac{m}{\rho}\quad  \end {align}

\begin {align} \text{Apparent weight} & = mg-\rho_w V_g \\ & = mg-\rho_w \bigg( \frac{m}{\rho} \bigg)g \quad (\text{using} )\\& = mg \bigg( 1-\frac{\rho_w}{\rho} \bigg) \\ & = (2.0 \times 10^{-15 \text{kg}})(9.8\text{m/s}^2)\bigg( 1 - \frac{1000\; \text{kg/m}^3}{1050 \; \text{kg/m}^3} \bigg) \\ &= 9.3 \times 10^{-16}\; \text N \end {align}

#### Exercise 13-27

(a) $\uparrow + y\\ \sum F_y = m\;a_y\\ \therefore \; m_w g-m_i g = m_ia_y \\ \text{Write}\; m_w = \rho _wV\; \text{and} \; m_i = \rho_i V, \\ \text{where}\; V = \text{volume of ice.} \\ \therefore\; \rho_wV_g- \rho_iV_g = \rho_iVa_y\\ \div V \rightarrow\quad \rho_wg-\rho_ig = \rho_ia_y\\ \div \rho_i \rightarrow \quad \frac{\rho_w}{\rho_i} g-g= a_y \\ \therefore a_y = g\bigg( \frac{\rho_w}{\rho_i}-1 \bigg) = (9.80\; \text{m/s}^2)\bigg( \frac{1000}{920}-1 \bigg) = 0.85\; \text{m/s}^2$

(b) At terminal speed: $v_y = \text{constant}\\ \therefore a_y = 0\\ \therefore \sum F_y = ma_y = 0\\ m_wg-m_ig- \mathcal{F} v_t=0\quad  \\ g, m_i, \rho_i, \rho_w, \text{and}\;\mathcal F \; \text{are known, but}\;m_w\; \text{is not known}.$

$\therefore \text{write}\; m_w = \rho_w V \quad (\text{where V = volume of ice}) \\ \text{But} \; V = \frac{m_i}{\rho_i} \; \therefore m_w = \rho_w \frac{m_i}{\rho_i} \; \cdot \; \text{Subst. in} .\\ \therefore \rho_w \frac{m_i}{\rho_i}g - m_ig = \mathcal F v_t$

\begin{align} \therefore v_t & = \frac{m_ig}{\mathcal F} \bigg( \frac{\rho_w}{\rho_i}- 1 \bigg) \\ &= \frac{(2.0\; \text{kg})(9.8\; \text{m/s}^2)}{3.0\; \text{kg/s}} \bigg( \frac{1000\; \text{kg/m}^3}{920\; \text{kg/m}^3}-1 \bigg)\\ & = 0.57\; \text{m/s} \end {align}

#### Exercise 13-28

(a)

\begin {align} v_t & = \frac{8\pi ^2 f^2 Rr^2(\rho-\rho_{\ell})}{9 \eta} \quad (\text{Eq. 13-19}) \\ \text{or}\; v_t & =\frac{2r^2 "g"}{9 \eta} (\rho-\rho_\ell) \quad (\text{Eq. 13-16}), \text{where} "g" = 4\pi^2f^2R \end {align}

All quantities are known except $r$.

\begin {align} \text{molar mass}\; M & = m\; N_A \quad (m=\text{mass of 1 molecule})\\ & = \rho V N_A \\ & = \rho \frac{4}{3}\pi r^3 N_A \end {align}

\begin {align} \therefore r & = \sqrt{\frac{3M}{4\rho \pi N_A}}\\ & = \sqrt{\frac{3(4.5\times 10^3 \text{kg/mol})}{4(1.33\times 10^3\text{kg/m}^3)\pi (6.02 \times 10^{23}\text{mol}^{-1})}} \\ & = 1.10 \times 10^{-8} \text m \\ \text{Then,}\; v_x & = \frac{8\pi ^2 f^2Rr^2 (\rho-\rho_\ell)}{9 \eta} \\ & = \frac{8 \pi^2(1.2 \times 10^3Hz)^2(0.14\text m)(1.10 \times 10^{-8}\text m)^2[(1330-1000)\text{kg/m}^3]}{9(0.0010 \; \text N \cdot g/\text m^2)} \\ & = 7.1 \times 10^{-5} \text{m/s} \quad (7.0_6 \times 10^{-5}\text{m/s}) \end {align}

(b)

\begin {align} v_t & = \frac{d}{t} \quad (\text{since} \; v_t\; \text{is constant}) \\ \therefore \; t & = \frac{d}{v_t} = \frac{1.5 \times 10^{-3} \text m}{7.0_6 \times 10^{-5}\text{m/s}} = 21 \; \text s \end {align}

#### Exercise 13-29

Subscripts: $l$arge, $s$mall

Surface areas:

\begin {align} \frac{A_l}{A_s} & = \frac{4 \pi r_l{^2}}{4 \pi r_s{^2}} \\ \therefore \; 2.0 & = \frac{r_l{^2}}{r_s{^2}}\\ \therefore \; \frac{r_l}{r_s} & = \sqrt{2.0} = 1.4 \end {align}

Diffusion coefficients:

$\frac{D_s}{D_l} = \frac{\frac{k_BT}{6\pi \eta r_s}}{\frac{k_BT}{6 \pi \eta r_L}} = \frac{r_L}{r_s} = 1.4$

#### Exercise 13-30

Let time $t = 0$ be the time at which the odour is created.

Let time $t_1$, be the time at which the first smell is detected.

\begin {align} \overline {R_1{^2}} & = 6Dt_1 \\ \therefore (3.0\; \text m)^2 & = 6Dt_1 \\ \therefore \; 9.0 & = 6Dt_1 \quad  \end {align}

At time $t_2$ the second smell is detected, where $t_2 = t_1 + 60\;\text s.$

\begin {align} \overline{R_2{^2}} & = 6Dt_2 \\ \therefore (5.0\;\text m)^2 & = 6D(t_1+ 60)\\ \therefore 25 & = 6Dt_1 + 360D \\ \text{But}\; 6Dt_1 & = 9.0, \; \text{from} . \\ \therefore \; 25 & = 9.0 + 360 D \\ \Rightarrow D & = 0.044\; \text m^2/ \text s \end {align}

#### Exercise 13-31

Dry molar mass $M_D$ can be determined from Eq. 13-48:

$M_D = \frac{s N_Ak_BT}{D(1-\rho_\ell/\rho)}$

Hydrated molar mass $M_H$ can be determined from Eq. 13-46:

\begin {align} M_H & = \frac{4}{3} \pi \rho N_A \bigg( \frac{k_BT}{6 \pi \eta D} \bigg)^3 \\ \therefore \frac{M_H}{M_D} & = \frac{\frac{4}{3}\pi \rho N_A \bigg( \frac{k_BT}{6 \pi \eta D} \bigg)^3}{\bigg( \frac{sN_Ak_BT}{D(1-\rho_\ell/\rho)} \bigg)} \\ & = \frac{4 \pi \rho N_Ak_B{^3}T^3}{3(6 \pi \eta)^3 D^3} \times \frac{D(1 - \rho_\ell /\rho)}{sN_Ak_BT} \\ & = \frac{4 \pi k_B{^2}T^2(\rho-\rho_\ell)}{3(6 \pi \eta)^3D^2s} \end {align}

\begin{align} \therefore \; \frac{M_H}{M_D} & = \frac{4\pi (1.38\times 10^{-23}\text{J/K})^2(293 \; \text K)^2((1420-1000)\text{kg/m}^3)}{3(6\pi(0.0010\; \text N \cdot \text s/ \text m^2))^3(6.67 \times 10^{-12}\text m^2/\text s)^2(435 \times 10^{-13}\text s)} \\ & = 2.2 \end {align}