Biophysics Textbook Questions - Chapter 13

\(C_1 = 0.50 \frac{\text{mol}}{\text{dm}^3}\)

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

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

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

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

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

But the membrane is only \(5.0\; \%\) pores. 

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

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:

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

(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 \(4^\circ \text C\). 

\(\begin {align} F_{Buoyant}= \rho_wV_g, \text{(where w is water)}, \text{where}\; V &= \text{volume of bacterium} \\ & = \frac{m}{\rho}\quad [1] \end {align}\)

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

(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:

\(v_y = \text{constant}\\ \therefore a_y = 0\\ \therefore \sum F_y = ma_y = 0\\ m_wg-m_ig- \mathcal{F} v_t=0\quad [1] \\ g, m_i, \rho_i, \rho_w, \text{and}\;\mathcal F \; \text{are known, but}\;m_w\; \text{is not known}.\)

\(\therefore \text{write}\; m_w = \rho_w V \quad (\text{where V = volume of ice}) \\ \text{But} \; V = \frac{m_i}{\rho_i} \; \therefore m_w = \rho_w \frac{m_i}{\rho_i} \; \cdot \; \text{Subst. in} [1].\\ \therefore \rho_w \frac{m_i}{\rho_i}g - m_ig = \mathcal F v_t \)

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

(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 \(r\).

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

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

(b)

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

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:

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

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.

\(\begin {align} \overline {R_1{^2}} & = 6Dt_1 \\ \therefore (3.0\; \text m)^2 & = 6Dt_1 \\ \therefore \; 9.0 & = 6Dt_1 \quad [1] \end {align}\)

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

\(\begin {align} \overline{R_2{^2}} & = 6Dt_2 \\ \therefore (5.0\;\text m)^2 & = 6D(t_1+ 60)\\ \therefore 25 & = 6Dt_1 + 360D \\ \text{But}\; 6Dt_1 & = 9.0, \; \text{from} [1]. \\ \therefore \; 25 & = 9.0 + 360 D \\ \Rightarrow D & = 0.044\; \text m^2/ \text s \end {align}\)

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:

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

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