The coefficient of dynamic viscosity $\eta$ is defined by the relation
$$F_{fr} = \eta \frac{\Delta v}{\Delta r}S,$$Thus $\eta$ has the units:

Let us write down the dimensions of the quantities:

$$ [Q] = L^3 T^{-1},\qquad [R]=L,\qquad [G]=\frac{[\Delta P]}{[L]}=M L^{-2} T^{-2},\qquad [\eta]=M L^{-1} T^{-1}.$$ We require that the dimensions on the right-hand side match those of $[Q]$: $$ L^3 T^{-1} = L^{a} (M L^{-2} T^{-2})^{b} (M L^{-1} T^{-1})^{c}. $$ Comparing the exponents for the fundamental dimensions:

1. for mass $M$: $b + c = 0 \;\Rightarrow\; c = -b$;
2. for time $T$: $-1 = -2b - c$;
3.  for length $L$: $3 = a - 2b - c$.

Substituting $c=-b$ into the equation for time, we get: $$-1 = -2b -(-b) = -b \Rightarrow b=1,~c=-1.$$

Substituting $b=1$ and $c=-1$ into the equation for length: $$3 = a -2 -(-1)  \Rightarrow a=4.$$

Thus, we obtain:

Consider a selected cylinder of fluid with radius $r$ and length $L$. The pressure $P+\Delta P$ acts on the left end face, and $P$ acts on the right end face, so the total pressure difference $\Delta P$ acting on the element is directed from the left end face to the right end face.
The pressure force acting on the end face of the cylinder (projection onto the axis) is equal to the pressure multiplied by the cross-sectional area:

For a stationary flow, the sum of the longitudinal forces acting on the considered cylinder is zero. The longitudinal forces are the pressure force acting on the end ($+\Delta P\pi r^2$), and the total viscous friction force acting on the lateral surface of the cylinder. Consequently:
$$
F_p - F_{fr} = 0.
$$The friction force on the lateral surface of a thin layer:
$$
F_{fr} = \eta\,\frac{-\Delta v}{\Delta r}\cdot 2\pi r L.
$$Substituting $F_p=\Delta P\pi r^2$ and $F_{fr}$ into the equilibrium equation:
$$
\Delta P\,\pi r^2 = \eta\,\frac{-\Delta v}{\Delta r}\, 2\pi r L.
$$Hence:
$$
\frac{-\Delta v}{\Delta r} = - g(r) = \frac{\Delta P}{2\eta L}\, r.
$$Denote $k=\dfrac{\Delta P}{2\eta L}$, then
$$
g(r)=-k r.
$$

Ответ:

We obtained that $g(r)=-kr$ is a direct proportionality.
The decrease in speed from $r$ to the wall $R$ is the sum of small changes in speed; geometrically, this is the area of the trapezoid under the line $g$ on the interval $[r,R]$:
$$
v(R)-v(r)=\frac{(g(r)+g(R))}{2}\cdot(R-r)=\frac{-(kr+kR)}{2}(R-r).
$$Since $v(R)=0$:

The maximum speed is achieved at the tube axis at $r=0$. Substituting $r=0$ into the expression for $v(r)$:

Note that
\[
v(r) = \frac{\Delta P}{4\eta L}\,(R^2 - r^2)
\]
is a linear function of the variable $x=r^2$:
\[
v(x) = \frac{\Delta P}{4\eta L}\,(R^2 - x),\qquad x=r^2.
\]
Therefore, the graph of $v$ as a function of $x=r^2$ is a straight line, which has a value of $v_{\max}$ at $x=0$ and zero at $x=R^2$.

Ответ:

Since $v(x)$ is linear on the interval $0\le x\le R^2$, the area under the graph is
\[
\text{Area} = \frac{v(0)+v(R^2)}{2}\cdot (R^2 - 0) = \frac{v_{\max}+0}{2}\,R^2 = \frac{v_{\max}\,R^2}{2}.
\]

For a thin annular layer
$$\Delta Q = 2\pi r\,\Delta r\cdot v(r) = \pi v(r)\,\Delta(r^2),$$that is, the area element under the graph of $v(r^2)$ multiplied by $\pi$.
Then the total volumetric flow rate is:
\[
Q = \pi \cdot \text{Area} = \pi \cdot \frac{v_{\max}\,R^2}{2}.
\]
Substituting $v_{\max}=\dfrac{\Delta P}{4\eta L}R^2$:

By definition, $Z = \dfrac{\Delta P}{Q}$. Using the expression for $Q$:

In a parallel connection of tubes, the total pressure difference is the same for each branch, and the flow rate is distributed among the branches. If all $N$ narrow tubes are identical and the total flow rate through them is $Q$, then the flow rate through each tube is:

From the expression for the fluid flow rate in a single tube:
\[
\frac{\pi \Delta P_2 (\alpha R)^4}{8\eta \beta L}=\frac{\pi \Delta P_1 R^4}{8\eta LN}.
\]
Hence:

Using the initial data and the table, we calculate the resistance of each level:
$$Z = \frac{8\eta L}{\pi R^4n}.$$

Similarly, we calculate the pressure difference for each tube:
$$\Delta P = Q_0 \cdot Z$$

The resistance depends on the radius as \(Z \propto 1/R^4\).
\[
\frac{Z_{\rm new}}{Z_{\rm old}} = \frac{1}{1.2^4} \approx 0.482
\]

New resistance of arterioles: $Z_{\rm new} = 1.51\times10^8 \cdot 0.482 \approx 7.28\times10^7$ Pa$\cdot$s/m${}^3$
Total resistances of the network:
$Z_{\rm tot,\;new} = Z_{\rm aorta} + Z_{\rm major\;art.} + Z_{\rm new} + Z_{\rm capillaries} \approx 8.78\times10^7$ Pa$\cdot$s/m${}^3$
$Z_{\rm tot,\;old} = Z_{\rm aorta} + Z_{\rm major\;art.} + Z_{\rm old} + Z_{\rm capillaries} \approx 16.6\times10^7$ Pa$\cdot$s/m${}^3$
Then:
$$\frac{\Delta P_{\rm new}}{\Delta P_0}=4\frac{Z_{\rm tot,\;new}}{Z_{\rm tot,\;old}},$$where $\Delta P_0$ is the total pressure difference at rest.

Let's calculate the power:
$$W=F\cdot v=\Delta P S v=Q\Delta P$$Then:
$$
W_0 = Q_0 \Delta P_0;~W_{\rm new} = Q \Delta P_{\rm new} $$

The figure below schematically depicts the circulatory systems of the following vertebrates:

1. Amphibians;
2. Mammals;
3. Fishes;

Match the numbers 1-3 in the list with the letters A-C in the diagram. Each letter can be used only once.

Estimate the effective surface area of the lungs $S$.

Use $D = 10^{-11}$ m${}^2$/s, $d = 1$ μm.

The number of moles of oxygen consumed per unit time can be obtained from
\[
\dfrac{\Delta N}{\Delta t} = jS \Rightarrow \dfrac{\Delta \nu}{\Delta t} = \dfrac{jS}{N_A}.
\]
The difference in oxygen concentration between the alveolar air and the capillaries is given by
\[
\Delta n = \Delta p /kT.
\]
Thus, the power generated is equal to
\[
W = Q\dfrac{\Delta \nu}{\Delta t} = \dfrac{QjS}{N_A} = \dfrac{QS}{N_A}\cdot D\dfrac{\Delta n}{d} = \dfrac{QSD\Delta p}{N_AdkT} \Rightarrow S = \dfrac{WRTd}{QD\Delta p} \approx 70 \text{m}^2.
\]
We used the relation $R = kN_A$ at the final step.

• (A) The equilibrium equation in the system: \[ \text{H}_2\text{CO}_3\rightleftharpoons \text{H}\text{CO}_3^- + \text{H}^+ \] Let's create a table of molarities of the reacting substances.

\[ K_{a_1} = \dfrac{[\text{H}^+][\text{H}\text{CO}_3^-]}{[\text{H}_2\text{CO}_3]} = \dfrac{x^2}{0.15 - x} \Rightarrow x = 6.22\cdot 10^{-3}~\text{M},~\text{pH}_A~2.20. \]

• (B) For the buffer solution
  \[
  K_{a_1} = \dfrac{[\text{H}^+][\text{H}\text{CO}_3^-]}{[\text{H}_2\text{CO}_3]}\approx\dfrac{[\text{H}^+]C_{salt}}{C_{acid}} \Rightarrow \text{pH}_B 3.57.
  \]

• (A) It is necessary to take the dilution of the solutions into account. New molarities:

$$[\text{H}_2\text{CO}_3] = 0.15 \cdot 1/1.2 = 0.125~\text{M},\ [\text{HCl}] = 0.1 \cdot 0.2/1.2 \approx 0.017~\text{M}.$$

 

\[
K_{a_1} = \dfrac{[\text{H}^+][\text{H}\text{CO}_3^-]}{[\text{H}_2\text{CO}_3]} = \dfrac{x(0.017 + x)}{0.125 - x} \Rightarrow x = 1.77\cdot 10^{-3} \text{M}, [\text{H}^+]_A = 0.018 \text{M}, \text{pH}_A 1.73.
\]
\[
\Delta \text{pH}_A = -0.47
\]

• (B) A reaction occurs between the strong acid and the salt:
  \[
  \text{NaHCO}_3 + \text{HCl} \rightarrow \text{H}_2\text{CO}_3 + \text{NaCl}.
  \]
  Thus, the new molarity of the weak acid is
  \[
  (0.15 \cdot 1 + 0.1 \cdot 0.2)/1.2 \approx 0.142   \text{M},
  \]
  and the molarity of the salt is
  \[
  (0.15 \cdot 1 - 0.1 \cdot 0.2)/1.2 \approx 0.108   \text{M}.
  \]
  Then
  \[
  [\text{H}^+]_B \approx \dfrac{K_{a_1} C_{acid}}{C_{salt}}
  = \dfrac{10^{-3.57} \cdot 0.142}{0.108}
  \approx 3.54 \cdot 10^{-4}   \text{M}
  \Rightarrow \text{pH}_B   3.45. \\ \Delta \text{pH}_B = -0.12.
  \]
  Thus, the buffer solution performs well in maintaining a constant $\text{pH}$.
   

According to Henry's law, $[\text{CO}_2] = kp(\text{CO}_2) = 1.22\cdot10^{-3}~\text{M}$.

$[\text{H}_2\text{CO}_3] = K_h[\text{CO}_2] = 3.66 \cdot 10^{-6}~\text{M}$.
$[\text{HCO}_3^-] = \dfrac{K_{a_1} [\text{H}_2\text{CO}_3]}{[\text{H}^+]} = \dfrac{K_{a_1} K_h[\text{CO}_2]}{[\text{H}^+]} \approx 0.025~\text{M}$.

\[
s = [\text{CO}_2] + [\text{H}_2\text{CO}_3] + [\text{HCO}_3^-] \approx 0.026 \text{M}.
\]
Alternatively, one can notice that $[\text{CO}_2] \ll [\text{HCO}_3^-]$ and $[\text{H}_2\text{CO}_3] \ll [\text{HCO}_3^-]$, so approximately
\[
s \approx [\text{HCO}_3^-] = 0.025 \text{M}.
\]

\[
K_a = \dfrac{[\text{H}^+][\text{HbO}_2^-]}{[\text{HbO}_2\text{H}]} \Rightarrow [\text{HbO}_2^-] = \dfrac{K_a}{[\text{H}^+]} [\text{HbO}_2\text{H}]
\]
\[
\alpha_{O_2} = \dfrac{[\text{HbO}_2\text{H}]}{[\text{HbO}_2\text{H}] + [\text{HbO}_2^-]} = \dfrac{1}{1+\dfrac{K_a}{[\text{H}^+]}} = 0.86.
\]
Similarly $\beta = 0.14$.

Mass fraction of ferrum:
\[
w(\text{Fe}) = 100\% - (w(\text{C}) + w(\text{H}) + w(\text{O}) + w(\text{N})) =9.06\%.
\]
From this information the total molecular mass can be derived:
\[
M = \dfrac{M(\text{Fe})}{w(\text{Fe})}\approx 616.4 \text{g}/\text{mol}.
\]
For unknown numbers we have:
\[
w(\text{C}) = \dfrac{aM(\text{C})}{M} \Rightarrow a \approx 34.
\]
Another approach without calculation the total molecular mass:
\[
\dfrac{a}{1} = \dfrac{w(\text{C})/M(\text{C})}{w(\text{Fe})/M(\text{Fe})} \approx 34.
\]
Similarly, $b = 32,~c = 4,~d = 4$.

Below is a list of chemical bonds that appear in the structures mentioned above. Write down the letters representing the chemical bonds in the appropriate cells of the table below. Note that the same letters may be used in different cells.

A. Disulfid bridges (–S–S–).

B. Ionic bonds.

C. Hydrogen bonds within a molecule.

D. Hydrophobic interactions.

E. Peptide bonds.

F. Intermolecular hydrogen bonds.

\[
K_a = \dfrac{[\text{PL}]}{[\text{P}][\text{L}]} \Rightarrow \theta = \dfrac{[\text{PL}]}{[\text{PL}] + [\text{P}]} = \dfrac{K_a [\text{P}][\text{L}]}{K_a [\text{P}][\text{L}] + [\text{P}]} = \dfrac{[\text{L}]}{[\text{L}] + 1/K_a}.
\]
From the obtained expression, it is seen that $\theta = 0.5$ when $[\text{L}] = [\text{L}]_{0.5} = 1/ K_a\Rightarrow K_a = 1 / [\text{L}]_{0.5}$.

The Van't Hoff equation:
\[
\ln K_a = -\dfrac{\Delta_r H^\circ}{RT} + \text{const}.
\]
Thus, to answer the question, it is necessary to plot the dependence $\ln K_a(1/T)$ and determine its slope $\alpha$:
\[
\Delta_r H^\circ = -R\dfrac{\Delta \ln K_a}{\Delta (1/T)} = -\alpha R.
\]

According to the table, the dependence $\ln K_a (1/ T)$ can be plotted.

Ответ:

From the graph, we determine:
\[
\alpha \approx 6600 \text{К}\Rightarrow \Delta_r H^\circ \approx -55 \text{kJ}/\text{mol}.
\]

Similarly to point G4, we have
\[
\theta = \dfrac{[\text{L}]^n}{[\text{L}]^n + 1/K_a} \Rightarrow \dfrac{\theta}{1 - \theta} = K_a[\text{L}]^n\Rightarrow \lg\left(\dfrac{\theta}{1 - \theta}\right) = n\log_{10}[\text{L}] + \log_{10} K_a.
\]

Ответ:

From the slope of the tangent line, we find $n_H\approx 2.7.$ The upper theoretical limit corresponds to full cooperativity and is equal to the number of binding sites $(n_{H})_{max}= 4$.

Give the reason why myoglobin cannot be used as an effective oxygen carrier (efective binding and release of oxygen molecules) from the lungs to the tissues.

A. The myoglobin molecule has a hyperbolic oxygen saturation curve.

B. The concentration of myoglobin in the blood is significantly lower than the concentration of hemoglobin.

C. The myoglobin molecule is lighter, which makes it too mobile.

D. The myoglobin molecule is too small, which can cause it to enter other tissues.

The sigmoidal nature of the oxygen dissociation curve of hemoglobin provides optimal conditions for oxygen release during the transition from areas of high partial pressure in the pulmonary capillaries to zones of low partial pressure in peripheral tissues. Under physiological conditions (temperature 37°C, pH 7.4), the partial pressure of oxygen is about 100 mm Hg in arterial blood and decreases to 40 mm Hg in venous blood. In this range, hemoglobin releases about 23% of the bound oxygen, which corresponds to a decrease in the degree of oxygenation from 98% to 75%. In contrast to hemoglobin, monomeric myoglobin, lacking cooperative interactions, demonstrates a significantly lower oxygen release efficiency — no more than 5%. This fundamental difference determines the physiological role of myoglobin as a reserve oxygen depot, which is mobilized only during a critical decrease in tissue oxygenation during pronounced hypoxia. Thus, the correct answer is A.

Choose the correct statement:

1. Curve A corresponds to $\text{pH}~7.2$, curve B corresponds to $\text{pH}~7.6$.


2. Curve A corresponds to $\text{pH}~7.6$, curve B corresponds to $\text{pH}~7.2$.

An increase in pH leads to a decrease in proton concentration, thus, according to Le Chatelier's principle, the equilibrium shifts towards the products, i.e., the degree of hemoglobin saturation increases.

Which curve corresponds to hemoglobin in the lungs, and which corresponds to hemoglobin in tissues (mark with an "X" in the table in the answer sheets).

Due to the increased concentration of carbon dioxide, the environment in the peripheral tissues is more acidic than in the lungs. Considering the previous point, we find that curve A corresponds to the lungs, and B corresponds to the peripheral tissues. This explains hemoglobin's high ability to bind oxygen in the lungs and release it in the tissues.

In certain conditions, increased level of myoglobin in blood could be a result of:

A. State of alcohol intoxication.

B. Myocardial infarction.

C. Use of sleeping pills.

D. Alzheimer's disease.

Of all the above, only in myocardial infarction does damage to the cells of the muscle tissue (cardiomyocytes of the heart muscle) occur, as a result of which myoglobin molecules enter the bloodstream.