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Chapter 25: Problem 31

(a) What pressure difference is required to inject a sample equal to \(1.0 \%\) of the length of a \(60.0-\mathrm{cm}\) capillary in \(4.0 \mathrm{~s}\) if the diameter is \(50 \mu \mathrm{m}\) ? Assume that the viscosity of the solution is \(0.0010 \mathrm{~kg} /(\mathrm{m} \cdot \mathrm{s})\) (b) The pressure exerted by a column of water of height \(h\) is \(h \rho g\), where \(\rho\) is the density of water and \(g\) is the acceleration of gravity \(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right) .\) To what height would you need to raise the sample vial to create the necessary pressure to load the sample in \(4.0 \mathrm{~s}\) ? Is it possible to raise the inlet of this column to this height? How could you obtain the desired pressure?

Short Answer

Expert verified
Calculate pressure difference using Poiseuille's law, then use the pressure-height formula to find required height. If impractical, use a pressure pump.

Step by step solution

01

Understand the Problem

We need to find the pressure difference required to push a sample equal to 1% of a 60 cm capillary in 4 seconds. The capillary diameter is given as 50 碌m, and the viscosity of the solution is 0.0010 kg/(m路s). We also need to calculate the height of a water column that would create this pressure difference using \( h \rho g \).
02

Calculate the Sample Volume

To calculate the length of the sample to be injected, we first determine 1% of the capillary鈥檚 length: \[ L = 0.01 imes 60 ext{ cm} = 0.6 ext{ cm} = 0.006 ext{ m} \] Then, calculate the volume of the capillary that corresponds to this length: \[ V = \pi (r^2) L \]where \( r = \frac{50 \mu m}{2} = 25 \times 10^{-6} \text{ m} \). Substituting the values gives:\[ V = \pi \times (25 \times 10^{-6})^2 \times 0.006 \text{ m} \]
03

Apply Poiseuille's Law

Poiseuille's equation for flow rate through a capillary is:\[ Q = \frac{\pi \Delta P r^4}{8 L \eta} \]Solving for \( \Delta P \), the pressure difference, we have:\[ \Delta P = \frac{8 Q L \eta}{\pi r^4} \]Using the flow rate \( Q = \frac{V}{t} \) and substituting into the equation, calculate \( \Delta P \) using the previously found volume, \( t = 4 \, \text{s} \), and given \( \eta = 0.0010 \, \text{kg}/(\text{m} \cdot \text{s}) \).
04

Calculate Pressure Using Column Formula

The pressure exerted by a column of water is calculated using:\[ \Delta P = h \rho g \] where \( \rho = 1000 \, \text{kg/m}^3 \) and \( g = 9.8 \, \text{m/s}^2 \). Solve for \( h \) to find the height needed to create the calculated pressure from Step 3.
05

Assess Feasibility and Alternatives

Evaluate if it is practical to raise the inlet to the calculated height. If it is impractical, consider alternatives, such as using a pressure pump to achieve the desired pressure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capillary Flow
Capillary flow refers to the movement of fluid through a tube with a very small diameter, or a capillary. It is an important phenomenon in fields like medicine, biology, and engineering, where fluids need to be moved through narrow spaces. In our problem, we deal with the flow through a capillary that is 60 cm long and has a diameter of 50 碌m. Such small dimensions mean that even a tiny movement of fluid requires careful consideration of various physical forces at play.

One of the key aspects of capillary flow is that it is significantly affected by the surface tension of the fluid and the adhesion between the fluid and the tube. Capillary action can cause liquids to rise without external forces, as seen in the absorption of water by a paper towel or the movement of water within the vessels of plants. However, in controlled environments like a laboratory setup, an external pressure difference is usually applied to regulate capillary flow, ensuring precise amounts of fluid movement required for experiments.
Pressure Difference
The pressure difference is a critical factor in driving fluid flow through a capillary. In our given problem, we need to generate an adequate pressure difference to push a specific amount of fluid through the capillary in a set time.

According to Poiseuille's Law, which describes the flow of a viscous fluid in a cylindrical pipe, the pressure difference across a capillary can be calculated using:
  • the viscosity of the fluid
  • the length and radius of the capillary
  • the desired flow rate
The formula \[\Delta P = \frac{8 Q L \eta}{\pi r^4}\]allows us to calculate the necessary pressure to achieve capillary flow. Here, the pressure drives the fluid, compensating for the resistance posed by the fluid鈥檚 viscosity. In practical applications, this may involve using a syringe or external pump, creating a uniform pressure throughout the fluid column.
Viscosity
Viscosity is a measure of a fluid's resistance to flow, often described as its "thickness." Imagine syrup flowing through a thin straw compared to water. Syrup, with higher viscosity, moves slower due to greater internal friction.

In the given example, the solution's viscosity is 0.0010 kg/(m路s). This value must be factored into the equations to determine how much pressure is needed to move the fluid through the capillary. Higher viscosity means the fluid offers more resistance and requires a larger pressure difference to maintain the same flow rate. This concept is akin to the difference in effort needed to stir honey versus water; more viscous substances demand more force to move at the same speed.
Fluid Dynamics
Fluid dynamics is the study of how fluids (liquids and gases) move and the forces acting upon them. It encompasses a wide range of phenomena from ocean currents to blood flow in arteries and is a critical field in engineering and physical sciences.

In analyzing fluid flow through a capillary, we apply principles of fluid dynamics, including Poiseuille's Law. This law helps in modeling the movement and calculating flow rates and pressure differences in cylindrical tubes. Fluid dynamics considers both the viscosity and geometry of the container through which the fluid is moving, aiding in predictions of flow behavior.
  • Fluid characteristics: density, viscosity, surface tension
  • Container characteristics: diameter, shape, texture
  • External forces: gravity, pressure differences
By understanding these factors, one can design experiments and machinery that effectively control and utilize fluid movement.

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Most popular questions from this chapter

Benzoic acid containing \({ }^{16} \mathrm{O}\) can be separated from benzoic acid containing \({ }^{18} \mathrm{O}\) by electrophoresis at a suitable \(\mathrm{pH}\) because they have slightly different acid dissociation constants. The difference in mobility is caused by the different fraction of each acid in the anionic form, \(\mathrm{A}^{-}\). Calling this fraction \(\alpha\), we can write $$ \begin{array}{ll} \mathrm{H}^{16} \mathrm{~A} \rightleftharpoons \mathrm{H}^{+}+{ }^{16} \mathrm{~A}^{-} & \mathrm{H}^{18} \mathrm{~A} \rightleftharpoons \mathrm{H}^{+}+{ }^{18} \mathrm{~N}^{18} \mathrm{~A}^{-} \\ { }^{16} \alpha=\frac{{ }^{16} K}{{ }^{16} K+\left[\mathrm{H}^{+}\right]} & { }^{18} \alpha=\frac{{ }^{18} K}{{ }^{18} K+\left[\mathrm{H}^{+}\right]} \end{array} $$ where \(K\) is the equilibrium constant. The greater the fraction of acid in the form \(\mathrm{A}^{-}\), the faster it will migrate in the electric field. It can be shown that, for electrophoresis, the maximum separation will occur when \(\Delta \alpha / \sqrt{\alpha}\) is a maximum. In this expression, \(\Delta \alpha={ }^{16} \alpha-{ }^{18} \alpha\), and \(\bar{\alpha}\) is the average fraction of dissociation \(\left[=\frac{1}{2}\left({ }^{16} \alpha+{ }^{18} \alpha\right)\right] .\) (a) Let us denote the ratio of acid dissociation constants as \(R={ }^{16} K /{ }^{18} K .\) In general, \(R\) will be close to unity. For benzoic acid, \(R=1.020\). Abbreviate \({ }^{16} K\) as \(K\) and write \({ }^{18} K=K / R\). Derive an expression for \(\Delta \alpha / \sqrt{\bar{\alpha}}\) in terms of \(K,\left[\mathrm{H}^{+}\right]\), and \(R\). Because both equilibrium constants are nearly equal ( \(R\) is close to unity), set \(\bar{\alpha}\) equal to \({ }^{16} \alpha\) in your expression. (b) Find the maximum value of \(\Delta \alpha / \sqrt{\alpha}\) by taking the derivative with respect to \(\left[\mathrm{H}^{+}\right]\)and setting it equal to 0 . Show that the maximum difference in mobility of isotopic benzoic acids occurs when \(\left[\mathrm{H}^{+}\right]=(K / 2 R)(1+\sqrt{1+8 R})\) (c) Show that, for \(R \approx 1\), this expression simplifies to \(\left[\mathrm{H}^{+}\right]=2 K\), or \(\mathrm{pH}=\mathrm{p} K-0.30\). That is, the maximum electrophoretic separation should occur when the column buffer has \(\mathrm{pH}=\mathrm{p} K-0.30\), regardless of the exact value of \(R .{ }^{63}\)

A gel-filtration column has a radius, \(r\), of \(0.80 \mathrm{~cm}\) and a length, \(l\), of \(20.0 \mathrm{~cm}\). (a) Calculate the volume, \(V_{\mathrm{t}}\), of the column, which is equal to \(\pi r^{2} l\). (b) The void volume, \(V_{\mathrm{o}}\), was \(18.1 \mathrm{~mL}\), and the total volume of mobile phase was \(35.8 \mathrm{~mL}\). Find \(K_{\mathrm{uv}}\) for a solute eluted at \(27.4 \mathrm{~mL}\).

(a) A particular solution in a particular capillary has an electroosmotic mobility of \(1.3 \times 10^{-8} \mathrm{~m}^{2} /(\mathrm{V} \cdot \mathrm{s})\) at \(\mathrm{pH} 2\) and \(8.1 \times\) \(10^{-8} \mathrm{~m}^{2} /(\mathrm{V} \cdot \mathrm{s})\) at \(\mathrm{pH} 12\). How long will it take a neutral solute to travel \(52 \mathrm{~cm}\) from the injector to the detector if \(27 \mathrm{kV}\) is applied across the \(62-\mathrm{cm}\)-long capillary tube at \(\mathrm{pH} 2 ?\) At \(\mathrm{pH} 12 ?\) (b) An analyte anion has an electrophoretic mobility of \(-1.6 \times\) \(10^{-8} \mathrm{~m}^{2} /(\mathrm{V} \cdot \mathrm{s})\). How long will it take to reach the detector at \(\mathrm{pH} 2\) ? At pH 12 ?

Molecular mass by sodium dodecylsulfate-gel electrophoresis. Ferritin is a hollow iron-storage protein \({ }^{61}\) consisting of 24 subunits that are a variable mixture of heavy (H) or light (L) chains, arranged in octahedral symmetry. The hollow center, with a diameter of \(8 \mathrm{~nm}\), can hold up to 4500 iron atoms in the approximate form of the mineral ferrihydrite \(\left(5 \mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\right)\). Iron(II) enters the protein through eight pores located on the threefold symmetry axes of the octahedron. Oxidation to Fe(III) occurs at catalytic sites on the H chains. Other sites on the inside of the \(\mathrm{L}\) chains appear to nucleate the crystallization of ferrihydrite. Migration times for protein standards and the ferritin subunits are given in the table. Prepare a graph of \(\log (\) molecular mass \()\) versus \(1 /\) (relative migration time), where relative migration time \(=\) (migration time)/(migration time of marker dye). Compute the molecular mass of the ferritin light and heavy chains. The masses of the chains, computed from amino acid sequences, are 19766 and \(21099 \mathrm{Da}\). $$ \begin{array}{lll} \text { Protein } & \begin{array}{l} \text { Molecular } \\ \text { mass }(\text { Da }) \end{array} & \begin{array}{l} \text { Migration } \\ \text { time }(\mathrm{min}) \end{array} \\ \hline \text { Orange G marker dye } & \text { small } & 13.17 \\ \alpha \text {-Lactalbumin } & 14200 & 16.46 \\ \text { Carbonic anhydrase } & 29000 & 18.66 \\ \text { Ovalbumin } & 45000 & 20.16 \\ \text { Bovine serum albumin } & 66000 & 22.36 \\ \text { Phosphorylase B } & 97000 & 23.56 \\ \beta \text {-Galactosidase } & 116000 & 24.97 \\ \text { Myosin } & 205000 & 28.25 \\ \text { Ferritin light chain } & & 17.07 \\ \text { Ferritin heavy chain } & & 17.97 \end{array} $$

Protein charge ladder. Electrophoretic mobility is proportional to charge. If members of a charge ladder (Figure 25-26) have the same friction coefficient (that is, the same size and shape), then the charge of the unmodified protein divided by its electrophoretic mobility, \(z_{0} / \mu_{0}\), is equal to the charge of the \(n\)th member divided by its electrophoretic mobility \(\left(z_{0}+\Delta z_{n}\right) / \mu_{n}\). Setting these two expressions equal to each other and rearranging gives $$ \Delta z_{n}=z_{0}\left(\frac{\mu_{n}}{\mu_{0}}-1\right) $$ where \(z_{0}\) is the charge of the unmodified protein, \(\Delta z_{n}\) is the charge difference between the \(n\)th modified protein and the unmodified protein, \(\mu_{n}\) is the electrophoretic mobility of the \(n\)th modified protein, and \(\mu_{0}\) is the electrophoretic mobility of the unmodified protein. The migration time of the neutral marker molecule in Figure 25-26 is \(308.5 \mathrm{~s}\). The migration time of the unmodified protein is \(343.0 \mathrm{~s}\). Other members of the charge ladder have migration times of \(355.4\), \(368.2,382.2,395.5,409.1,424.9,438.5,453.0,467.0,482.0,496.4 .\) \(510.1,524.1,536.9,551.4,565.1,577.4\), and \(588.5 \mathrm{~s}\). Calculate the electrophoretic mobility of each protein and prepare a plot of \(\Delta z_{n}\) versus \(\left(\mu_{n} / \mu_{0}\right)-1\). If the points lie on a straight line, the slope is the charge of the unmodified protein, \(z_{0}\). Prepare such a plot, suggest an explanation for its shape, and find \(z_{0}\).
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