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How long will it take to pass \(200 .\) mL of \(\mathrm{H}_{2}\) at \(273 \mathrm{K}\) through a \(10 .\) cm-long capillary tube of \(0.25 \mathrm{mm}\) if the gas input and output pressures are 1.05 and 1.00 atm, respectively?

Short Answer

Expert verified
The time required to pass 200 mL of hydrogen gas through the 10 cm long capillary tube with a diameter of 0.25 mm is approximately 63,492 seconds.

Step by step solution

01

Write down the given data

Given data includes: - Volume of gas, V = 200 mL = 0.2 L (1 mL = 0.001 L) - Temperature, T = 273 K - Length of capillary tube, L = 10 cm = 0.1 m - Diameter of capillary tube, d = 0.25 mm = 2.5 × 10^(-4) m - Input pressure, P1 = 1.05 atm - Output pressure, P2 = 1.00 atm
02

Find the pressure difference

Find the pressure difference (ΔP) between the input and output ends of the capillary tube: ΔP = P1 - P2 ΔP = \(1.05 - 1.00 = 0.05\) atm Now, let's convert the pressure difference ΔP from atm to Pascals (Pa) by multiplying with the conversion factor: ΔP = 0.05 atm × 101325 Pa/atm = 5066.25 Pa
03

Find the viscosity of the gas

We will use the Sutherland equation to calculate the viscosity (μ) of the hydrogen gas: μ = \(1.72 × 10^{-5} × (\frac{T^{3/2}}{T + 120})\), where T is the temperature in Kelvin. μ = \(1.72 × 10^{-5} × (\frac{273^{3/2}}{273 + 120})) = 8.81 × 10^{-6}\) Pa·s
04

Calculate the flow rate using Poiseuille's formula

According to Poiseuille's formula, the flow rate (Q) through a capillary tube is given by: Q = \(π \frac{ΔP d^4}{128 μ L}\), where ΔP is the pressure difference, d is the diameter, μ is the viscosity, and L is the length of the capillary tube. Plugging in the given values, we get: Q = \(π \frac{5066.25 (2.5 × 10^{-4})^4}{128 × 8.81 × 10^{-6} × 0.1}\) Q = 3.15 × 10^{-6} L/s
05

Calculate the time required to pass the gas volume

Finally, we can determine the time (t) needed to pass the 0.2 L volume of gas using the flow rate Q: t = \(\frac{V}{Q}\) t = \(\frac{0.2}{3.15 × 10^{-6}}\) = 63492.06 s The time required to pass 200 mL of hydrogen gas through the 10 cm long capillary tube with a diameter of 0.25 mm is approximately 63,492 seconds.

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Key Concepts

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

Understanding Capillary Tube Gas Flow
When it comes to gas transport in microfluidics, the understanding of capillary tube gas flow is crucial. A capillary tube, often used in medical and research settings, is characterized by its narrow diameter, which allows gases to flow through it due to a difference in pressure.

In our textbook solution, we examined the time it takes for hydrogen gas to pass through a small-diameter capillary tube. It's noteworthy that because a capillary tube has such a small diameter, the interaction between the gas molecules and the tube's walls has a significant effect on the flow. This interaction is largely due to the viscous properties of the gas, which we will explore further. The flow rate in such tubes can be described using a refined form of Poiseuille's law, which was originally defined for liquids, and is adapted for gases in capillary tubes.
The Role of Pressure Difference in Gases
In any system where gases flow from one area to another, a pressure difference is a driving force behind this movement. The concept of pressure difference is analogous to a hill where water flows downwards — a high-pressure area corresponds to the top of the hill, and a low-pressure area is like the bottom.

When we consider the movement of gas through a capillary tube, it's the pressure difference between the input and output of the tube that propels the gas forward. This pressure differential determines the flow rate of the gas, as shown in our textbook solution. When paired with Poiseuille's law for gas dynamics, an accurate computation of the flow rate can be achieved.
Viscosity of Gases and Its Influence on Flow
Viscosity, often described as the 'thickness' or internal friction of a fluid, is a measure of its resistance to deformation or flow. In gases, viscosity plays a critical role as it influences how gases interact with surfaces and with each other at a molecular level.

Unlike liquids, the viscosity of gases increases with temperature, which can be counterintuitive since we usually associate 'thinness' with heat and 'thickness' with cold. The Sutherland's equation, as used in our solution, provides a way to calculate the viscosity of gases at different temperatures, factoring in the complex intermolecular forces at play. Understanding the viscosity of a gas is essential when applying Poiseuille's law to determine the flow rate, particularly as it affects the amount of force required to move the gas through a tube.
Calculating Viscosity with Sutherland's Equation
Sutherland's equation is a formula used to calculate the viscosity of gases as a function of temperature. It stands out for accounting for the temperature dependence of gas viscosity, incorporating constants that are specific to each gas.

In the context of the exercise, Sutherland's equation was used to find the viscosity of hydrogen gas at a certain temperature. By having an accurate value for viscosity, calculations involving gas flow, such as using Poiseuille's law, become much more reliable. This equation underscores the intricate relationship between temperature and the kinetic energy of gas molecules, which directly impacts the gas's behavior during flow through capillary tubes or any other conduits.

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Most popular questions from this chapter

A current of 2.00 A is applied to a metal wire for 30. s. How many electrons pass through a given point in the wire during this time?

You are interested in purifying a sample containing the protein alcohol dehydrogenase obtained from horse liver; however, the sample also contains a second protein, catalase. These two proteins have the following transport properties at \(298 \mathrm{K}\). $$\begin{array}{lll} &{\text { Catalase }} & \text { Alcohol Dehydrogenase } \\ \hline \overline{\mathrm{s}}(\mathrm{s}) & 11.3 \times 10^{-13} & 4.88 \times 10^{-13} \\ D\left(\mathrm{m}^{2} \mathrm{s}^{-1}\right) & 4.1 \times 10^{-11} & 6.5 \times 10^{-11} \\ \bar{V}\left(\mathrm{cm}^{3} \mathrm{g}^{-1}\right) & 0.715 & 0.751 \end{array}$$ a. Determine the molecular weight of catalase and alcohol dehydrogenase. b. You have access to a centrifuge that can provide angular velocities up to 35,000 rpm. For the species you expect to travel the greatest distance in the centrifuge tube, determine the time it will take to centrifuge until a \(3 \mathrm{cm}\) displacement of the boundary layer occurs relative to the initial \(5 \mathrm{cm}\) location of the boundary layer relative to the centrifuge axis. c. To separate the proteins, you need a separation of at least \(1.5 \mathrm{cm}\) between the boundary layers associated with each protein. Using your answer to part (b), will it be possible to separate the proteins by centrifugation?

As mentioned in the text, the viscosity of liquids decreases with increasing temperature. The empirical equation \(\eta(T)=A e^{E / R T}\) provides the relationship between viscosity and temperature for a liquid. In this equation, \(A\) and \(E\) are constants, with \(E\) being referred to as the activation energy for flow. a. How can one use the equation provided to determine \(A\) and \(E\) given a series of viscosity versus temperature measurements? b. Use your answer in part (a) to determine \(A\) and \(E\) for liquid benzene given the following data: $$\begin{array}{rr}\mathbf{T}\left(^{\circ} \mathbf{C}\right) & \boldsymbol{\eta}(\mathbf{c P}) \\\\\hline 5 & 0.826 \\\40 . & 0.492 \\\80 . & 0.318 \\\120 . & 0.219 \\\160 . & 0.156\end{array}$$

a. The diffusion coefficient of sucrose in water at \(298 \mathrm{K}\) is \(0.522 \times 10^{-9} \mathrm{m}^{2} \mathrm{s}^{-1} .\) Determine the time it will take a sucrose molecule on average to diffuse an rms distance of \(1 \mathrm{mm}\) b. If the molecular diameter of sucrose is taken to be \(0.8 \mathrm{nm}\) what is the time per random walk step?

The Reynolds' number (Re) is defined as \(\mathrm{Re}=\rho\left\langle\mathrm{v}_{x}\right\rangle d / \eta,\) where \(\rho\) and \(\eta\) are the fluid density and viscosity, respectively; \(d\) is the diameter of the tube through which the fluid is flowing; and \(\left\langle\mathrm{v}_{x}\right\rangle\) is the average velocity. Laminar flow occurs when \(\operatorname{Re}<2000\), the limit in which the equations for gas viscosity were derived in this chapter. Turbulent flow occurs when \(\mathrm{Re}>2000 .\) For the following species, determine the maximum value of \(\left\langle\mathrm{v}_{x}\right\rangle\) for which laminar flow will occur: a. \(\mathrm{Ne}\) at \(293 \mathrm{K}(\eta=313 \mu \mathrm{P}, \rho=\) that of an ideal gas) through a 2.00 -mm-diameter pipe. b. Liquid water at \(293 \mathrm{K}\left(\eta=0.891 \mathrm{cP}, \rho=0.998 \mathrm{g} \mathrm{mL}^{-1}\right)\) through a 2.00 -mm-diameter pipe.

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