/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 A large tank is filled with meth... [FREE SOLUTION] | 91Ó°ÊÓ

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A large tank is filled with methane gas at a concentration of \(0.650 \mathrm{kg} / \mathrm{m}^{3} .\) The valve of a \(1.50-\mathrm{m}\) pipe connecting the tank to the atmosphere is inadvertently left open for twelve hours. During this time, \(9.00 \times 10^{-4} \mathrm{kg}\) of methane diffuses out of the tank, leaving the concentration of methane in the tank cssentially unchanged. The diffusion constant for methane in air is \(2.10 \times 10^{-5} \mathrm{m}^{2} / \mathrm{s} .\) What is the cross-sectional area of the pipe? Assume that the concentration of methane in the atmosphere is zero.

Short Answer

Expert verified
The cross-sectional area of the pipe is approximately 0.106 m².

Step by step solution

01

Understand the Problem

We need to find the cross-sectional area of a pipe through which a certain amount of methane diffuses over a specified time, with given concentration and diffusion constant.
02

Use Fick's Law of Diffusion

Fick's first law of diffusion describes the rate of transfer of particles: \[ \dot{m} = -D \cdot A \cdot \frac{dC}{dx} \]where \( \dot{m} \) is the mass flow rate, \( D \) is the diffusion constant, \( A \) is the cross-sectional area, and \( \frac{dC}{dx} \) is the concentration gradient.
03

Set Up the Equation

First, find the concentration gradient. Since the tank's concentration \( C_1 \) is given as \( 0.650 \, \mathrm{kg/m}^3 \) and the concentration at the other end \( C_2 \) (atmosphere) is zero, the gradient is:\[ \frac{dC}{dx} = \frac{C_2 - C_1}{L} = \frac{0 - 0.650}{1.50} = -0.4333 \, \mathrm{kg/m^4} \]Now, set up the equation with the known mass flow \( \dot{m} = \frac{9.00 \times 10^{-4}}{12 \times 3600} \, \mathrm{kg/s} \). Substitute the known values into Fick's law:\[ \frac{9.00 \times 10^{-4}}{43200} = -2.10 \times 10^{-5} \cdot A \cdot (-0.4333) \]
04

Solve for the Cross-Sectional Area

Now solve the equation for \( A \):\[ A = \frac{9.00 \times 10^{-4}}{43200 \times 2.10 \times 10^{-5} \times 0.4333 } \]Calculate to find the value of \( A \).
05

Calculate the Result

Perform the calculation:\[ A = \frac{9.00 \times 10^{-4}}{392.86 \times 10^{-5}} \]\[ A \approx 0.106 \, \mathrm{m^2} \]

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Key Concepts

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

Diffusion Constant
In the study of diffusion, the diffusion constant plays a key role by determining how quickly substances can spread out over a distance. The diffusion constant, often denoted as \( D \), represents the ease with which a gas or other substance diffuses, under given conditions. It's unique to every pair of diffusing substances and their surrounding circumstances, like temperature and pressure.
  • The diffusion constant is typically measured in square meters per second \( \text{m}^2/\text{s} \).
  • It serves as a proportionality factor in Fick's law of diffusion, which relates the rate of diffusion to the concentration gradient and the cross-sectional area where diffusion occurs.
  • A higher diffusion constant means faster diffusion.
In our methane diffusion problem, the diffusion constant for methane in air is given as \( 2.10 \times 10^{-5} \text{m}^2/\text{s} \). This value indicates how quickly methane can spread through the air under these specific conditions.
Concentration Gradient
The concentration gradient is a fundamental concept in diffusion that describes the change in concentration of a substance over a distance. In simple terms, it refers to how the concentration of a substance varies from one point to another.
  • This gradient is calculated as the difference in concentration between two points, divided by the distance between those points.
  • A steeper concentration gradient results in a faster rate of diffusion, as particles naturally move from areas of high concentration to areas of low concentration.
In the given exercise, the concentration gradient \( \frac{dC}{dx} \) between the methane gas inside the tank and the surrounding atmosphere is calculated to be \( -0.4333 \, \text{kg/m}^4 \). The negative sign indicates that the concentration decreases as we move away from the tank, driving the diffusion of methane outwards.
Methane Gas Diffusion
Methane gas diffusion is explained by the movement of methane molecules from a region of higher concentration to a region of lower concentration. Over time, this diffusion process will work towards equalizing the concentration of methane across the available space. The phenomenon is well-predicted by Fick's law of diffusion, which takes into account the diffusion constant, the concentration gradient, and the area through which diffusion occurs.
  • Methane, being a gas, spreads rapidly in the air, but its rate is controlled by the diffusion constant and the existing concentration gradient.
  • In the problem at hand, methane diffuses from the tank via a pipe to the atmosphere, where its concentration is considered to be zero, meaning the gradient is maximized.
This open pipe situation creates optimal conditions for diffusion, leading to a notable amount of methane escaping from the tank. This real-world example illustrates how gas diffusion concepts are applied to calculate practical outcomes, like the cross-sectional area of the pipe necessary to maintain this rate of diffusion.

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