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Chapter 10: Problem 38

The ideal gas law can be recast in terms of the density of a gas. (a) Use dimensional analysis to find an expression for the density \(\rho\) of a gas in terms of the number of moles \(n\), the volume \(V\), and the molecular weight \(M\) in kilograms per mole. (b) With the expression found in part (a), show that $$ P=\frac{\rho}{M} R T $$ for an ideal gas. (c) Find the density of the carbon dioxide atmosphere at the surface of Venus, where the pressure is \(90.0 \mathrm{~atm}\) and the temperature is \(7.00 \times 10^{2} \mathrm{~K}\). (d) Would an evacuated steel shell of radius \(1.00 \mathrm{~m}\) and mass \(2.00 \times 10^{2} \mathrm{~kg}\) rise or fall in such an atmosphere? Why?

Short Answer

Expert verified
The density of carbon dioxide in Venus's atmosphere can be found using the formula \(\rho = \frac{MP}{RT}\), where \(M\) is the molecular weight of carbon dioxide, \(P\) is the atmospheric pressure, \(R\) is the ideal gas constant, and \(T\) is the temperature. Whether a steel shell would rise or fall in Venus's atmosphere depends upon the comparison between the buoyant force due to the displaced carbon dioxide gas and the weight of the steel shell.

Step by step solution

01

Part a: Dimensional Analysis

The density of a substance is defined as its mass per unit volume. For a gas, \(\rho = \frac{m}{V}\), where \(m\) is the mass of the gas and \(V\) is its volume. We know that the number of moles \(n\) for a sample of a gas with mass \(m\) can be calculated using the molecular weight \(M\), \(n = \frac{m}{M}\). Substituting this in the equation for density we get, \(\rho = \frac{nM}{V}\). This is the equation for the density of a gas in terms of the number of moles, volume, and molecular weight.
02

Part b: Ideal gas equation in terms of density

Now we are to show that \(P=\frac{\rho}{M} RT\). The ideal gas law can be written in terms of moles, so \(PV = nRT\). Replacing \(n = \frac{\rho V}{M}\) in this, we get \(P = \frac{\rho RT}{M}\), which is equivalent to the given expression, demonstrating that it holds for an ideal gas.
03

Part c: Calculating Density of Carbon Dioxide on Venus

Given the pressure \(P = 90.0 \space atm = 90.0 \times 1.013\times10^5 \space Pa\) and the temperature \(T = 7 \times 10^2 \space K\). Since \(\rho = \frac{MP}{RT}\), and using \(R = 8.314 \space J/(K.mol)\) and molecular weight of carbon dioxide \(M = 44 \times 10^{-3} \space kg/mol\) and substituting the given values, we can find the density.
04

Part d: Buoyancy in Carbon Dioxide Atmosphere

The buoyant force acting on the steel shell is equal to the weight of the displaced gas. Calculate the volume of the steel shell \(\frac{4}{3} \pi r^3\) and multiply it by the density of carbon dioxide obtained in part (c) and \(g = 9.8 \space m/s^2\) to find the buoyant force. Compare this with the weight of the steel shell to find out whether it would rise or fall. If the buoyant force is larger, the shell will rise; otherwise, it will fall.

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Key Concepts

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

Density of Gas
The density of a gas is an essential property that combines the concepts of mass and volume. When discussing gases, density (\( \rho \)) is defined as the mass (\( m \)) of the gas per unit volume (\( V \)). For an ideal gas, this can be expressed in terms of the number of moles (\( n \)), and molecular weight (\( M \)). Using the relationship \( n = \frac{m}{M} \), where "m" stands for mass and "M" is the molecular weight, we can reformulate this to describe density: \( \rho = \frac{nM}{V} \).

In this equation, "n" represents the number of moles, "M" the molecular weight in kilograms per mole, and "V" the volume in cubic meters. The density expression showcases how molecular weight and the number of moles together determine how compact or dense the gas is in a given space. Thus, when determining whether a specific gas is light or heavy in a specific volume, this equation is quite helpful.
Dimensional Analysis
Dimensional analysis involves analyzing the dimensions of physical quantities involved in an equation to extract meaningful relationships. It's an essential skill in ensuring equations are dimensionally consistent, meaning they make sense in terms of units.

For gases, the dimension of density, \( \rho \), demonstrates how mass and volume relate and is usually expressed in units of kg/m鲁.
In the equation \( \rho = \frac{nM}{V} \), replacing the number of moles "n" with the ratio of mass to molecular weight (\( n = \frac{m}{M} \)) verifies its consistency through dimensional analysis:
  • Mass (\( m \)) has dimensions of [M].
  • Molecular weight (\( M \)) has dimensions of [M][N]鈦宦�, where [N] is the amount of substance in moles.
  • Volume (\( V \)) has dimensions of [L]鲁.
When combined in the formula \( \rho = \frac{m}{V} \), the dimension of density becomes [M][L]鈦宦�, aligning with our understanding of density. Dimensional analysis thus helps confirm this mathematical model of density is coherent and effectively reflects the physical situation.
Buoyancy
Buoyancy determines whether an object will float or sink in a fluid, and it directly relates to a substance's density. This principle is often applied in different scenarios, such as whether a balloon rises in air or a ship floats in water.

The principle of buoyancy dictates that an object immersed in a fluid is subject to an upward force, called the buoyant force, equal to the weight of the fluid it displaces. For an object like a steel shell placed in a carbon dioxide atmosphere, we can compare the buoyant force (\( F_b \)) to its weight (\( W \)) to determine if it will rise or fall.
The buoyant force is calculated as:
  • The volume of the displaced gas times the density of the gas obtained from "Density of Gas" concepts.
  • The acceleration due to gravity (\( g \) - standard value: 9.8 m/s虏).
If \( F_b > W \), the shell will lift, whereas, if \( F_b < W \), it will not. This simple relationship helps predict behavior in varied environments, such as trying to understand conditions like the atmosphere of Venus.

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Most popular questions from this chapter

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