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The atmosphere of Mars is mostly CO\(_2\) (molar mass 44.0 g/mol) under a pressure of 650 Pa, which we shall assume remains constant. In many places the temperature varies from 0.0\(^\circ\)C in summer to -100\(^\circ\)C in winter. Over the course of a Martian year, what are the ranges of (a) the rms speeds of the CO\(_2\) molecules and (b) the density (in mol/m\(^3\)) of the atmosphere?

Short Answer

Expert verified
(a) RMS speeds vary from 304 m/s to 394 m/s. (b) Density changes from 0.286 mol/m\(^3\) to 0.451 mol/m\(^3\).

Step by step solution

01

Understand the Concept of RMS Speed

The root-mean-square (rms) speed is a measure of the speed of particles in a gas, which is derived from the kinetic theory of gases. The formula for rms speed is \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant \( 1.38 \times 10^{-23} \text{ J/K} \), \( T \) is the absolute temperature in Kelvin, and \( m \) is the mass of a single molecule in kg.
02

Convert Temperatures to Kelvin

To use the rms speed formula, first convert temperatures from Celsius to Kelvin: \( T_{summer} = 0.0\circ C = 273.15 \text{ K} \)\( T_{winter} = -100\circ C = 173.15 \text{ K} \).
03

Calculate Molar Mass to Kilograms

Convert the molar mass of CO\(_2\) from g/mol to kg by dividing by 1000: \( M = \frac{44.0 \text{ g/mol}}{1000} = 0.044 \text{ kg/mol} \).
04

Calculate Mass per Molecule

Using Avogadro's number \( N_A = 6.022 \times 10^{23} \text{ molecules/mol} \), calculate the mass of a single CO\(_2\) molecule: \( m = \frac{M}{N_A} = \frac{0.044}{6.022 \times 10^{23}} \approx 7.31 \times 10^{-26} \text{ kg/molecule} \).
05

Calculate RMS Speed for Summer

Use \( T = 273.15 \text{ K} \) in the rms speed formula: \( v_{rms, summer} = \sqrt{ \frac{3 \cdot 1.38 \times 10^{-23} \cdot 273.15}{7.31 \times 10^{-26}}} \approx 394 \text{ m/s} \).
06

Calculate RMS Speed for Winter

Use \( T = 173.15 \text{ K} \) in the rms speed formula: \( v_{rms, winter} = \sqrt{ \frac{3 \cdot 1.38 \times 10^{-23} \cdot 173.15}{7.31 \times 10^{-26}}} \approx 304 \text{ m/s} \).
07

Calculate Density Using Ideal Gas Law

The ideal gas law \( PV = nRT \) can be rearranged to find the density in mol/m\(^3\):\( n = \frac{P}{RT} \). Use \( R = 8.314 \text{ J/(mol}\cdot\text{K)} \) and \( P = 650 \text{ Pa} \).
08

Calculate Density for Summer

At \( T = 273.15 \text{ K} \), density is: \( n_{summer} = \frac{650}{8.314 \cdot 273.15} \approx 0.286 \text{ mol/m}^3 \).
09

Calculate Density for Winter

At \( T = 173.15 \text{ K} \), density is:\( n_{winter} = \frac{650}{8.314 \cdot 173.15} \approx 0.451 \text{ mol/m}^3 \).

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Key Concepts

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

Root-Mean-Square Speed
The root-mean-square (RMS) speed of a gas molecule is a way to describe the average speed of particles within a gas. This concept, rooted in the kinetic theory of gases, provides insight into how molecular speed changes with temperature. The formula for calculating the RMS speed is given by:\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]Here:
  • \( k \) is the Boltzmann constant and it equals \( 1.38 \times 10^{-23} \) J/K.
  • \( T \) is the absolute temperature in Kelvin, which is crucial since the speed of a molecule relies heavily on the temperature.
  • \( m \) is the mass of a single molecule in kilograms.
For gases like carbon dioxide (CO\(_2\)) on Mars, the RMS speed will vary with the seasons since the temperature shifts, from 0°C in summer (273.15 K) to -100°C in winter (173.15 K). This variation in temperature directly influences the speed. A higher temperature results in a higher RMS speed, evident in the differences between the speeds calculated for summer (about 394 m/s) and winter (about 304 m/s). Remember, this speed represents an average as not all molecules will move at the RMS speed. Instead, it gives a useful snapshot of behavior at the molecular scale.
Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry and physics that describes the relationship between pressure, volume, temperature, and the number of moles of a gas. It's expressed as:\[ PV = nRT \]where:
  • \( P \) is the pressure in pascals (Pa).
  • \( V \) is the volume in cubic meters (m²).
  • \( n \) is the number of moles of gas.
  • \( R \) is the ideal gas constant, 8.314 J/(mol·K).
  • \( T \) is the temperature in Kelvin.
For gases like CO\(_2\) on Mars, assuming the pressure is constant, the temperature changes significantly impact the density of the gas. According to this law, as the temperature decreases during the Martian winter, the gas density increases because cooler temperatures mean kinetic energy of molecules is lower, allowing molecules to pack closer together. In summer, higher temperatures lead to decreased density. Using the ideal gas law, you calculate this density by rearranging the formula to:\[ n = \frac{P}{RT} \]This expression allows the calculation of density at different temperatures, showing a range from approximately 0.286 mol/m³ in summer to 0.451 mol/m³ in winter.
Molecular Mass Conversion
Converting molecular mass, or molar mass, to a usable form in calculations is crucial, especially when computing properties like the RMS speed of molecules. Molar mass is commonly given in grams per mole (g/mol), but for computations involving kinetic theory or other equations that require individual molecular masses, conversion to kilograms per mole (kg/mol) is necessary. To convert, divide the molar mass by 1000, turning grams into kilograms. For example, the molar mass of CO\(_2\) is 44.0 g/mol. Hence, in kilograms per mole, it becomes:\[ M = \frac{44.0 \text{ g/mol}}{1000} = 0.044 \text{ kg/mol} \]This step ensures all units are consistent, which is vital in calculations that involve other physical constants as they are typically in SI units. Afterward, using Avogadro's number (\( 6.022 \times 10^{23} \text{ molecules/mol} \)), we can find the mass of a single molecule by dividing the molar mass in kg/mol by this number. This gives:\[ m = \frac{M}{N_A} \approx 7.31 \times 10^{-26} \text{ kg/molecule} \]By converting the molar mass properly, you enable accurate and meaningful calculations in problems involving gases, like predicting their behavior under varied Martian atmospheric conditions.

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Most popular questions from this chapter

A balloon of volume 750 m\(^3\) is to be filled with hydrogen at atmospheric pressure (1.01 \(\times\) 10\(^5\) Pa). (a) If the hydrogen is stored in cylinders with volumes of 1.90 m\(^3\) at a gauge pressure of 1.20 \(\times\) 10\(^6\) Pa, how many cylinders are required? Assume that the temperature of the hydrogen remains constant. (b) What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if both the gas in the balloon and the surrounding air are at 15.0\(^\circ\)C? The molar mass of hydrogen (H\(_2\)) is 2.02 g/mol. The density of air at 15.0\(^\circ\)C and atmospheric pressure is 1.23 kg/m\(^3\). See Chapter 12 for a discussion of buoyancy. (c) What weight could be supported if the balloon were filled with helium (molar mass 4.00 g/mol) instead of hydrogen, again at 15.0\(^\circ\)C?

How many moles are in a 1.00-kg bottle of water? How many molecules? The molar mass of water is 18.0 g/mol.

The size of an oxygen molecule is about 2.0 \(\times\) 10\({^-}{^1}{^0}\) m. Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from ideal gas behavior at ordinary temperatures (\(T\) = 300 K).

A flask with a volume of 1.50 \(L\), provided with a stopcock, contains ethane gas (C\(_2\)H\(_6\)) at 300 K and atmospheric pressure (1.013 \(\times\) 10\(^5\) Pa). The molar mass of ethane is 30.1 g/mol. The system is warmed to a temperature of 550 \(K\), with the stopcock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?

During a test dive in 1939, prior to being accepted by the U.S. Navy, the submarine \(Squalus\) sank at a point where the depth of water was 73.0 m. The temperature was 27.0\(^\circ\)C at the surface and 7.0\(^\circ\)C at the bottom. The density of seawater is 1030 kg/m\(^3\). (a) A diving bell was used to rescue 33 trapped crewmen from the \(Squalus\). The diving bell was in the form of a circular cylinder 2.30 m high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: Ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.) (b) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?

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