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At a height of \(300 \mathrm{~km}\) above Earth's surface, an astronaut finds that the atmospheric pressure is about \(10^{-8} \mathrm{mmHg}\) and the temperature is \(500 \mathrm{~K}\). How many molecules of gas are there per milliliter at this altitude?

Short Answer

Expert verified
Approximately \(1.93 \times 10^8\) molecules per milliliter.

Step by step solution

01

- Understand the Ideal Gas Law

The ideal gas law is given by the equation: \[ PV = nRT \]where, - P is the pressure in pascals (Pa),- V is the volume in cubic meters (m³),- n is the number of moles,- R is the universal gas constant (8.314 J/(mol·K)),- T is the temperature in kelvin (K).
02

- Convert Given Pressure

Convert the given pressure from mmHg to pascals (Pa). We use the conversion factor:\[ 1 \text{ mmHg} = 133.322 \text{ Pa} \]Given, Pressure \( P = 10^{-8} \text{ mmHg} \):\[ P = 10^{-8} \text{ mmHg} \times 133.322 \text{ Pa/mmHg} = 1.33322 \times 10^{-6} \text{ Pa} \]
03

- Volume Conversion

Convert the given volume from milliliters to cubic meters. We know that:\[ 1 \text{ mL} = 1 \times 10^{-6} \text{ m}^3 \]Thus, for 1 milliliter:\[ V = 1 \times 10^{-6} \text{ m}^3 \]
04

- Find Number of Moles

Rearrange the ideal gas law to solve for the number of moles (n):\[ n = \frac{PV}{RT} \]Substitute the known values:\[ n = \frac{(1.33322 \times 10^{-6} \text{ Pa}) \times (1 \times 10^{-6} \text{ m}^3)}{8.314 \text{ J/(mol·K)} \times 500 \text{ K}} \]Calculate the number of moles:\[ n ≈ 3.20 \times 10^{-16} \text{ mol} \]
05

- Convert Moles to Number of Molecules

Use Avogadro's number to convert moles to molecules. Avogadro's number is \(6.022 \times 10^{23}\text{ molecules/mol}\):\[ \text{Number of molecules} = (3.20 \times 10^{-16} \text{ mol}) \times (6.022 \times 10^{23} \text{ molecules/mol}) \]Calculate the number of molecules:\[ \text{Number of molecules} ≈ 1.93 \times 10^8 \text{ molecules/mL} \]

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Key Concepts

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

Ideal Gas Law Application
Let's start by understanding the ideal gas law, a fundamental principle in chemistry. The ideal gas law is represented by the equation: \[ PV = nRT \]where: - **P** is the pressure in pascals (Pa) - **V** is the volume in cubic meters (m³) - **n** is the number of moles - **R** is the universal gas constant (8.314 J/(mol·K)) - **T** is the temperature in kelvin (K) This equation helps us relate the pressure, volume, temperature, and amount (moles) of an ideal gas. An ideal gas perfectly follows this law without any deviations. Although real gases exhibit slight deviations, they are minimal under many conditions.
Pressure Conversion
Pressure conversion is crucial when working with the ideal gas law. In our exercise, the pressure is given in millimeters of mercury (mmHg), which needs converting to pascals (Pa) to use in the ideal gas law.To convert mmHg to Pa:\[ 1 \text{ mmHg} = 133.322 \text{ Pa} \]Given pressure:\[ P = 10^{-8} \text{ mmHg} \]Thus, converting to pascals:\[ P = 10^{-8} \text{ mmHg} \times 133.322 \text{ Pa/mmHg} = 1.33322 \times 10^{-6} \text{ Pa} \]This conversion is necessary because pascals are the SI unit for pressure and are compatible with the gas constant R in our calculations.
Volume Conversion
Volume conversion is also essential. Here, the volume is given in milliliters (mL), which must be converted to cubic meters (m³). This ensures consistency with the gas constant R in the ideal gas law.To convert mL to m³: \[ 1 \text{ mL} = 1 \times 10^{-6} \text{ m}^3 \]In our problem, the volume was given as 1 milliliter, so:\[ V = 1 \times 10^{-6} \text{ m}^3 \]Using cubic meters aligns with the ideal gas law equation, making calculations straightforward and accurate.
Avogadro's Number
Avogadro's number (6.022 \times 10^{23} \text{ molecules/mol}) is a key concept in chemistry that helps convert moles of a substance to the number of molecules. In our exercise, we first find the number of moles of gas and then convert it to molecules.The number of moles ( n ) is calculated using the rearranged ideal gas law:\[ n = \frac{PV}{RT} \]After calculating the moles, we convert them to molecules using Avogadro's number:\[ \text{Number of molecules} = n \times 6.022 \times 10^{23} \text{ molecules/mol} \]For instance, if we have 3.20 \times 10^{-16} \text{ mol}, then:\[ \text{Number of molecules} = 3.20 \times 10^{-16} \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \ \text{Number of molecules} = 1.93 \times 10^8 \text{ molecules/mL} \]By using Avogadro's number, we can see the microscopic world of molecules and their interactions through calculations.

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Most popular questions from this chapter

Dilute aqueous hydrogen peroxide is used as a bleaching agent and for disinfecting surfaces and small cuts. Its concentration is sometimes given as a certain number of "volumes hydrogen peroxide," which refers to the number of volumes of \(\mathrm{O}_{2}\) gas, measured at STP, that a given volume of hydrogen peroxide solution will release when it decomposes to \(\mathrm{O}_{2}\) and liquid \(\mathrm{H}_{2} \mathrm{O}\). How many grams of hydrogen peroxide are in \(0.100 \mathrm{~L}\) of \(" 20\) volumes hydrogen peroxide" solution?

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