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II (a) Calculate the mass of nitrogen present in a volume of \(3000 \mathrm{~cm}^{3}\) if the temperature of the gas is \(22.0^{\circ} \mathrm{C}\) and the absolute pressure is \(2.00 \times 10^{-13} \mathrm{~atm},\) a partial vacuum easily obtained in laboratories. The molar mass of nitrogen \(\left(\mathrm{N}_{2}\right)\) is \(28.0 \mathrm{~g} / \mathrm{mol}\). (b) What is the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of the \(\mathrm{N}_{2} ?\)

Short Answer

Expert verified
Mass of nitrogen is approximately \(6.92 \times 10^{-13}\) g, and density is approximately \(2.31 \times 10^{-13}\) kg/m³.

Step by step solution

01

Convert the Volume

Convert the volume from cm³ to m³. Since there are 1,000,000 cm³ in a m³, calculate:\[ V = 3000 \text{ cm}^3 = \frac{3000}{1000000}\text{ m}^3 = 0.003\text{ m}^3 \]
02

Convert Temperature to Kelvin

The temperature in Kelvin is calculated by adding 273.15 to the Celsius temperature:\[ T = 22.0 + 273.15 = 295.15 \text{ K} \]
03

Use Ideal Gas Law

The ideal gas law \( PV = nRT \) can be rearranged to solve for the number of moles, \( n \):\[ n = \frac{PV}{RT} \]Use the gas constant \( R = 0.0821 \text{ atm} \cdot \text{L/mol} \cdot \text{K} \). Convert volume to liters (1 m³ = 1000 L):\[ V = 0.003 \text{ m}^3 = 3 \text{ L} \]Substitute values:\[ n = \frac{(2.00 \times 10^{-13} \text{ atm})(3 \text{ L})}{(0.0821 \text{ atm} \cdot \text{L/mol} \cdot \text{K})(295.15 \text{ K})} \approx 2.47 \times 10^{-14} \text{ mol} \]
04

Calculate Mass of Nitrogen

The mass \( m \) is found using the molar mass of nitrogen:\[ m = n \times \text{Molar mass} \]\[ m = 2.47 \times 10^{-14} \text{ mol} \times 28.0 \text{ g/mol} \approx 6.92 \times 10^{-13} \text{ g} \]
05

Calculate Density

Density \( \rho \) is mass per unit volume. First, convert mass to kg:\[ 6.92 \times 10^{-13} \text{ g} = 6.92 \times 10^{-16} \text{ kg} \]Then calculate density using the volume in m³:\[ \rho = \frac{m}{V} = \frac{6.92 \times 10^{-16} \text{ kg}}{0.003 \text{ m}^3} = 2.31 \times 10^{-13} \text{ kg/m}^3 \]

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Key Concepts

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

Mass of Nitrogen
The mass of nitrogen is an essential parameter when studying gases, particularly in laboratory settings where precise measurements are needed. Calculating the mass of nitrogen involves understanding its relationship with other quantities like volume, pressure, and temperature. This relationship is best expressed through the Ideal Gas Law.To find the mass, we first need to determine the number of moles of nitrogen gas present using the Ideal Gas Law formula:\( PV = nRT \).Here, \( P \) represents the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. Each of these terms must be in the correct units to achieve accurate results:
  • Pressure must be in atmospheres (atm).
  • Volume should be in liters (L).
  • Temperature should be in Kelvin (K).
Once we have the number of moles, \(n\), we can calculate the mass \(m\) using:\[ m = n \times \text{Molar Mass of Nitrogen (N}_2) \].The molar mass of nitrogen \(N_2\) is \(28.0 \, \text{g/mol}\). Substituting the calculated number of moles and the molar mass gives the mass of nitrogen in grams. Accurate conversion of mass from grams to kilograms may be necessary, particularly if further calculations demand it.
Density Calculation
Density is a measure of mass per unit volume, and for gases, it's a vital property that reflects how much of the gas is contained in a given space. In this exercise, calculating the density of nitrogen involves both the derived mass of the nitrogen gas and the volume it occupies.First, make sure the mass is in kilograms, as the result will be given in \( \text{kg/m}^3 \). After determining the mass, use the formula for density:\[ \rho = \frac{m}{V} \],where \(\rho\) is the density, \(m\) is the mass in kilograms, and \(V\) is the volume in cubic meters.
Since gases are often measured in different volume units (like liters or cubic centimeters), ensure you use the correct conversion factors:
  • 1 liter equals 0.001 cubic meters.
  • 1 cubic meter equals 1000 liters.
By correctly converting and organizing your data, you can derive a meaningful value for the density of nitrogen under given conditions. This calculated density not only serves as an essential parameter in scientific calculations but also aids in practical applications such as monitoring gas usage and storage.
Convert Units
In scientific calculations involving gases, converting units accurately ensures that all quantities work harmoniously within formulas, especially when applying laws like the Ideal Gas Law.The conversion of units often involves changing:
  • Volume from \(\text{cm}^3\) to \(\text{m}^3\): Divide the \(\text{cm}^3\) volume by 1,000,000 to get \(\text{m}^3\).
  • Temperature from Celsius to Kelvin: Add 273.15 to the Celsius temperature.
  • Pressure may sometimes need conversion depending on the units used for the gas constant \(R\). Ensure it matches well with the \(R\) value used.
For example, to convert 3000 \(\text{cm}^3\) to \(\text{m}^3\), use:\[ V = \frac{3000}{1000000} = 0.003 \text{ m}^3 \].Conversion of temperature from 22.0°C to Kelvin is:\[ T = 22.0 + 273.15 = 295.15 \text{ K} \].Unit conversions maintain consistency across calculations, yielding accurate, understandable results. These conversions are vital in converting the derived mass to density or any other form separately measured from the original setup, promoting precise and efficient scientific communication.

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Most popular questions from this chapter

An ideal gas at 4.00 atm and \(350 \mathrm{~K}\) is permitted to expand adiabatically to 1.50 times its initial volume. Find the final pressure and temperature if the gas is (a) monatomic with \(C_{p} / C_{V}=\frac{5}{3},\) (b) diatomic with \(C_{p} / C_{V}=\frac{7}{5}\).

One type of gas mixture used in anesthesiology is a \(50 \% / 50 \%\) mixture (by volume) of nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) and oxygen \(\left(\mathrm{O}_{2}\right),\) which can be premixed and kept in a cylinder for later use. Because these two gases don't react chemically at or below 2000 psi, at typical room temperatures they form a homogeneous single- gas phase, which can be considered an ideal gas. If the temperature drops below \(-6^{\circ} \mathrm{C},\) however, \(\mathrm{N}_{2} \mathrm{O}\) may begin to condense out of the gas phase. Then any gas removed from the cylinder will initially be nearly pure \(\mathrm{O}_{2}\). As the cylinder empties, the proportion of \(\mathrm{O}_{2}\) will decrease until the gas coming from the cylinder is nearly pure \(\mathrm{N}_{2} \mathrm{O}\). In another test, the gas is put into a cylinder having a movable piston. The gas is then allowed to expand adiabatically, thereby reducing its pressure from 2000 psi to 50 psi. Why might some \(\mathrm{N}_{2} \mathrm{O}\) condense during this process? A. This is an isochoric process in which the pressure decreases, so the temperature also decreases. B. Because of the expansion, heat is removed from the system, so the internal energy and temperature of the gas decrease. C. This is an isobaric process, so as the volume increases, the temperature decreases proportionally. D. The expanding gas does work with no heat input, so the internal energy and temperature of the gas decrease.

A gas in a cylinder expands from a volume of \(0.110 \mathrm{~m}^{3}\) to \(0.320 \mathrm{~m}^{3} .\) Heat flows into the gas just rapidly enough to keep the pressure constant at \(1.80 \times 10^{5} \mathrm{~Pa}\) during the expansion. The total heat added is \(1.15 \times 10^{5} \mathrm{~J}\). (a) Find the work done by the gas. (b) Find the change in internal energy of the gas.

How many atoms are you? Estimate the number of atoms in the body of a \(65 \mathrm{~kg}\) physics student. Note that the human body is mostly water, which has molar mass \(18.0 \mathrm{~g} / \mathrm{mol},\) and that each water molecule contains three atoms.

The surface of the sun. The surface of the sun has a temperature of about \(5800 \mathrm{~K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is \(1.67 \times 10^{-27} \mathrm{~kg} .\) ) (b) What would be the mass of an atom that had half the rms speed of hydrogen?

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