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Chapter 10: Problem 66

A plasma-screen TV contains thousands of tiny cells filled with a mixture of \(\mathrm{Xe}, \mathrm{Ne}\) , and He gases that emits light of specific wavelengths when a voltage is applied. A particular plasma cell, \(0.900 \mathrm{mm} \times 0.300 \mathrm{mm} \times 10.0 \mathrm{mm},\) contains 4\(\%\) Xe in a 1: Ne: He mixture at a total pressure of 500 torr. Calculate the number of Xe, Ne, and He atoms in the cell and state the assumptions you need to make in your calculation.

Short Answer

Expert verified
The volume of the plasma cell is \(2.7 \times 10^{-6} m^3\), and the total pressure in atm is 0.6579 atm. Partitioning the pressure, we find that the partial pressures for Xe, Ne, and He are 0.0263 atm, 0.3023 atm, and 0.3023 atm, respectively. Using the Ideal Gas Law, we calculate the number of moles for Xe, Ne, and He. Finally, by multiplying the moles by Avogadro's number, we find the number of Xe, Ne, and He atoms in the cell. The assumptions made during the calculation are that the Ideal Gas Law is applicable, the room temperature is 298 K, and the mixture contains gases in a 1:1:1 ratio.

Step by step solution

01

Calculate the volume of the cell

Given the dimensions of the plasma cell, we can simply multiply them to get the volume of the cell. \(Volume = 0.900 mm \times 0.300 mm \times 10.0 mm\) Note that we have to convert the dimensions to meters to use in the Ideal Gas Law. \(Volume = (0.900 \times 10^{-3} m) \times (0.300 \times 10^{-3} m) \times (10.0 \times 10^{-3} m)\) Calculate the volume: \(Volume = 2.7 \times 10^{-6} m^3\)
02

Convert total pressure to atm

Given the total pressure is 500 torr, we need to convert it to atm. \(500 \, torr \times \frac{1 \, atm}{760 \, torr} = 0.6579 \, atm\)
03

Partition the pressure of each gas

The gases are mixed in a 1:1:1 ratio, so they will have equal parts of the total pressure. Since Xe is 4% of the total pressure, we can determine the percentage of Ne and He in the mixture. For Ne and He: \(50\% - 4\% = 46\%\) of the total pressure each. Calculate partial pressures: \(P_{Xe} = 0.6579 \, atm \times 0.04\\ P_{Ne} = 0.6579 \, atm \times 0.46\\ P_{He} = 0.6579 \, atm \times 0.46\)
04

Calculate the number of moles for each gas using Ideal Gas Law

Using the Ideal Gas Law (PV = nRT), we can calculate the number of moles for each gas. Rearrange the equation to solve for n: \(n = \frac{PV}{RT}\), where R is the ideal gas constant (0.0821 L atm K^-1 mol^-1) and T is the temperature in Kelvin (assumed to be the room temperature, 298 K). We also need to convert the volume from m³ to liters (1 m³ = 1000 L): \(Volume = 2.7 \times 10^{-6} m^3 \times \frac{1000 \, L}{1 \, m^3} = 2.7 \times 10^{-3} L\) Calculate moles: \(n_{Xe} = \frac{P_{Xe} \times Volume}{R \times T} \\ n_{Ne} = \frac{P_{Ne} \times Volume}{R \times T} \\ n_{He} = \frac{P_{He} \times Volume}{R \times T}\)
05

Calculate the number of atoms for each gas

The number of atoms can be found by multiplying the moles of each gas by Avogadro's number (\(6.022 \times 10^{23}\)). Calculate atoms: \(Xe \, atoms = n_{Xe} \times 6.022 \times 10^{23}\\ Ne \, atoms = n_{Ne} \times 6.022 \times 10^{23}\\ He \, atoms = n_{He} \times 6.022 \times 10^{23}\) Now, we have the number of atoms for Xe, Ne, and He in the plasma cell. Assumptions made during the calculation: 1. The Ideal Gas Law is applicable. 2. The room temperature is 298 K. 3. The mixture contains gases in a 1:1:1 ratio.

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Most popular questions from this chapter

As discussed in the "Chemistry Put to Work" box in Section 10.8 , enriched uranium can be produced by effusion of gaseous \(\mathrm{UF}_{6}\) across a porous membrane. Suppose a process were developed to allow effusion of gaseous uranium atoms, U(g). Calculate the ratio of effusion rates for \(^{235} \mathrm{U}\) and \(^{238} \mathrm{U},\) and compare it to the ratio for \(\mathrm{UF}_{6}\) given in the essay.

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Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? \begin{equation}\begin{array}{l}{\text { (a) Because Xe is a noble gas, there is less tendency for the Xe }} \\ {\text { atoms to repel one another, so they pack more densely in }} \\ {\text { the gaseous state. }} \\ {\text { (b) Xe atoms have a higher mass than } \mathrm{N}_{2} \text { molecules. Because }} \\ {\text { both gases at STP have the same number of molecules per }} \\ {\text { unit volume, the Xe gas must be denser. }}\\\\{\text { (c) The Xe atoms are larger than } \mathrm{N}_{2} \text { molecules and thus take }} \\ {\text { up a larger fraction of the space occupied by the gas. }} \\\ {\text { (d) Because the Xe atoms are much more massive than the }} \\\ {\mathrm{N}_{2} \text { molecules, they move more slowly and thus exert }} \\\ {\text { less upward force on the gas container and make the gas }} \\ {\text { appear denser. }}\end{array}\end{equation}

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