Question

A plasma-screen TV contains thousands of tiny cells filled with a mixture of Xe, Ne, and He gases that emits light of specific wavelengths when a voltage is applied. A particular plasma cell, 0.900 mm×0.300mm×10.0mm, contains 4.00%Xe in a 1:1 Ne : He mixture at a total pressure of 500. torr. Assumptions: In order to calculate total moles of gas and total atoms, we assumed a reasonable room temperature. Since '4.00% Xe' was not defined, we conveniently assumed mole percent. The 1:1 relationship of Ne to He is assumed to be by volume and not by mass.Part A) Calculate the number of Xe atoms in the cell. Part B) Calculate the number of Ne atoms in the cell. Part C) Calculate the number of He atoms in the cell.

Answer 1

Answer:

(a) 1.77*10^15 Xe atoms

(b) 2.10*10^16 Ne atoms

(c) 2.10*10^16 He atoms

Explanation:

Total V = 0.900 mm * 0.300mm * 10.0mm

Total V = 2.7mm cubed

Converting cubed to litres:

2.7mm cubed = 2.7 * 10^-6 Liters

Converting pressure from torr to atm,

(500 torr) * (1 atm ÷ 760 torr) = 0.6578947368 atm

Room temperature is assumed to be 21 degrees celcius, or 294.15 Kelvin

Applying Ideal Gas Equation:

PV = nRT

Therefore, n = PV ÷ RT,

n = [(2.7*10^-6 Liters)(.6578947368 atm)] ÷ [(0.08206)(294.15)]

n = 7.359017079 * 10^-8

Therefore, the total moles present in plasma is 7.359017079 * 10^-8.

(a) To find Xe atoms,

Multiply the total moles present by 4 percent. Therefore,

(7.359017079 * 10^-8)(.04) = 2.943606832*10^-9,

this value gives you Xe moles.

Then multiply by Avogadro's constant:

[(2.943606832*10^-9)(6.022*10^23)] = 1.77*10^15 Xe atoms

(b) and (c) are the same.

Using the total moles number we found in the beginning and minus the number of moles for Xe.

We have:

7.359017079 * 10^-8 - 2.943606832*10^-9 = 7.064656396*10^-8

And Since Ne and He are in a 1:1 ratio, divide this number by two.

We have:

7.064656396*10^-8 ÷ 2 = 3.532328198*10-8 moles

By multiplying this number by Avogadro's number we get atoms:

[(3.532328198*10-8 moles)(6.022*10^23)] = 2.10*10^16

Therefore, the answer for both (b) and (c) is 2.10*10^16.