Chemistry, 01.02.2021 21:30 moomoofower

The first red (632.8 nm) Helium Neon gas laser was reported by White and Rigdon from Bell Labs in 1962. The laser was a discharge tube that was 1.2 m long with a 7 mm diameter. A He-Ne ratio of 10:1 was used with a total pressure of 0.00092 atm. This is a really low pressure with a really small amount of He and Ne. We want to try making a new laser with the same discharge tube with 5 mol Helium and 1 mol Neon. For the new laser with 5 mol He and 1 mol Ne, calculate the following quantities. Assume room temperature (298.15 K) and that the gasses are ideal. a. What is the total volume (cm^3)?
b. What is the mole fraction of He?
c. What is the mole fraction of Ne?
d. What is the partial pressure of He (in atm)?
e. What is the partial pressure of Ne (in atm)?
f. What is the total pressure of the new laser tube (in atm)?

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Answer from: dinarussell74

a) Total Volume = 46.18 cm³

b) Mole $_{fraction}$ (He) = 0.8333

c) Mole $_{fraction}$ (Ne) = 0.1666

d) Partial Pressure P $_{He}$ = 2648.68 atm

e) Partial Pressure P $_{Ne}$ = 529.73 atm

f) total pressure of the new laser tube = 3178.41 atm

Explanation:

Given that;

height of of discharge tube h = 1.2 m = 1.2 × 100 = 120cm

diameter of the tube D = 7 mm = 0.7cm

Radius = D/2 = 0.7/2 = 0.35 cm

a) What is the total volume (cm³)?

Volume = πr²h

we substitute

Volume = π × (0.35)² × 120

Volume = 46.18 cm³

b) What is the mole fraction of He?

Mole fraction of Helium will be;

Mole $_{fraction}$ (He) = Moles of He / Total Moles ( He + Ne)

given that; 5 mol He and 1 mol Ne

Mole $_{fraction}$ (He) = 5 / ( 5 + 1 )

Mole $_{fraction}$ (He) = 5 / 6

Mole $_{fraction}$ (He) = 0.8333

c) What is the mole fraction of Ne?

Mole fraction of Neon will be;

Mole $_{fraction}$ (Ne) = Moles of Ne / Total Moles ( He + Ne)

given that; 5 mol He and 1 mol Ne

Mole $_{fraction}$ (Ne) = 1 / ( 5 + 1 )

Mole $_{fraction}$ (Ne) = 1 / 6

Mole $_{fraction}$ (Ne) = 0.1666

d) d. What is the partial pressure of He (in atm)?

using ideal gas equation;

P $_{He}$ = nRT/V

where n is number or amount of moles( 5 )

R is the universal gas constant ( 0.08205 L-atm.mol⁻¹K⁻¹

T is temperature ( 298.15 K)

And V is the volume; ( 46.18 cm³ = 46.18/1000 = 0.04618 L )

so we substitute

P $_{He}$ = (5 × 0.08205 × 298.15) / 0.04618

P $_{He}$ = 122.316 / 0.04618

P $_{He}$ = 2648.68 atm

e) What is the partial pressure of Ne (in atm)?

P $_{Ne}$ = (1 × 0.08205 × 298.15) / 0.04618

P $_{Ne}$ = 24.463 / 0.04618

P $_{Ne}$ = 529.73 atm

f) What is the total pressure of the new laser tube (in atm)?

total pressure of the new laser tube will be;

= P $_{He}$ + P $_{Ne}$

= 2648.68 atm + 529.73 atm

= 3178.41 atm

total pressure of the new laser tube = 3178.41 atm

Answer from: Quest

Ans. : frostbite is when exposure to cold temperatures causes freezing to the skin or other tissues. it most commonly affects the hands, feet, and face. the longer areas are exposed to cold, typically the worse the frostbite. the water temperature used for rewarming a frostbitten part should be : (102 -105) f degrees.

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

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The first red (632.8 nm) Helium Neon gas laser was reported by White and Rigdon from Bell Labs in 19...

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