Question

A 1.00 l sample of a gas at 25.0◦c and 1.00 atm contains 50.0 % helium and 50.0 % neon by mass. what is the partial pressure of the neon. express your answer in atmospheres.

Answer 1

V = Volume of sample of gas = 1.00 L = 0.001 m³

T = temperature of the gas = 25.0 ⁰C = 25 + 273 = 298 K

P = pressure = 1.00 atm = 101325 Pa

n = number of moles of gas

using ideal gas equation

PV = n RT

101325 (0.001) = n (8.314) (298)

n = 0.041

n₁ = number of moles of helium

n₂ = number of moles of neon

m₁ = mass of helium = n₁ (4) = 4 n₁

m₂ = mass of neon = n₂ (20.2) = 20.2 n₂

given that :

m₁ = m₂

4 n₁ = 20.2 n₂

n₁ = 5.05 n₂

also

n₁ +  n₂ = n

5.05 n₂ +  n₂ = 0.041

n₂ = 0.0068

mole fraction of neon is given as

mole fraction = n₂ /n = 0.0068/0.041 = 0.166

P₂ = partial pressure of neon = (mole fraction) P

P₂ = (0.166) (1)

P₂ = 0.166 atm

Answer 2

The partial pressure of the neon is  0.16 atm.

Explanation:  

 A 1.00 l sample of a gas at 25.0◦c and 1.00 atm contains 50.0 % helium and 50.0 % neon by mass. what is the partial pressure of the neon. express your answer in atmospheres.

The partial pressure is the notional pressure of constituent gas if it is occupied alone by the entire volume of the original mixture at the same temperature.

50.0 % helium and 50.0 % neon by mass.

$P = 1 atm$

Mass of the gases = $M$

$Moles of helium = \frac{0.5M}{4 g/mol}\\Moles of neon = \frac{0.5M}{20 g/mol}$

By using ideal gas equation we get:

$PV=nRT\\n=\frac{PV}{RT}=\frac{1.00 L\times 1.00 atm}{0.0820 L atm/mol K\times 298 K}=0.0409 mol$

where $n =$total number of moles of gases

$0.0409 mol=\frac{0.5M}{4 g/mol}+\frac{0.5M}{20 g/mol}$

$M = 0.2728 g$

Then we can find the mass and mole of gases

The gas molar volume is the volume of one mole of the gas at a given temperature and pressure.

Mass of helium gas = $\frac{0.2728g}{2}=0.1364 g$

Moles of helium gas $=n_1=\frac{0.1364 g}{4 g/mol}=0.0341 mole$

Mass of neon gas  $= \frac{0.2728g}{2}=0.1364 g$

Moles of neon gas $= n_2=\frac{0.1364 g}{20 g/mol}=0.00682mole$

Partial pressure of the neon gas:$P\times \chi_2=P\times \frac{n_2}{n_1+n_2}=1 atm\times \frac{0.00682}{0.00682+0.0341}=0.16 atm$

Therefore the partial pressure of the neon gas is 0.16 atm.

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