Question

A gas mixture in a 1.55-L container at 298 K contains 10.0 g of Ne and 10.0 g of Ar. Calculate the partial pressure (in atm) of Ne and Ar in the container.

Answer 1

Answer:

partial pressure (in atm) of Ne and Ar in the container is 7.82 atm and 3.94atm respectively

Explanation:

see the attached file

Answer 2

This question involves the concepts of the general gas equation and partial pressures.

a) The partial pressure of Ne in the container is "7.9 atm".

b) The partial pressure of Ar in the container is "3.94 atm"

a)

First, we will calculate the no. of moles of Ne gas:

$n_{Ne}=\frac{m}{M}$

where,

$n_{Ne}$ = no. of moles of Ne = ?

m = mass of Ne = 10 g

M = molecular mass of Ne = 20.2 g/mol

Therefore,

$n_{Ne}=\frac{10\ g}{20.2\ g/mol}\\\\n_{Ne}=0.5\ mol$

Now, using the general gas equation to find the partial pressure of Ne:

$P_{Ne}V=n_{Ne}RT$

where,

$P_{Ne}$ = Partial pressure of Ne = ?

V = volume = 1.55 L = 1.55 x 10⁻³ m³

R = General gas constant = 8.314 J/mol.k

T = absolute temperature = 298 k

Therefore,

$P_{Ne}=\frac{(0.5\ mol)(8.314\ J/mol.k)(298\ k)}{1.55\ x\ 10^{-3}\ m^3}$

$P_{Ne}=(799216.77\ Pa)(\frac{1\ atm}{101325\ Pa})\\\\P_{Ne}=7.9\ atm$

b)

First, we will calculate the no. of moles of Ne gas:

$n_{Ar}=\frac{m}{M}$

where,

$n_{Ar}$ = no. of moles of Ar = ?

m = mass of Ne = 10 g

M = molecular mass of Ne = 39.95 g/mol

Therefore,

$n_{Ar}=\frac{10\ g}{39.95\ g/mol}\\\\n_{Ar}=0.25\ mol$

Now, using the general gas equation to find the partial pressure of Ne:

$P_{Ar}V=n_{Ar}RT$

where,

$P_{Ne}$ = Partial pressure of Ne = ?

V = volume = 1.55 L = 1.55 x 10⁻³ m³

R = General gas constant = 8.314 J/mol.k

T = absolute temperature = 298 k

Therefore,

$P_{Ar}=\frac{(0.25\ mol)(8.314\ J/mol.k)(298\ k)}{1.55\ x\ 10^{-3}\ m^3}$

$P_{Ar}=(399608.39\ Pa)(\frac{1\ atm}{101325\ Pa})\\\\P_{Ar}=3.94\ atm$

Learn more about the general gas equation here:

brainly.com/question/24645007?referrer=searchResults

The attached picture shows the general gas equation.