A sample containing 1.20 moles of Ne gas has an initial volume of 7.50 L. What is the final volume of the gas, in liters, when each of the following changes occur in the quantity of the gas at constant pressure and temperature? 1) a leak allows one half of the Ne atoms to escape
2) a sample of 3.10 moles of Ne is added to the 1.20 moles of Ne gas in the container
3) a sample of 35.0 g of Ne is added to the 1.20 moles of Ne gas in the container.

1 = 3.75L

2 = 26.91L

3 = 18.37L

Explanation:

Hello,

The question above can be solved when we know Avogadro's law which states that the volume of a fixed mass of gas is directly proportional the number of moles present provided temperature and pressure are kept constant.

Mathematically,

N = kV k = n/v

N1/V1 = N2/V2 = N3/V3 ==Nn/Vn

N1 = 1.20 moles

V1 = 7.50L

1) if half of the Ne atoms escaped what would be the final volume.

1 mole of Ne = 6.022×10²³ atoms

½(1.20) moles of Ne = ?

0.6 moles of Ne = ?

1 mole = 6.022×10²³ atom

1.2 moles = ?

x = 1.2 × 6.022×10²³ atoms

x = 7.23×10²³atoms

If ½ of 7.23×10²³ atoms escaped, how many would be left

½ × 7.23×10²³ atoms = 3.61×10²³atoms

Now we have to find the number of moles and then use our equation.

1 mole = 6.022×10²³ atoms

y mole = 3.61×10²³ atom

y = 0.6 mole

N2 = 0.6 mole

N1 / V1 = N2 / V2

Make V2 the subject of formula,

V2 = (N2 × V1) / N1

V2 = (0.6 × 7.50) / 1.20

V2 = 3.75L

The volume after half of the Ne atoms escaped is 3.75L

2)

When a sample of 3.10 mole is added to 1.20 moles present

N1 = 1.20

V1 = 7.51

N2 = (1.20 + 3.10) = 4.30L

V2 = ?

N1 / V1 = N2 / V2

V2 = (N2 × V1) / N1

V2 = (4.30 × 7.51) / 1.20

V2 = 26.91L

The volume of Ne gas if 3.10 moles is added to it is 26.91L

3)

A sample of 35g is added to the 1.20 mole Ne in the container.

We need to convert the mass (35g) to moles. This can be done using mass-molarmass relationships

Number of moles = mass / molar mass

Molar mass of Ne = 20.17g/mol

Number of moles = 35 / 20.17

Number of moles = 1.735 moles

N2 = 1.20 moles + 1.735 moles

N2 = 2.935 moles

N1 / V1 = N2 / V2

V2 = (N2 × V1) / N1

V2 = (2.935 × 7.51) / 1.20

V2 = 18.37L

On addition of 35g of Ne gas to the container, the volume is increased to 18.37L

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