A 3.90 L sealed vessel under vacuum is filled with 6.40 g of fluorine gas. What is the pressure (in kPa) of the vessel at 25.0 ⁰

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2 years ago

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2 answers:

kondor19780726 [428] · Answer 1 · 2023-02-21T23:30:31+03:00

The pressure of the vessel at a temperature of 25 °C is 106.73 KPa

We'll begin by calculating the number of mole of fluorine gas. This can be obtained as follow:

Mass of F₂ = 6.40 g

Molar mass of F₂ = 19 × 2 = 38 g/mol

Mole = mass / molar mass

Mole of F₂ = 6.40 / 38

Finally, we shall determine the pressure. This can be obtained as follow:

Volume (V) = 3.90 L

Temperature (T) = 25.0 °C = 25 + 273 = 298 K

Mole of F₂ (n) = 0.168 mole

Gas constant (R) = 8.314 L.KPa/Kmol

<h3>Pressure (P) =? </h3>

PV = nRT

P × 3.9 = 0.168 × 8.314 × 298

P × 3.9 = 416.232096

Divide both side by 3.90

P = 416.232096 / 3.90

Therefore, the pressure of the vessel is 106.73 KPa

Leto [7] · Answer 2 · 2023-02-22T00:41:38+03:00

From the ideal gas law, the pressure of the gas is 106 kPa.

First we must obtain the number of moles of the gas;

Number of moles = mass/molar mass

Molar mass of fluorine gas = 38 g/mol

Number of moles of F2 = 6.40 g/38 g/mol = 0.168 moles

Using the ideal gas equation;

PV = nRT

P = ?

V = 3.90 L

n = 0.168 moles

T = 25.0 ⁰C + 273 = 296 K

R = 8.314 kPa L K-1 mol-1

Substituting values;

P = nRT/V

P = 0.168 moles × 8.314 kPa L K-1 mol-1 × 296 K/3.90 L

P = 106 kPa