Question

The pressure of an ideal gas is 782mmHg at 35°C in a box with a volume of 2,239mL. What is the number of moles? If this gas is oxygen, O2, what is the number of molecules present?

Answer 1

9.1 * 10^-2 moles of gas

5.49*10^22 molecules O2

Step-by-step explanation:

Step 1: Data given

Pressure of the gas = 782 mmHg = 782 / 760 atm = 1.02895 atm

Temperature = 35 °C = 308 K

Volume = 2.239 L

Step 2: Calculate moles of gas

p*V = n*R*T

n = (p*V)/(R*T)

⇒with n = the number of moles of gas = TO BE DETERMINED

⇒with p = the pressure of the gas = 1.03026 atm

⇒with V = the volume of the gas = 2.239 L

⇒with R = the gas constant = 0.08206 L*atm / mol *K

⇒with T = the temperature = 35 °C = 308 K

n = (p*V)/ (R*T)

n = (1.03026 * 2.239) / (0.08206*308)

n = 0.091268 moles of gas = 9.1 * 10^-2 moles of gas

Step 3: Calculate molecules O2

Number of molecules = moles * number of Avogadro

Number of molecules = 0.091268 moles * 6.02 * 10^23

Number of molecules = 5.49*10^22 molecules O2

There are 5.49*10^22 molecules O2

Answer 2

The vessel has 0.0913 moles of O₂. Therefore it has 5.49ₓ10²² molecules of O₂

Step-by-step explanation:

Let's apply the Ideal Gases Law to solve this.

First of all, we convert the pressure from mmHg to atm

782 mmHg . 1 atm / 760 mmHg = 1.03 stm

We convert the volume from mL to L

2239 mL . 1L / 1000mL = 2.239L

and finally we convert T° C to K → 35°C + 273 = 308K

Formula is: P . V = n . R . T

1.03 atm . 2.239L = n . 0.082 L.atm/mol.K . 308K

n = 1.03 atm . 2.239L / 0.082 L.atm/mol.K . 308K → 0.0913 moles

Let's count the molecules → 0.0913 mol . 6.02ₓ10²³ molecules / 1 mol =

5.49ₓ10²² molecules