Question

A 250 mL container holds 16.00 g of oxygen gas at 45 Celsius what is the pressure in the container

Answer 1

`P=52.2atm`

Step-by-step explanation:

Hello,

In this case, we consider the oxygen as an ideal gas; for that reason, we use the ideal gas equation:

`PV=nRT`

So we compute he moles of oxygen gas (O₂ that is diatomic) as we are given mass:

`n=16.00gO_2*(1molO_2)/(32gO_2)=0.5000molO_2`

Next, we compute the required pressure form the ideal gas equation:

`P=(nRT)/(V)=(0.5000mol*0.082(atm*L)/(mol*K)*(45+273.15)K)/(250mL*(1L)/(1000L)) \\\\P=52.2atm`

Best regards.

Answer 2

The pressure in the container is 52.2 atm

Step-by-step explanation:

Step 1: Data given

Volume of container = 250 mL = 0.250 L

Mass of oxygen gas (O2) = 16.00 grams

Temperature = 45.0 °C = 318.15 K

Molar mass of oxygen gass = 32.0 g/mol

Step 2: Calculate the moles of oxygen gas

Moles O2 = mass O2 / molar mass O2

Moles O2 = 16.00 grams / 32.00 g/mol

Moles O2 = 0.500 moles

Step 3: Calculate the pressure in the container

p*V = n*R*T

⇒with p = the pressure of the oxygen gas = TO BE DETERMINED

⇒with V = the volume of the container = 0.250 L

⇒with n = the moles of O2 gas = 0.500 moles

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

⇒with T = the temperature = 318.15 K

p = (n*R*T)/V

p = (0.500 mol * 0.08206 L*atm/mol*K * 318.15 K) / 0.250 L

p = 52.2 atm

The pressure in the container is 52.2 atm