Answer:
The pressure in the container is 52.2 atm
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
Answer:
[tex]P=52.2atm[/tex]
Explanation:
Hello,
In this case, we consider the oxygen as an ideal gas; for that reason, we use the ideal gas equation:
[tex]PV=nRT[/tex]
So we compute he moles of oxygen gas (O₂ that is diatomic) as we are given mass:
[tex]n=16.00gO_2*\frac{1molO_2}{32gO_2}=0.5000molO_2[/tex]
Next, we compute the required pressure form the ideal gas equation:
[tex]P=\frac{nRT}{V}=\frac{0.5000mol*0.082\frac{atm*L}{mol*K}*(45+273.15)K}{250mL*\frac{1L}{1000L}} \\\\P=52.2atm[/tex]
Best regards.
