How many moles of oxygen gas must be in a 12.2 L container to exert a pressure of 0.877 atm at a temperature of 35.0°C (the ideal gas constant in this case is 0.0821 L atm / mole K)?
Please show work
Your old friend PV = nRT
That's the ideal gas law equation. Memorize that.
@DrBob222
so how would I write out the equation?
It's PV = nRT
P = 0.877 atm
V = 12.2 L
n = ?
R = 0.0821 L*atm/mol*K
T = 273.15 + 35 = ?
Solve for n
so its 0.57?
No, I don't get 0.57.
PV = nRT
n = PV/RT
n = 0.877*12.2/0.0821*308
To solve this problem, we will use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
First, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 35.0 + 273.15
T(K) = 308.15 K
Now, we can rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Given values:
P = 0.877 atm
V = 12.2 L
R = 0.0821 L atm / mole K
T = 308.15 K
Substituting the values into the equation:
n = (0.877 atm * 12.2 L) / (0.0821 L atm / mole K * 308.15 K)
Now, we can solve for n:
n = 10.45844 / 25.33701515
n ≈ 0.4129 moles
Therefore, approximately 0.4129 moles of oxygen gas must be in the 12.2 L container to exert a pressure of 0.877 atm at a temperature of 35.0°C.