What mass of helium gas is needed to pressurize a 98.0 L tank to 165 atm at 21 °C? What mass of oxygen gas would be needed to pressurize a similar tank to the same specifications
See your post above. Same PV = nRT
To calculate the mass of helium gas needed to pressurize the tank, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P is the pressure in atmospheres (atm)
V is the volume in liters (L)
n is the number of moles of gas
R is the ideal gas constant = 0.0821 L·atm/(mol·K)
T is the temperature in Kelvin (K)
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15 = 21 + 273.15 = 294.15 K
Now we can rearrange the Ideal Gas Law equation to solve for the number of moles:
n = PV / (RT)
n = (165 atm) * (98.0 L) / (0.0821 L·atm/(mol·K) * 294.15 K)
n ≈ 52.09 moles
The molar mass of helium is approximately 4 g/mol. Therefore, the mass of helium needed can be calculated by multiplying the number of moles by the molar mass:
Mass = n * molar mass
Mass = 52.09 mol * 4 g/mol
Mass ≈ 208.36 g
Therefore, approximately 208.36 grams of helium gas is needed to pressurize the 98.0 L tank to 165 atm at 21 °C.
To determine the mass of oxygen gas needed for the second tank, we would follow the same steps but use the molar mass of oxygen, which is approximately 32 g/mol.
Mass = n * molar mass
Mass = 52.09 mol * 32 g/mol
Mass ≈ 1667.68 g
Therefore, approximately 1667.68 grams of oxygen gas would be needed to pressurize a similar tank to the same specifications.