A vessel with an internal volume of 10.0 L contains 2.80 g of nitrogen gas, 0.403 g of hydrogen gas, and 79.9 g of argon gas. At 25°C, what is the pressure (in atm) inside the vessel? (Hint: Nitrogen and Hydrogen exist as diatomic molecules, N2 and H2).
To find the pressure inside the vessel, we can use the ideal gas law equation:
PV = nRT
Where:
P: pressure (in atm)
V: volume of the vessel (in L)
n: number of moles of gas
R: ideal gas constant = 0.0821 L·atm/mol·K
T: temperature (in Kelvin)
First, we need to calculate the number of moles of each gas present in the vessel.
For nitrogen gas (N2):
Molar mass of N2 = 2 * molar mass of nitrogen = 2 * 14.01 g/mol = 28.02 g/mol
Number of moles of N2 = mass of nitrogen gas / molar mass of N2 = 2.80 g / 28.02 g/mol = 0.0999 mol (approximately)
For hydrogen gas (H2):
Molar mass of H2 = 2 * molar mass of hydrogen = 2 * 1.008 g/mol = 2.016 g/mol
Number of moles of H2 = mass of hydrogen gas / molar mass of H2 = 0.403 g / 2.016 g/mol = 0.200 mol
For argon gas (Ar):
Molar mass of Ar = 39.95 g/mol
Number of moles of Ar = mass of argon gas / molar mass of Ar = 79.9 g / 39.95 g/mol = 2.00 mol
Now, we can calculate the total number of moles of gas in the vessel:
Total moles of gas = moles of N2 + moles of H2 + moles of Ar = 0.0999 mol + 0.200 mol + 2.00 mol = 2.30 mol
The temperature is given as 25°C, which needs to be converted to Kelvin:
T(K) = T(°C) + 273.15 = 25 + 273.15 = 298.15 K
Now, we have all the values needed for the ideal gas law equation. Substituting the values in:
PV = nRT
P * 10.0 L = 2.30 mol * 0.0821 L·atm/mol·K * 298.15 K
P * 10.0 L = 57.82549 L·atm
Dividing both sides of the equation by 10.0 L:
P = 57.82549 L·atm / 10.0 L
P = 5.782549 atm
So, the pressure inside the vessel is approximately 5.782549 atm.