at what temperature does 16.3 g of nitrogen gas have a pressure of 1.25 atm in a 25.0 - L tank?
To determine the temperature at which 16.3 g of nitrogen gas has a pressure of 1.25 atm in a 25.0 L tank, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's calculate the number of moles (n) of nitrogen gas using its molar mass.
The molar mass of nitrogen (N₂) is 28.02 g/mol (14.01 g/mol × 2).
n = mass / molar mass = 16.3 g / 28.02 g/mol
n ≈ 0.582 mol
Now we can rearrange the ideal gas law equation to solve for temperature (T):
T = PV / (nR)
Substituting the given values:
T = (1.25 atm) * (25.0 L) / (0.582 mol * 0.0821 L·atm/(mol·K))
T ≈ 720 K
Therefore, at a temperature of approximately 720 Kelvin, 16.3 g of nitrogen gas will have a pressure of 1.25 atm in a 25.0 L tank.