At what temperature does a 16.3 g of nitrogen gas have a pressure of 1.25 atm in a 25.0-L tank?
Plz show me step by step how to do this, with using pressure as atm.
You don't need a step by step. Just substitute and solve for T.
Use PV = nRT
n = 16.3 g/molar mass
T in Kelvin.
V = 25.0 L
Post your work if you get stuck.
To determine the temperature at which a given amount of nitrogen gas exerts 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 (1.25 atm)
V = volume in liters (25.0 L)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
To solve for temperature (T), we need to rearrange the equation:
T = PV / (nR)
To find the number of moles of nitrogen gas (n), we can use the molar mass of nitrogen gas.
1. Calculate the number of moles of nitrogen gas (n):
- Find the molar mass of nitrogen gas (N2): 14.01 g/mol
- Divide the mass of nitrogen gas by its molar mass:
n = mass / molar mass = 16.3 g / 28.02 g/mol = 0.581 mol
2. Substitute the known values into the equation for temperature (T):
T = (1.25 atm) * (25.0 L) / (0.581 mol * 0.0821 L·atm/mol·K)
3. Calculate the temperature:
T = 64.02 K
Therefore, at a temperature of 64.02 K, 16.3 g of nitrogen gas will exert a pressure of 1.25 atm in a 25.0-L tank.
To find the temperature at which a 16.3 g of nitrogen gas will have a pressure of 1.25 atm in a 25.0-L tank, we can use the ideal gas law equation:
PV = nRT
P = pressure in atm
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
Step 1: Convert the pressure from atm to Pa (Pascal)
1 atm = 101325 Pa
So, 1.25 atm = 1.25 x 101325 Pa = 126656.25 Pa
Step 2: Convert the mass of nitrogen gas to moles
To convert mass (m) to moles (n), you can use the molar mass (M) of nitrogen gas (N2), which is 28.0134 g/mol.
n = m / M
n = 16.3 g / 28.0134 g/mol
n ≈ 0.581 mol
Step 3: Plug the known values into the ideal gas law equation and solve for T
PV = nRT
T = PV / (nR)
T = (126656.25 Pa) x (25.0 L) / ((0.581 mol) x (0.0821 L·atm/mol·K))
T ≈ 652.79 K
Therefore, the temperature at which a 16.3 g of nitrogen gas will have a pressure of 1.25 atm in a 25.0-L tank is approximately 652.79 K.