Calculate the density of Argon (Ar) gas at 400.0K and 1.23 atm.
P*molar mass = density*RT
density will be in g/L.
what is p and what is RT
Aw C'mon. P is pressure in atm, R is the gas constant in L and atm, T is kelvin temperature.
This is just a rearrangement of PV = nRT.
To calculate the density of a gas, you can use the ideal gas law equation, which states: PV = nRT. Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.08206 L.atm / K.mol), and T is the temperature in Kelvin.
To calculate the number of moles (n), we need to use the ideal gas law equation rearranged as n = PV / RT.
Given:
Pressure (P) = 1.23 atm
Temperature (T) = 400.0 K
Now, we need to find the volume of the gas, considering that gases expand to fill their container, so there is no specific volume given. In order to obtain the density, we can assume a fixed volume of 1 liter.
Volume (V) = 1 L
Now, we can substitute these values into the ideal gas law equation and calculate the number of moles (n):
n = (1.23 atm * 1 L) / (0.08206 L.atm / K.mol * 400.0 K)
n ≈ 0.037 mol
Next, we can calculate the molar mass of Argon (Ar), which is approximately 39.95 g/mol.
Finally, to calculate the density, we divide the molar mass by the volume:
Density = (molar mass) / (volume)
Density = 39.95 g/mol / 1 L
Density ≈ 39.95 g/L
Therefore, the density of Argon gas at 400.0 K and 1.23 atm is approximately 39.95 g/L.