- #1
Kayne
- 24
- 0
Hi All,
I would like to understand a little more about the following equation
[tex] moles = \frac{Mass(g)}{Molar Mass}[/tex]
I know that this is to calculate the Moles for use in the ideal gas law equation
[tex] PV=n \times R \times T [/tex]
Lets say I had to find two variable in the ideal gas law one being Pressure the other moles. If I know that the volume of a tank is 120L of Air and the molar mass is 16 is this enough to work out this equation??
[tex] moles = \frac{Mass(g)}{Molar Mass}[/tex]
[tex] moles = \frac{120000}{16}= 7500[/tex]
Now can this be substituted back into the ideal gas equation with the following
V = 120L
n = 7500
R = 0.0821 L.Atm/mol.K
T = 303 K
[tex] P=\frac{7500 \times 8.314 \times 303}{120} = 1554 kPa [/tex]
Is this correct?
I would like to understand a little more about the following equation
[tex] moles = \frac{Mass(g)}{Molar Mass}[/tex]
I know that this is to calculate the Moles for use in the ideal gas law equation
[tex] PV=n \times R \times T [/tex]
Lets say I had to find two variable in the ideal gas law one being Pressure the other moles. If I know that the volume of a tank is 120L of Air and the molar mass is 16 is this enough to work out this equation??
[tex] moles = \frac{Mass(g)}{Molar Mass}[/tex]
[tex] moles = \frac{120000}{16}= 7500[/tex]
Now can this be substituted back into the ideal gas equation with the following
V = 120L
n = 7500
R = 0.0821 L.Atm/mol.K
T = 303 K
[tex] P=\frac{7500 \times 8.314 \times 303}{120} = 1554 kPa [/tex]
Is this correct?