"Under constant conditions of P(pressure) and T(temperature) (and R [gas constant]), write an equation showing d(density) written as a function of M(molecular mass)."
I don't understand this, really! Please help! Thanks in advance!
I'm not sure what you are looking for but here is the general gas law written to include density.
PV = nRT
n = mols = grams/molar mass.
PV = (g/molar mass) RT
P = gRT/(molar mass*V)
density = g/V; therefore,
P = density*RT/molar mass.
Depending upon how I interpret your question your question may be asking for the above equation. However, if I go further, and show P, T, and R as constant, then constant = density/molar mass
If this is not exactly what you want please clarify.
Remember P V = n R T
now n is the number of moles
so n/V = P /R T
that is the number of moles per unit volume
what we want is the mass per unit volume
So multiply both sides by the mass per mole
This is pretty much what I'm looking for! Thanks!
Sure! I'd be happy to help.
To derive an equation showing the density (d) as a function of the molecular mass (M) under constant conditions of pressure (P) and temperature (T), we can use the ideal gas law equation:
PV = nRT,
where:
- P is the pressure,
- V is the volume,
- n is the number of moles of the gas,
- R is the gas constant, and
- T is the temperature.
We can rearrange this equation to solve for n/V:
n/V = P/RT.
Now, let's consider the definition of molar density (dm):
dm = n/V.
Since molecular mass (M) is the mass per mole of the gas, we can write:
dm = m/M,
where m is the mass of the gas.
Now, let's combine the equations:
dm = n/V = P/RT.
Substituting dm = m/M, we get:
m/M = P/RT.
Finally, to solve for density (d), which is the mass per unit volume, we divide both sides of the equation by the volume (V):
d = m/V = (m/M) / (V/V) = (m/M) / d.
Therefore, the equation showing the density (d) as a function of the molecular mass (M) under constant conditions of pressure (P) and temperature (T) is:
d = (m/M) / d.
I hope this helps! Let me know if you have any further questions.