- #1
cwill53
- 220
- 40
- Homework Statement
- With the assumption that the air temperature is a uniform ##0.0^{\circ}C##, what is the density of the air at an altitude of 1.00 km as a percentage of the density at the surface?
- Relevant Equations
- $$pV=nRT$$
$$\frac{\partial p}{\partial z}=-\rho g$$
I got the correct answer to this question with the following calculations, but I do need some correction in terms of what units I'm integrating across.
ρ##\rho ## is density.
mtot=n##m_{tot}=nM##, where n is the number of moles of a substance and M is the molar mass of the substance.
$$pV=nRT\rightarrow pV=\frac{m_{tot}}{M}RT\Rightarrow \rho =\frac{pM}{RT}$$
pV=nRT→pV=mtotMRT⇒ρ=p$$p\propto \rho ,\frac{p_{1}}{p_{2}}=\frac{\rho _{1}}{\rho _{0}}$$
$$\rho =\frac{(101325Pa)(0.02896kg/mol)}{(8.31446\frac{J}{(mol\cdot K)})(273.15K)}=1,2921kg/m^3\propto 1.2921Pa$$
The molar mass of air, composed of nitrogen, oxygen, argon, carbon dioxide, and some other elements is 28.96g/mol. I'm not exactly sure if the above is correct in terms of the proportionality between the density and pressure, but I guessed that this was correct because you can't integrate pressure across density.
So,
$$\frac{dp}{dy}=-\rho g=-\frac{pM}{RT}g$$
$$\int_{p_{0}}^{p_{1}}\frac{dp}{p}=\int_{1.2921Pa}^{p_{1}}\frac{1}{p}dp=\frac{-Mg}{RT}\int_{0m}^{1000m}dz$$
$$ln(p)\rvert_{1.2921Pa}^{p_{1}}=ln(p_{1})-ln(1.2921Pa)=ln(p_{1})-0.25627Pa$$
$$\frac{-Mg}{RT}[z]\rvert_{0m}^{1000m}=-\frac{(0.02896kg/mol)(9.81m/s^2)(1000m)}{(8.31446J/(mol\cdot K))(273.15K)}=-0.1250927Pa\Rightarrow ln(p_{1})=0.1311773Pa\Rightarrow p_{1}=e^{0.1311773Pa}=1.14016Pa\Rightarrow p_{1}=.88241p_{0}\Leftrightarrow \rho _{1}=.88241\rho_{0}$$
Now this is correct numerically, even though some of the units for the bounds might be off (correct me please).
I also wanted to add another parameter. Say the 1.00 km in the question was replaced with 6.00 km, and the temperature was varying at a rate:
$$\frac{\partial T}{\partial z}= -6.5^{\circ}C/km$$
How would I go about solving the problem then?
ρ##\rho ## is density.
mtot=n##m_{tot}=nM##, where n is the number of moles of a substance and M is the molar mass of the substance.
$$pV=nRT\rightarrow pV=\frac{m_{tot}}{M}RT\Rightarrow \rho =\frac{pM}{RT}$$
pV=nRT→pV=mtotMRT⇒ρ=p$$p\propto \rho ,\frac{p_{1}}{p_{2}}=\frac{\rho _{1}}{\rho _{0}}$$
$$\rho =\frac{(101325Pa)(0.02896kg/mol)}{(8.31446\frac{J}{(mol\cdot K)})(273.15K)}=1,2921kg/m^3\propto 1.2921Pa$$
The molar mass of air, composed of nitrogen, oxygen, argon, carbon dioxide, and some other elements is 28.96g/mol. I'm not exactly sure if the above is correct in terms of the proportionality between the density and pressure, but I guessed that this was correct because you can't integrate pressure across density.
So,
$$\frac{dp}{dy}=-\rho g=-\frac{pM}{RT}g$$
$$\int_{p_{0}}^{p_{1}}\frac{dp}{p}=\int_{1.2921Pa}^{p_{1}}\frac{1}{p}dp=\frac{-Mg}{RT}\int_{0m}^{1000m}dz$$
$$ln(p)\rvert_{1.2921Pa}^{p_{1}}=ln(p_{1})-ln(1.2921Pa)=ln(p_{1})-0.25627Pa$$
$$\frac{-Mg}{RT}[z]\rvert_{0m}^{1000m}=-\frac{(0.02896kg/mol)(9.81m/s^2)(1000m)}{(8.31446J/(mol\cdot K))(273.15K)}=-0.1250927Pa\Rightarrow ln(p_{1})=0.1311773Pa\Rightarrow p_{1}=e^{0.1311773Pa}=1.14016Pa\Rightarrow p_{1}=.88241p_{0}\Leftrightarrow \rho _{1}=.88241\rho_{0}$$
Now this is correct numerically, even though some of the units for the bounds might be off (correct me please).
I also wanted to add another parameter. Say the 1.00 km in the question was replaced with 6.00 km, and the temperature was varying at a rate:
$$\frac{\partial T}{\partial z}= -6.5^{\circ}C/km$$
How would I go about solving the problem then?
Last edited: