# Thread: Gas/Liquid pressure from temperature and density

1. ## Gas/Liquid pressure from temperature and density

How can I calculate a gas' or a liquid's pressure from it's density and it's temperature? For example, for air, I suppose the pressure $\displaystyle p\$ and the density $\displaystyle \rho$ is proportional, $\displaystyle p\ \propto\ \rho$. Is this true? That would mean that since the unit of pressure is $\displaystyle Pa$ and the unit for density is $\displaystyle kg/m^3$, the unit for pressure per density would be $\displaystyle \frac{Pa}{kg/m^3}\ =\ \frac{Pa\cdot m^3}{kg}$, or $\displaystyle \frac{Nm^3/m^2}{kg}\ =\ \frac{Nm}{kg}$, or $\displaystyle \frac{m^2}{s^2}$. Im just thinking a little bit. And since the atmospherical pressure is $\displaystyle 101\ 325\ Pa$, and the air density in one atm pressure is $\displaystyle 1.2$ to heaped $\displaystyle 1.3\ kg/m^3$ in normal temperatures (here in Sweden). So then we can say that $\displaystyle \frac{p}{\rho}\ =\ \frac{101\ 325 Pa}{1.25\ kg/m^3}\ =\ 81\ 060\ Pa\cdot m^3/kg$

Now I don't know if this is true. And this is only for the temperature $\displaystyle 11^\circ\ C\ \sim\ 52^\circ\ F$.

Then I guess It is not the same thing at all with liquids, not water at least. Since the density for water (in $\displaystyle 4^\circ\ C$) is always $\displaystyle 1\ kg/dm^3$ but the pressure can change. Okay I guess that is not completely true, but almost.

2. Originally Posted by TriKri
How can I calculate a gas' or a liquid's pressure from it's density and it's temperature? For example, for air, I suppose the pressure $\displaystyle p\$ and the density $\displaystyle \rho$ is proportional, $\displaystyle p\ \propto\ \rho$. Is this true? That would mean that since the unit of pressure is $\displaystyle Pa$ and the unit for density is $\displaystyle kg/m^3$, the unit for pressure per density would be $\displaystyle \frac{Pa}{kg/m^3}\ =\ \frac{Pa\cdot m^3}{kg}$, or $\displaystyle \frac{Nm^3/m^2}{kg}\ =\ \frac{Nm}{kg}$, or $\displaystyle \frac{m^2}{s^2}$. Im just thinking a little bit. And since the atmospherical pressure is $\displaystyle 101\ 325\ Pa$, and the air density in one atm pressure is $\displaystyle 1.2$ to heaped $\displaystyle 1.3\ kg/m^3$ in normal temperatures (here in Sweden). So then we can say that $\displaystyle \frac{p}{\rho}\ =\ \frac{101\ 325 Pa}{1.25\ kg/m^3}\ =\ 81\ 060\ Pa\cdot m^3/kg$

Now I don't know if this is true. And this is only for the temperature $\displaystyle 11^\circ\ C\ \sim\ 52^\circ\ F$.

Then I guess It is not the same thing at all with liquids, not water at least. Since the density for water (in $\displaystyle 4^\circ\ C$) is always $\displaystyle 1\ kg/dm^3$ but the pressure can change. Okay I guess that is not completely true, but almost.
Start with the Idea Gas Law (see here). If you need something more follow
the link at the bottom of the linked page labled "Equation of State".

RonL

3. Also your temperatures should be in Kelvin (K) if in metric or Rankine (R) if in English units. Both systems are based with a 0 temperature at absolute zero.

-Dan

4. Thanks! That was just what I was looking for.

Edit: topsquark: Yes, I supose it should. I prefer kelvin since it is more close to the temperature system in Sweden (Celsius). By the way, did you know Anders Celsius was from Sweden?

5. Originally Posted by TriKri
By the way, did you know Anders Celsius was from Sweden?
I would have guessed he was from France.

6. Originally Posted by TriKri
Thanks! That was just what I was looking for.

Edit: topsquark: Yes, I supose it should. I prefer kelvin since it is more close to the temperature system in Sweden (Celsius). By the way, did you know Anders Celsius was from Sweden?
Originally Posted by ThePerfectHacker
I would have guessed he was from France.
I also would have guessed France. I'm not surprised you would choose to do the temperature in K. I don't know anyone who uses the R scale anymore. (In fact I'm one of the few I know that have ever heard of it!)

-Dan

7. Originally Posted by topsquark
I don't know anyone who uses the R scale anymore. (In fact I'm one of the few I know that have ever heard of it!)
Rankine scale right?
I think the magic number is 84 R (for boiling).

I use the C scale, in fact, I need to convert from F to C because, my parents use C , for they come from the Mother Country.

8. Originally Posted by ThePerfectHacker
Rankine scale right?
I think the magic number is 84 R (for boiling).

I use the C scale, in fact, I need to convert from F to C because, my parents use C , for they come from the Mother Country.
84 R for boiling what? 1 degree R is equal to 1 degree F, so the freezing temp of water is about 459 R, then add 212 R to get to boiling water gives about 671 R for boiling water.

-Dan

9. Why is not possible to get absolute zero again? I heard of this law.

10. Originally Posted by topsquark
84 R for boiling what? 1 degree R is equal to 1 degree F, so the freezing temp of water is about 459 R, then add 212 R to get to boiling water gives about 671 R for boiling water.

-Dan
Ahem, freezing temp of water is 491,67 R, and you have to add 180 R to get boiling temperature. So 671 is right at least.

- Kristofer

11. Originally Posted by TriKri
Ahem, freezing temp of water is 491,67 R, and you have to add 180 R to get boiling temperature. So 671 is right at least.

- Kristofer
This is what happens when I use my Swiss-cheese-like memory instead of looking it up like I should have. Thanks for the spot!

-Dan