How do you calculate oxygen temperature as atm pressure changes? [closed]

Question

I know that when gases are decompressed the atoms suddenly have a bunch of space and absorb all the heat around them thereby giving the effect that it is cool but how do you calculate this for oxygen for example? And how do you calculate the reverse when say oxygen is being compressed what is the temperature at 200 atmosphere or whatever the best unit is for calculating pressure I don't know ?

I tried Mathematica but I think I need a pro version to get that info.

Please explain what each symbol stands for as I only know basic symbols like power volts amps .

Answer 1

This should be a comment, but currently I can't comment.

Have you read the Wikipedia section on ideal gas law at https://en.wikipedia.org/wiki/Ideal_gas_law? It seems to me that this article answers your question very well, together with the information that the standard atomic weight of oxygen is $M_{\mathrm{O_2}} \approx 16 \mathrm{\frac{g}{mol}}$, i.e. the specific gas constant of oxygen is $R_{\mathrm{specific,O_2}} = R/M_{\mathrm{O_2}} \approx 520 \mathrm{\frac{J}{kg\cdot K}}$.

Answer 2

When you have a gas in a container, and you decrease the volume of the container, you can see that the molecules will speed up with the following analogy (which comes from statistical thermodynamics - the study of the thermal properties of gases by looking at the behavior of an ensemble of molecules).

If you have a ball bouncing back and forth between two walls, with perfectly elastic collisions each time, it will maintain its speed indefinitely. That is a "gas in equilibrium".

Now imagine that you move the walls towards each other at a velocity $v$. If the ball is initially moving towards the moving wall at velocity $v'$, then after the collision it will be moving in the opposite direction with a velocity $v'+2v$. At every round trip, it will pick up more speed - and the round trips will happen faster and faster as the ball is going faster and the walls are closer together.

The mean force that the wall feels depends on the number of collisions that it encounters, and the speed of the ball when it hits. Both of these go up - that means that the force on the wall (the pressure) goes up.

Similarly, the velocity of the ball goes up: in a gas, this means that the temperature goes up.

Now we're ready for an equation. We write for the volume of the gas $V$, pressure $P$ and temperature $T$; then we need the ratio of the heat capacity "at constant volume" $c_V$ and "at constant pressure" $c_P$; we call this ratio $\gamma = \frac{c_P}{c_V}$. For a diatomic molecule like oxygen, this number is very close to $\frac{7}{5} = 1.4$ (see wikipedia)

Now we can write an expression for what happens to the pressure and temperature when we compress a gas without allowing heat in or out - we call this an "adiabatic" process:

$$PV^\gamma=\rm{const}$$

So if you know the initial pressure and volume, you can compute the final pressure from the above equation. Similarly, if you want to know the final temperature, you can find it from

$$\frac{PV}{T}=\rm{const}$$

and the above equation.

Answer 3

If you take a closed cylinder of air and measure the pressure at various temperatures, you'll find that the pressure, $p$, and temperature, $T$, are linearly related: $P={\rm const}\times T$. This is usually referred to as Gay-Lussac's law.

Combining this law with other laws of thermodynamics (specifically Boyle's law and Charles' law), one can derive the ideal gas law: $$ pV=nRT $$ where $V$ is the volume, $R$ is the gas constant ($\sim8.314\,\text{J/K/mol}$), and $n$ the number of moles of the gas considered (see also this Wikipedia link for more on the amount of substance).

With a little rearranging, the temperature of an ideal gas becomes, $$ T=\frac{pV}{nR} $$ which, if you know the volume of gas and how much of the gas is inside the volume, you can determine the temperature at a certain pressure.