Originally Posted by **TyrannosaurusWreck** So I had an 'interactive' problem where you're given a simulation and you have to plug in your calculations to get the simulation to work. In this one I was given
"a small chamber containing an ideal gas consisting of 26 gas particles held at a constant temperature of 400 K. The gas has a pressure of 897 Pa. The chamber volume is currently set at 1.6×10−22 m3. You can change the volume of the chamber, within a range of 0.5×10−22 m3 to 2.5×10−22 m3. You also can vary the number of gas particles, from 10 to 30. Your task is to set the number of particles and the volume of gas so that the gas pressure is reduced to 552 Pa. The temperature of the gas in this simulation remains fixed at 400 K."
I did get working answers- 1.2e-22m3 and 12 particles, but I just guessed them. My method was- PV=NkT; T is constant, and initial and final values are proportional so P1V1/N1=P2V2/N2. I ended up with 5.38e-21*Nf = 552*Vf but I'm stuck here, how do I work it with two unknowns? |

We don't know if the gas is an ideal gas, but we can approximate the gas using an ideal-gas-like formula. So... let's use this one:

$\displaystyle PV = cNkT$

where we expect c will be a constant value of order unity. c =1 for an ideal gas, c = 3/2 for a diatomic gas with only translational modes, c = 5/2 for a diatomic gas with translational and rotational modes activated, etc...

What's c for the current situation?

$\displaystyle c = \frac{PV}{NkT} = \frac{897 \times 1.6 \times 10^{-22}}{26 \times 1.38 \times 10^{-23} \times 400} = 1$

Since $\displaystyle c = 1$, we know this is an ideal gas.

We have two variables we can change, volume and number of particles. Our target is a pressure of 552 Pa. Therefore, the formula becomes

$\displaystyle \frac{V}{N} = \frac{kT}{P} = \frac{1.3 \times 10^{-23} \times 400}{552} = 10^{-23}$

Just to make it a bit easier to pick values of V and N, let's factor out the scale of the volume:

Let $\displaystyle V = a \times V_0$

where $\displaystyle V_0 = 10^{-22}$ and a is just a number that describes the volume, but its value is now of order unity. This is nothing particularly interesting; we're just splitting a number into the two parts as it would if we described it using standard form.

Consequently...

$\displaystyle \frac{a}{N} = \frac{10^{-23}}{10^{-22}} = 0.1$

So... we can vary:

- $\displaystyle a$ between 0.5 and 2.5; and

- N between 10 and 30

Since the number of particles is an integer, there are 15 possible answers in the form of (a ,N) pairs:

(1.0, 10), (1.1, 11), (1.2, 12), ..., (2.4, 24), (2.5, 25)

the answer you gave is (1.2, 12) above.