Create a conceptual image illustrating the principle of an ideal gas behavior. Depict a cylinder containing gas with a volume indication of 3.0 L. Next to it, show another cylinder with gas and a volume indication that's double the size emphasizing the concept of temperature and number of moles being doubled. Make sure the depiction of pressure remains constant in both cylinders. Do not include any textual explanations in this depiction. Instead, use visual queues like arrows and scaling to denote change.

An ideal gas has a volume of 3.0 L. If the number of moles of gas and the temperature are doubled, while the pressure remains constant, what is the new volume?

I thought it would be simpler and you could see how much it increased from one set of data to another.

To use 3L, Use PV = nRT
Set P =1, V = 3, solve for n, use R as 0.08206 and T = 200 (any T will do).
solve for n, the double n, and double T, and solve for V.

The easiest way to do it, logically, is
PV = nRT
use P = 1 and V = 3, then
V = nRT.
3 = nRT. Call n = 1, R is a constant, call T = 1, then 3 = R. Now double n to 2, T to 2 and substitute
V = 2*R*2 = 4V
V was 3 so it is now 12.
Now matter how you set up the numbers, you will always come back to 12 L for the answer.

I just pick any? Don't I have to put 3.0 L in the equation?

Pick a T (in Kelvin), number of moles, and pressure, calculate V using PV = nRT. Then double N and T and recalculate V.

Well, if the number of moles and the temperature are doubled, you could say that the gas is having a real "moley" experience! As for the new volume, remember that when the pressure is constant, the volume is directly proportional to the number of moles of gas. Since both the number of moles and the temperature are doubled, we can say that the new volume would also be doubled. So the new volume would be 6.0 L. Voila!

To find the new volume when the number of moles and the temperature are doubled while the pressure remains constant, we can use the combined gas law. The combined gas law states:

(P1 * V1) / (n1 * T1) = (P2 * V2) / (n2 * T2)

where:
- P1 and P2 are the initial and final pressures, respectively
- V1 and V2 are the initial and final volumes, respectively
- n1 and n2 are the initial and final number of moles, respectively
- T1 and T2 are the initial and final temperatures in Kelvin, respectively

In this case, the pressure is constant, so P1 = P2. Also, since the number of moles and temperature are doubled, we have n2 = 2 * n1 and T2 = 2 * T1.

We can substitute these values in the combined gas law to solve for V2:

(P1 * V1) / (n1 * T1) = (P2 * V2) / (n2 * T2)

Since P1 = P2, we can cancel out the pressure terms:

V1 / (n1 * T1) = V2 / (2 * n1 * 2 * T1)

Simplifying further:

V1 / (n1 * T1) = V2 / (4 * n1 * T1)

Cross-multiplying:

V1 * (4 * n1 * T1) = V2 * (n1 * T1)

Multiplying:

4 * V1 * n1 * T1 = V2 * n1 * T1

Cancelling n1 * T1 from both sides:

4 * V1 = V2

Therefore, the new volume, V2, is four times the initial volume, V1.

In this case, the initial volume is given as 3.0 L, so the new volume would be:

V2 = 4 * V1 = 4 * 3.0 L = 12.0 L

So, the new volume when the number of moles and temperature are doubled while the pressure remains constant is 12.0 L.