How do I solve this question

Question

15.0 L of an ideal gas at 298 K and 3.36 ATM are heated to 383 K with a new pressure of 5.30 8 AM. What is the new volume in liters?

Answer 1

Hello Yesenia! Let's get rolling.

Information given:

1. Volume 1 = 15.0 L
2. Temperature 1 = 298 K ---------> Temperature 2 = 383 K
3. Pressure 1 = 3.36 Atm -----------> Pressure 2 = 5.308 Atm

Information they're asking: New Volume 2?

There are several equations when it comes to ideal gases questions. One that comes to mind is the combined gas law, which essentially has all the gas laws together in a single equation. It goes as follows: P₁V₁ / T₁ = P₂V₂ / T_2.

We have agreed we are to find Volume # 2, or the change in volume that happens when the temperature is increased (as well as the noticeable change in pressure). Thus, the next step is algebraic in order to leave V2 alone, at the right side of the equation. Once we do this, the equation now looks like this:

P₁V₁T₂ / T₁P₂ = V_2.

Substituting it with numbers (Yay! we were provided Kelvins, that's always a plus), we get;

3.36 Atm * 15.0 L * 383 K = V₂ = 12.2 L

298 K * 5.308 Atm