- This is a Boyle’s law question, so you can use the P1V1=P2V2 equation; since temperature does not change, it is not a factor. You’re looking for the volume after a change, so it would be the second V (V2). Isolate it to create the equation: V2=(P1V1)/P2. Plug in the numbers: V2=(3atm*3.0L)/(1atm)=9.0L.
To intuitively check yourself, Boyle’s law states that if the pressure decreases, the volume increases (and vice versa, assuming a fixed mass and temp). In this case, your pressure decreased from 3atm to 1atm and thus your volume went from 3.0L to 9.0L.
- This is a Charles’s Law problem which utilizes the (V1/T1) = (V2/T2) equation. You’re looking for V2 again, so you rearrange the equation to: V2 = (V1T2)/T1= (3.5L*373K)/(293K) = 4.5L
Again, you can check intuitively. Charles’s Law states that when temp goes up, so does volume (if mols and pressure are the same). The temp increased from 20 to 100 degrees C, so your volume increased from 3.5L to 4.5 L.
- This problem utilizes all 3 elements: pressure, volume and temperature, so you can use the combined gas law: (P1V1)/T1=(P2V2)/T2, rearrange to isolate V2.
V2=(P1V1T2)/(P2T1) = (720torr325mL333K)/(900torr*303K)=286mL
Key things to consider when doing these problems: as a chem tutor, I see a lot of students make mistakes by first plugging in the numbers and solving for the unknown, so I recommend three things:
- Write your “given” and label them correctly on the side. Because if you mix your P1 with your P2, you’ll get the wrong answer, for example.
- Solve for the unknown using the variables first. Once you have isolated the unknown, plug in your values. Moving a lot of numbers increases your chances of making a mistake.
- Write your units every time and cross them out when they cancel. If your question asks for a volume, but your answer is left with a torr unit, you know you made a mistake somewhere along the way.