Since

,If we adjust one the the variables, (P, V, n, or T) then one or much more oftheother variables must also change. Thisleads come the equation or if the number of moles continues to be thesame .You are watching: P1v1=p2v2 solve for p2

**Boyle"s Law:**

Boyle"s law examines the impact of changing volume on Pressure.To isolate this variables, temperature need to remain constant.We can remove temperature indigenous both sides of the equation andwe areleft v P1V1= P2V2

Sample Problem: A pistonwith a volume of gas of 1.0 m3 in ~ 100 kPa is compressed come afinalvolume the 0.50 m3. Whatis the last pressure?

P1 is 100 kPa

V1 is 1.0 m3

V2 is 0.50 m3

P2 is unknown

P1V1=P2V2 becomes**See an animation on rearranging the merged gas lawCharles"s Law**

Charles"s legislation examines the impact of changingtemperatureon volume. To isolate these variables, pressure need to remain constant.

so Charles"s regulation isSample problem: A piston through a volume that gas of1.0 m3at 273 K is cooled come a temperature that 136.5 K. Whatis the last volume? (Assume pressure is maintained constant.)

T1 is 273 K

V1 is 1.0 m3

V2 is unknown

T2 is 136.5 K

Thesolution i do not care**Charleslaw Applet** watch what happens as soon as you increasetemperature.Increasing temperature __________ pressure.

GUY-LUSSAC"S LAWNear the revolve of the 19th century, Guy-Lussac investigated therelationship in between pressure and also temperature if the volume was heldconstant. When the temperature goes increase the push inside arigid container likewise goes up. For example, your car tires, wheninflated, are basically rigid, the volume will certainly not change. Didyou an alert that as soon as the temperature goes increase the press inside yourtires additionally increases?We can again usage the merged gas regulation to quantify this relationship.

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Sample Problem: If yourtire is 2 liters and the initial push is 2 atm, what is the finalpressure when the temperature goes indigenous 0 degrees celcius (273 K) to100 degrees celcius (373 K)?

T1 is 273 KP1 is 2 atmP2 is unknownT2 is 373 K

First, begin with the combinedgas law and cancel the end the volumes because they execute not change.

After removed the volumes, Rearranging the equation:so the final pressure P2, is (2.00atm)(373K)/(273 K) = 2.73 atm