1 Mr. ShieldsRegents Chemistry U05 L05

2 Robert Boyle Robert Boyle (1627 – 1691) -Boyle’s law (1662) Expresses the relationship between Pressure and Volume for gases PV=k (constant T and n) (constant T and n)

3 Developing Boyles Law If volume is held constant how can we inc. Pressure? Recall that the KMT assumes Pressure is created by the Frequent collision of gas molecules with the wall of the Container since … Assumption 5 states that KE of a gas is directly proportional to Temperature … If T then KE and since P = F/A (F = KE) Then as KE so does P

4 Developing Boyles Law Molecules moving with a higher velocity (v) hit the container walls more frequently and with higher force than molecules with lower velocity and that’s what creates Pressure (P=F/A). Think of the force you would feel in a collision if The car you’re in hits a wall while moving at 5 mph vs. 30 mph Since KE (1/2 mv 2 ) increases with increasing T then so Does molecular velocity

5 Developing Boyles Law So, The frequency of collision per unit area, and the force of the collision will determine the final pressure. So if P is increased by increasing the Temperature how Can we explain Boyles Law which holds T constant as Pressure is made to change by changing volume? Let’s see how this works.

6 Developing Boyles Law What do you think happens to the number of collisions Per unit area of the container walls if we take the same molecules from the larger cube and put them into the smaller cube? So what happens to P?

7 Developing Boyles Law The relationship between P and V is known as BOYLE’S LAW. PV=k So Pressure x volume is always equal to the same value To keep k constant P must increase the same number of Times V decreases (and vice versa). This is known as an inverse relationship P, V = variablesT, n = Constants

8 Developing Boyles Law Let’s look at an example: If P1 = 2 atm and the vol = 64cm 3 then PV=K2atm x (64cm 3 )= k sok=128 What is P when the volume dec. to 8cm3 ? PV =k P(8) = 128so P increases to 16 atm (i.e V decreased 1/8th and P increases 8x)

9 Below is a typical data table comparing P,V measurements: Trial P (atm) V (cm 3 ) PV (atm*cm 3 ) A graph of this data looks like this An inverse relationship P V Boyles law What does a plot of this data looks like?

10 Developing Boyles Law If P1V1=K and P2V2=k (i.e. different p and v values) Then P1V1 = P2V2 This is the form of Boyles law we will use to solve Problems If we know three of the variables we can calculate the fourth: For example if we wanted to find the new volume in the image to the right it would be P1V1/P2 = V2 Note: units on both sides of the equation MUST BE the same!)

11 Boyles Law Problem In the above figure what would P 2 be if P 1 is 1 atm & V 1 is 5L (V 2 is still 0.5L)? P 1 V 1 =P 2 V 2 ; 1atm(5L) = P 2 (0.5L); P 2 = (1x5)/0.5 = 10 atm (P 1 & V 1 ) (P 2 & V 2 ) P 1 V 1 = P 2 V 2 1atm x 1L = 2atm x V2 V2 = 0.5L

12 Problem 1: A 250 ml sample of gas at 26 KPa is expanded to 1000 ml at Constant T. What is the pressure exerted by the gas at the new volume? (note: n is not mentioned. This is common. If n is not Mentioned assume it is constant) P1v1 = P2V226KPa(250ml) = P2(1000ml) P2 = 26KPa(250ml)/1000ml = 6.5KPa

13 Problem 2: A 5.5L sample of gas at 700 torr is compressed to ml at Constant T. What is the pressure exerted by the gas at the new volume? P 1 V 1 = P 2 V 2 700torr(5.5L) = P 2 (0.25L) P 2 = 700torr(5.5L)/0.25L = 15,400 torr NOTE: Make sure you keep units the SAME! How many atm is this? 20.26atm

14 PROBLEM 3: A discarded spray paint can contains only a small volume of the propellant gas at a pressure of 34,470 Pa. The volume of the can is ml. If the can is run over by a garbage truck and flattened to a volume of ml what is the pressure (in Pa & atm) in the can Assuming there’s no leaks? P 1 V 1 = P 2 V 2 34,470Pa x ml = P 2 x 13.16ml P 2 = / = 1,239,401 Pa = atm