Ideal Gas Laws. How to make calculations when the gas under consideration is ideal

Transcription

1 Ideal Gas Laws How to make calculations when the gas under consideration is ideal

2 A bit of explanation about the assumption involved on the molecular level about an ideal gas: It is assumed that the molecules are very small (even point-like) and 1) Do not collide with each other and 2) Do not take up any space Given these assumptions, one can derive the ideal gas equation - which is:

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4 If 3 of the 4 variables are known, Then this law may be used as is to solve for the fourth variable.

5 If 3 of the 4 variables are known, Then this law may be used as is to solve for the fourth variable. Otherwise, it can be used parametrically, that is, comparing two conditions. Rearranging this equation and labeling the two conditions by subscripts 1 and 2: PV 1 1 PV 2 2 R and R nt 1 1 nt 2 2 since they both equal R then: PV 1 1 PV 2 2 nt nt

6 To use this equation P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 parametrically, simply cross out those variables which are constant from condition 1 to 2.

7 To use this equation P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 parametrically, simply cross out those variables which are constant from condition 1 to 2. For example if n and T are kept constant, then: P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 so: P 1 V 1 P 2 V 2 which is Boyle s law.

8 To use this equation P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 parametrically, simply cross out those variables which are constant from condition 1 to 2. For example if n and P are kept constant, then: P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 so: V 1 V 2 T 1 T 2 which is Charle s law.

9 To use this equation P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 parametrically, simply cross out those variables which are constant from condition 1 to 2....and if T and P are kept constant, then: P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 so: V 1 V 2 n 1 n 2 which is Avogadro s law.

10 PV nrt The first examples will use this equation as is, that is, given 3 of the variables, solve for the fourth.

11 Example 1: PV nrt Calculate the pressure for 2.5 mol of a gas confined to 14.0 L at 25EC. Given: P? V 14.0 L n 2.5 mol t 25EC but needs to be converted to K so: T ( ) K or: T 298 K

12 Example 1: PV nrt Calculate the pressure for 2.5 mol of a gas confined to 14.0 L at 25EC. Given: P? V 14.0 L n 2.5 mol T 298 K Substituting: P(14.0 L) (2.5 mol)( L atm K -1 mol -1 )(298 K)

13 Example 1: PV nrt Calculate the pressure for 2.5 mol of a gas confined to 14.0 L at 25EC. Given: P? V 14.0 L n 2.5 mol T 298 K Solving: P 4.4 atm

14 Example 2: PV nrt How many grams of He is require to fill a balloon that has a volume of 52.0 L to a pressure of 1.25 atm at 15EC?

15 Example 2: PV nrt How many grams of He is require to fill a balloon that has a volume of 52.0 L to a pressure of 1.25 atm at 15EC? P 1.25 atm V 52.0 L n? T ( ) K 288 K

16 Example 2: PV nrt How many grams of He is require to fill a balloon that has a volume of 52.0 L to a pressure of 1.25 atm at 15EC? P 1.25 atm V 52.0 L n? T ( ) K 288 K Substituting: (1.25 atm)(52.0 L) n( L atm K -1 mol -1 )(288 K)

17 Example 2: PV nrt How many grams of He is require to fill a balloon that has a volume of 52.0 L to a pressure of 1.25 atm at 15EC? P 1.25 atm V 52.0 L n? T ( ) K 288 K P(1.25 atm)(52.0 L) n( L atm K-1 mol-1)(288 K) evaluating: n 2.75 mol Is this the answer?

18 Example 2: PV nrt How many grams of He is require to fill a balloon that has a volume of 52.0 L to a pressure of 1.25 atm at 15EC? P 1.25 atm V 52.0 L n? T ( ) K 288 K (1.25 atm)(52.0 L) ( L atm K-1 mol-1)n(288 K) evaluating: n 2.75 mol No! The question is How many grams?

19 Example 2: PV nrt How many grams of He is require to fill a balloon that has a volume of 52.0 L to a pressure of 1.25 atm at 15EC? P 1.25 atm V 52.0 L n? T ( ) K 288 K n 2.75 mol m Mn so: m (4.002 g mol-1)(2.75 mol)

20 Example 2: PV nrt How many grams of He is require to fill a balloon that has a volume of 52.0 L to a pressure of 1.25 atm at 15EC? P 1.25 atm V 52.0 L n? T ( ) K 288 K n 2.75 mol m Mn and evaluating: m 11.0 g which is the answer.

21 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 The following examples may be solved using the parametric equation given above. These questions will have a before condition labeled with the subscript 1" and an after condition labeled 2".

22 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 Example 3: A gas in a rigid (leak-proof) container is initially at 25EC and 50.0 atm. The container is heat to 300EC. What is the pressure at this temperature?

23 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 Example 3: A gas in a rigid (leak-proof) container is initially at 25EC and 50.0 atm. The container is heat to 300EC. What is the pressure at this temperature? n and V remain constant. so: and: P 1 P 2 T 1 T 2 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2

24 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 Example 3: A gas in a rigid (leak-proof) container is initially at 25EC and 50.0 atm. The container is heat to 300EC. What is the pressure at this temperature. P 1 P 2 T 1 T 2 P atm P 2? T 1 ( ) K 298 K T 2 ( ) K 573 K Substituting: 50.0 atm P K 573 K

25 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 Example 3: A gas in a rigid (leak-proof) container is initially at 25EC and 50.0 atm. The container is heat to 300EC. What is the pressure at this temperature. P 1 P 2 T 1 T 2 P atm P 2? T 1 ( ) K 298 K T 2 ( ) K 573 K Solving: P atm

26 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 Example 4: A gas is in a leak-proof cylinder that can expand. Initially the cylinders volume is 15.0 qt. and is expanded to 55.0 qt. If the gas pressure was originally 14.0 psi, what would its final pressure in psi be at the same temperature?

27 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 Example 4: A gas is in a leak-proof cylinder that can expand. Initially the cylinders volume is 15.0 qt. and is expanded to 55.0 qt. If the gas pressure was originally 14.0 psi, what would its final pressure in psi be at the same temperature? Here T and n are held constant. So: P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 and: P 1 V 1 P 2 V 2 (i.e. Boyle s law)

28 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 Example 4: A gas is in a leak-proof cylinder that can expand. Initially the cylinders volume is 15.0 qt. and is expanded to 55.0 qt. If the gas pressure was originally 14.0 psi, what would its final pressure in psi be at the same temperature? P 1 V 1 P 2 V 2 V qt V qt P psi P 2?

29 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 Example 4: A gas is in a leak-proof cylinder that can expand. Initially the cylinders volume is 15.0 qt. and is expanded to 55.0 qt. If the gas pressure was originally 14.0 psi, what would its final pressure in psi be at the same temperature? P 1 V 1 P 2 V 2 V qt V qt P psi P 2? Notice that for a parametric equation, the units do not matter provided they are consistent from one side of the equation to the other.

30 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 Example 4: A gas is in a leak-proof cylinder that can expand. Initially the cylinders volume is 15.0 qt. and is expanded to 55.0 qt. If the gas pressure was originally 14.0 psi, what would its final pressure in psi be at the same temperature? P 1 V 1 P 2 V 2 Substituting: (14.0 psi)(15.0 qt) P 2 (55.0 qt) V qt V qt P psi P 2?

31 P 1 V 1 P 2 V 2 n 1 T 1 n 2 T 2 Example 4: A gas is in a leak-proof cylinder that can expand. Initially the cylinders volume is 15.0 qt. and is expanded to 55.0 qt. If the gas pressure was originally 14.0 psi, what would its final pressure in psi be at the same temperature? P 1 V 1 P 2 V 2 (14.0 psi)(15.0 qt) P 2 (55.0 qt) V qt V qt P psi P 2? and evaluating: P psi

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