1. Boyle’s, Charles’, Gay Lussac’s and Combined Gas Laws
The GAS LAWS
Boyle’s, Charles’, Gay Lussac’s and Combined Gas Laws

2. Boyle’s Law P1 V1 = P2 V2
1.00 L of hydrogen at standard pressure is compressed to 473 mL. What is the new pressure of the hydrogen?

3. P1 V1 = P2 V2 P2 = 2.11 atm H2 (1atm) (1.00 L) = P2 (.473 L)
First we must convert mL to L
473mL= .473 L
(1atm) (1.00 L) = P2 (.473 L)
(.473 L) (.473 L)
P2 = 2.11 atm H2

4. Boyle’s Law P1 V1 = P2 V2
Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in her blood under a pressure of 250 atm, then rises instantaneously to a depth where her blood has a pressure of 50.0 atm, what will the volume of gas in her blood be? Do you think this will harm the diver?

5. V2 = 0.25 L, yes this will harm the diver. Ouch!
P1 V1 = P2 V2
(250 atm) (0.05 L) = (50.0 atm) V2
(50.0 atm) (50.0 atm)
V2 = 0.25 L, yes this will harm the diver. Ouch!

6. Charles’ Law V1/T1= V2/T2
On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car (which has a temperature of 600 C), what will the new volume of the bag be?

7. V1/T1= V2/T2 V2 = 285 mL or 0.285 L (250 mL)(333 K ) = V2 (292 K)
First we must convert 0C to Kelvins
T1= 19 0C + 273= 292K
T2= 60 0C + 273= 333K
(250 mL)(333 K ) = V2 (292 K)
(292 K) (292 K)
V2 = 285 mL or L

8. Charles’ Law V1/T1= V2/T2
A soda bottle is flexible enough that the volume of the bottle can change even without opening it. If you have an empty soda bottle (volume of 2 L) at room temperature (25 0C), what will the new volume be if you put it in your freezer (-4 0C)?

9. V1/T1= V2/T2 V2 =1.81 L (2 L) (269 K ) = V2 (298 K) (298 K) (298 K)
First we must convert 0C to Kelvins
T1= 25 0C + 273= 298K
T2= -4 0C + 273= 269K
(2 L) (269 K ) = V2 (298 K)
(298 K) (298 K)
V2 =1.81 L

10. Charles’ Law V1/T1= V2/T2
How hot will a 2.3 L balloon have to get to expand to a volume of 400 L? Assume that the initial temperature of the balloon is 25 0C.

11. V1/T1= V2/T2 T2 =51,800 K (2.3 L) T2 = (400 L) (298 K )
First we must convert 0C to Kelvins
T1= 25 0C + 273= 298K
(2.3 L) T2 = (400 L) (298 K )
(2.3 L) (2.3 L)
T2 =51,800 K

12. Gay Lussac’s Law P1/ T1 = P2/ T2
5.00 L of a gas is collected at 22.0°C and mmHg. When the temperature is changed to standard, what is the new pressure?

13. P2 = 689 mm Hg P1/ T1 = P2/ T2 (745 mm Hg) (273 K) = P2 (295 K)
First we must convert 0C to Kelvins
T1= 22 0C + 273= 295K
T2= 0 0C + 273= 273K
(745 mm Hg) (273 K) = P2 (295 K)
(295 K) (295 K)
P2 = 689 mm Hg

14. Gay Lussac’s Law P1/ T1 = P2/ T2
A container is initally at 47 mm Hg and 77 K (liquid nitrogen temperature.) What will the pressure be when the container warms up to room temperature of 25 oC?

15. P2 = 182 mm Hg P1/ T1 = P2/ T2 (47 mm Hg) (298 K) = P2 (77 K)
First we must convert 0C to Kelvins
T2= 25 0C + 273= 298K
(47 mm Hg) (298 K) = P2 (77 K)
(77 K) (77 K)
P2 = 182 mm Hg

16. Combined Gas Laws P1 V1 = P2 V2 T1 T2
If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

17. V2 = 29.6 L P1 V1 = P2 V2 T1 T2 (12 atm)(23 L)(300 K)=
(14 atm) V2 (200K)
(14 atm) (200 K)
(14 atm) (200K)
V2 = 29.6 L

18. Combined Gas Laws P1 V1 = P2 V2 T1 T2
If I have 17 liters of gas at a temperature of 67 0C and a pressure of atm, what will be the pressure of the gas if I raise the temperature to 94 0C and decrease the volume to 12 liters?

19. P2 = 136 atm P1 V1 = P2 V2 T1 T2 P2 (12 L) (340K) (12 L) (340 K)
First we must convert 0C to Kelvins
T1= 67 0C + 273= 340K
T2= 94 0C + 273= 367K
P2 (12 L) (340K)
(88.89 atm)(17 L)(367 K)=
(12 L) (340 K)
(12 L) (340K)
P2 = 136 atm