Gas laws(gay Lussac's and combined gas laws)pptx

The ppt is not mine, it's from my masteral classmate ma'am Karen Chavez

1. AMONTON’S LAW AND COMBINED GAS LAW Here starts the lesson! KARREN GRACE N. CHAVEZ

2. GAY-LUSSAC’S LAW OR AMONTON’S LAW

3. Joseph Gay-Lussac formulated the law in 1802, but the formal statement of the relationship between temperature and pressure described by French physicist Guillaume Amonton in the late 1600’s. Joseph- Gay Lussacs (1802) Guillame Amonton (1600)

4. Amonton proved the same law by making a thermometer where the measured pressure was a readout for the current temperature.

5. PRESSURE TEMPERATURE VOLUME AMOUNT( mol ) Most Common Units 1 atm = 101.325 Pa 1 atm = 760 mm Hg = 760 tons 1 atm =14.70 l b /in 2 1 bar= 100,000 Pa = 0.989 atm 1 L= 1,000 ml T (K) = T( C) + 273 Equivalent for Avogadro’s number or 6.02 x 10 23

6. PRESSURE ABSOLUTE TEMPERATURE

7. During travel the tire of your car heat up because of friction. The temperature inside your tires will increase but the volume of air does not increase because it contained in the tire, hence resulting to the increase of pressure. Example: Tire Pressure

8. It states that the pressure of a given gas varies directly with the absolute temperature of the gas, when the volume is kept constant.

9. P 1 T 2= P 2 T 1 or

10. Sample Problem! The pressure of a nitrogen gas inside a rigid tank is 1.7 atmosphere at 30 C. Compute for the resulting pressure if the tank is cooled to 0 C

11. P 1 = 1.7 atm P 2 = ? T 1 = 30 C + 273 K= 303 K T 2 = C + 273 K= 273 K P 2 = P 1 T 2 T 1 P 2 = (1.7 atm )(273K) 303K P 2 = 464.1 atm.k 303 k P 2 = 1.5 atm P 1 T 2 = P 2 T 1 P 1 T 2 = P 2 T 1 T 1 T 1 List the known and the unknown: Write the equation . Rearrange the equation . Substitute the known Values.

12. Try this! A container of a gas is at 25 C and 225 kPa . If can withstand a pressure of 1000 kPa , at what temperature will it rupture?

13. P 1 = 225 kPa P 2 = 1000 kPa T 1 = 25 C + 273K = 298K T 2 = ? 760 mm Hg= 1 atm Identify the given: Solution: P 1 T 2 = P 2 T 1 P 1 T 2 = P 2 T 1 P 1 P 1 T 2 = P 2 T 1 P 1 T 2 = (1000 kPa )(298K ) 225 kPa   T 2 = 298,000 kPa.K 225 kPa   T 2 = 1,324 K

14. COMBINED GAS LAW ( Charles’ Law, Boyle’s Law, and Gay-Lussac’s Law)

15. The combined gas law is a gas law that combines Charles’ Law, Boyle’s Law, and Gay- Lussacs Law. Boyle’s Law Charles’ Law Gay-Lussac’s Law

16. This law emphasizes the ratio of the product of pressure, volume, and absolute temperature of a gas is equal to a constant. There are a couple of common equations for calculating the combined gas law COMBINED GAS LAW PV/T=k

17. Formula:

18. Sample Problem! A gas occupies 40 L at 1 atm and 200k. How much pressure will the gas exert if the temperature increases to 400K and its volume decreases to 20 L?

19. Given: P 1 = 1 atm P 2 = x V 1 = 40L V 2 = 20 L T 1 = 200K T 2 = 400K P 2 = 4 atm P 2 = P 1 V 1 T 2 V 2 T 1 P 2 = (1 atm )(40 L)(400 K) ( 20 L)(200 K)   P 1 V 1 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = P 2 V 2 T 1 V 2 T 1 V 2 T 1 P 2 = 16,000 atm.L.K 4,000 L.K

20. Try this! A gas has an initial pressure of 400 atm , a temperature of 40K, and a volume of 200 mL. The volume is then increased to 300 mL and the pressure is decreased to 200 atm , what is the new temperature?

21. P 1 = 400 atm P 2 = 200 atm V 1 = 200 mL V 2 = 300 mL T 1 = 40 K T 2 = ? Identify the given: P 1 V 1 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = P 2 V 2 T 1 P 1 V 1 P 1 V 1 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = (200 atm )(300 mL)(40K ) (400 atm )(200 mL)   T 2 = 2, 400, 000 atm.mL.K 80, 000 atm. mL T 2 = 30 K