1. Gas Laws A review

2. Important Information What is STP? Standard Temperature and Pressure. 1 atm pressure and 273 Kelvin What are standard conditions? Pressure is still 1 atm, but the temperature is 25 degrees C or 298 K What is STP? Standard Temperature and Pressure. 1 atm pressure and 273 Kelvin What are standard conditions? Pressure is still 1 atm, but the temperature is 25 degrees C or 298 K

3. Units Units are very important in this chapter. One must be consistent or be incorrect. All temperature must be in Kelvin. °C + 273 = Kelvin Pressure may be in one of these choices: 1 atm760 mmHg 760 torr101.3 kPa Units are very important in this chapter. One must be consistent or be incorrect. All temperature must be in Kelvin. °C + 273 = Kelvin Pressure may be in one of these choices: 1 atm760 mmHg 760 torr101.3 kPa

4. Effusion and Diffusion Effusion Diffusion The rate at which a gas moves from area of high concentration to low concentration The passage of a gas through a tiny orifice.

5. Boyles Law http://en.wikipedia.org/wiki/Boyle's_law Pressure and Volume are inversely proportional when temperature is constant. As Pressure increases, Volume decreases and visa versa. 1P 1 V 1 =P 2 V 2 If a system with a pressure of 5 atm and 45 ml is compressed to 30 ml, what is the new pressure? http://en.wikipedia.org/wiki/Boyle's_law Pressure and Volume are inversely proportional when temperature is constant. As Pressure increases, Volume decreases and visa versa. 1P 1 V 1 =P 2 V 2 If a system with a pressure of 5 atm and 45 ml is compressed to 30 ml, what is the new pressure?

6. Boyle’s Law 1P 1 V 1 = P 2 V 2 5 atm 45 ml = X 30 ml X=(5atm*45ml)/30ml X= 7.5 atm 1P 1 V 1 = P 2 V 2 5 atm 45 ml = X 30 ml X=(5atm*45ml)/30ml X= 7.5 atm

7. Charles’ Law http://en.wikipedia.org/wiki/Charles's_law Volume and Temperature are directly proportional when pressure is constant. As Volume increases, so does Temperature. V1 = V2 T1 T2 What happens to a 1 liter balloon at 25 degrees Celsius if the temperature is reduced to 0 degrees Celsius? http://en.wikipedia.org/wiki/Charles's_law Volume and Temperature are directly proportional when pressure is constant. As Volume increases, so does Temperature. V1 = V2 T1 T2 What happens to a 1 liter balloon at 25 degrees Celsius if the temperature is reduced to 0 degrees Celsius?

8. Charles’ Law V1 1 liter = V2 X T1 25°C T2 0°C Temperature must be in Kelvin!!!! C+ 273=K V1 1 liter = V2 X T1 298 K T2 273 K X= (1 L*273 K)/298K X=0.916 L new volume V1 1 liter = V2 X T1 25°C T2 0°C Temperature must be in Kelvin!!!! C+ 273=K V1 1 liter = V2 X T1 298 K T2 273 K X= (1 L*273 K)/298K X=0.916 L new volume

9. Gay-Lussac’s Law Pressure and Temperature are directly proportional when the volume remains constant. The pressure increases as the temperature in kelvin increases. Pressure and Temperature are directly proportional when the volume remains constant. The pressure increases as the temperature in kelvin increases.

10. Gay-Lussac’s Law What is the new pressure if a ridged container at 1 atm and 25 °C is heated to 75° C? P1 1 atm = P2 X T1 25°C T2 75°C Temperature must be in Kelvin!!!! C+ 273=K P1 1 liter = P2 X T1 298 K T2 348 K What is the new pressure if a ridged container at 1 atm and 25 °C is heated to 75° C? P1 1 atm = P2 X T1 25°C T2 75°C Temperature must be in Kelvin!!!! C+ 273=K P1 1 liter = P2 X T1 298 K T2 348 K

11. Gay-Lussac’s Law X= (1 L*348 K)/298K X= 1.17 L X= (1 L*348 K)/298K X= 1.17 L

12. Combined Gas Law Pressure and volume are inversely proportional. Both Pressure and Volume are directly proportional to temperature. We can combine Charles’, Boyle’s and Gay- Lussac’s Law to form the Combined gas law. Pressure and volume are inversely proportional. Both Pressure and Volume are directly proportional to temperature. We can combine Charles’, Boyle’s and Gay- Lussac’s Law to form the Combined gas law.

13. Combined Gas Law You have an expandable syringe holding a volume of 20ml at STP. What is the new volume at Standard Conditions? P1=1 atm * V1 =20mL P2=1 atm * V2 =X T1 = 273 K = T2 = 298 K X= 21.8 ml You have an expandable syringe holding a volume of 20ml at STP. What is the new volume at Standard Conditions? P1=1 atm * V1 =20mL P2=1 atm * V2 =X T1 = 273 K = T2 = 298 K X= 21.8 ml

14. Ideal Gas Law The main difference with the ideal gas law is that now we take the number of moles of gas into consideration. All previous gas laws had a constant amount of gas. In the ideal gas law, the number of moles of a gas is also a variable. To do this the equation requires a conversion factor called, R. The main difference with the ideal gas law is that now we take the number of moles of gas into consideration. All previous gas laws had a constant amount of gas. In the ideal gas law, the number of moles of a gas is also a variable. To do this the equation requires a conversion factor called, R.

15. Ideal Gas Law PV=nRT Pressure times Volume equals the number of moles times the gas constant times the Temperature in Kelvin. There are many versions of the Gas constant, R. We will use only two. R= 0.0821 (L*atm)/(mol*Kelvin) R= 8.31 (L*kPa)/(mol*Kelvin) PV=nRT Pressure times Volume equals the number of moles times the gas constant times the Temperature in Kelvin. There are many versions of the Gas constant, R. We will use only two. R= 0.0821 (L*atm)/(mol*Kelvin) R= 8.31 (L*kPa)/(mol*Kelvin)

16. Units and the Ideal Gas Law The units of your problem must match the units in your gas constant, R. If they do not you need to convert those units that do not match. Example: I have 20 grams of neon gas at 760 mmHg and 0° C. What is the volume of this gas? The units of your problem must match the units in your gas constant, R. If they do not you need to convert those units that do not match. Example: I have 20 grams of neon gas at 760 mmHg and 0° C. What is the volume of this gas?

17. Answer Which R? I chose R=0.08206 (L*atm)/(mole*Kelvin) First we need moles of gas not grams 20 g/(20g/mol)=1 mole of Neon Pressure from mmHg to atm 760 mmHg*(1 atm/760 mmHg)= 1 atm Temperture from Celsius to Kelvin 0° C +273=273 Kelvin Which R? I chose R=0.08206 (L*atm)/(mole*Kelvin) First we need moles of gas not grams 20 g/(20g/mol)=1 mole of Neon Pressure from mmHg to atm 760 mmHg*(1 atm/760 mmHg)= 1 atm Temperture from Celsius to Kelvin 0° C +273=273 Kelvin

18. Answer continued PV=nRT 1atm(V)= [1 mol*0.0821(L*atm)/(mol*K)]273K V=[1mol*0.0821 (L*atm/mol*K)*273K]/1atm Cancel your units and do the math: V=0.0821 L/273 V= 22.4 L In fact, all gases at STP occupy 22.4 L/mol PV=nRT 1atm(V)= [1 mol*0.0821(L*atm)/(mol*K)]273K V=[1mol*0.0821 (L*atm/mol*K)*273K]/1atm Cancel your units and do the math: V=0.0821 L/273 V= 22.4 L In fact, all gases at STP occupy 22.4 L/mol

19. How to determine which to use Ideal or Combined? Does the problem mention a change in either Temperature, Pressure or Volume? Combined Gas Law! Does the problem give or ask for the number of grams of a gas or the number of moles of a gas? Ideal Gas law! Ideal or Combined? Does the problem mention a change in either Temperature, Pressure or Volume? Combined Gas Law! Does the problem give or ask for the number of grams of a gas or the number of moles of a gas? Ideal Gas law!

20. Dalton’s Law of Partial Pressures In a mixture of gases, each gas has its own partial pressure. The total pressure in the container is the sum of the partial pressures. P T =P 1 +P 2 +P 3 …Pn It is important that all the pressures are in the same units before adding them together! In a mixture of gases, each gas has its own partial pressure. The total pressure in the container is the sum of the partial pressures. P T =P 1 +P 2 +P 3 …Pn It is important that all the pressures are in the same units before adding them together!

21. Graham’s Law of Effusion Larger molecules will migrate slower than a smaller molecule under a constant temperature. This is in direct violation of the KMT. REAL GASES have significant size Larger molecules will migrate slower than a smaller molecule under a constant temperature. This is in direct violation of the KMT. REAL GASES have significant size

22. Graham’s Law Which of these gases will effuse faster, He, Ne, Xe? When all gases are at the same temperature, they have the same kinetic energy. Therefore the smallest gas, Helium, will effuse faster. Which of these gases will effuse faster, He, Ne, Xe? When all gases are at the same temperature, they have the same kinetic energy. Therefore the smallest gas, Helium, will effuse faster.