Presentation is loading. Please wait.

Presentation is loading. Please wait.

Gas Laws 1.00 atm = 760 torr = 760mmHg = 14.7psi = 101.3kPa Standard Temperature and Pressure (STP) = 1.00 atm at 273K.

Similar presentations


Presentation on theme: "Gas Laws 1.00 atm = 760 torr = 760mmHg = 14.7psi = 101.3kPa Standard Temperature and Pressure (STP) = 1.00 atm at 273K."— Presentation transcript:

1 Gas Laws 1.00 atm = 760 torr = 760mmHg = 14.7psi = 101.3kPa Standard Temperature and Pressure (STP) = 1.00 atm at 273K

2 TOP OF BACK COVER Gas laws are based on 1. Volume 2. Temperature 3. Pressure 4. Amount of gas

3 BOTTOM OF BACK COVER How to solve gas law problems: 1. Write your givens. 2. Choose the correct formula based on givens. 3. Convert temperature to Kelvin if necessary. 4. Convert pressures to same unit if necessary. 5. Solve for unknown. 6. Check for sig figs.

4 Boyle’s Law TOP FLIP: Definition: the volume of a given mass of gas at a constant temperature varies inversely with the pressure pressure increases, volume decreases pressure decreases, volume increases P 1 V 1 = P 2 V 2

5 Boyle’s Law BOTTOM FLIP: Practice problem: The volume of a gas at 99.0 kPa is 300.0 mL. If the pressure is increased to 188 kPa, what will the new volume be? Given:

6 Charles’ Law Definition: the volume of a given mass of gas is directly proportional to its Kelvin temperature at constant pressure Kelvin = °C + 273 volume decreases, temperature decreases volume increases, temperature increases V 1 = V 2 T1T2T1T2

7 Charles’ Law Practice Problem: A gas at 89°C occupies a space of 0.67 L. At what °C will the volume increase to 1.12L? Given:

8 Gay-Lussac’s Law  Definition: the pressure of a given mass of a gas varies directly with the Kelvin temperature when the volume remains constant  P, T P 1 = P 2 T 1 T 2

9 Gay-Lussac’s Law Practice problem: A gas in a sealed container has a pressure of 125 kPa at a temp of 30.0°C. If the pressure increases to 201 kPa, what is the new temp?

10 Combined Gas Law Single law combining Boyle’s, Charles’ and Gay-Lussac’s laws and states the relationship between pressure, volume and temperature of a fixed amount of gas P 1 V 1 = P 2 V 2 T 1 T 2

11 Combined Gas Law Problem: A gas at 110. kPa and 30.0°C fills a flexible container with an initial volume of 2.00L. If the temp is raised to 80.0°C and the pressure is increased to 440. kPa, what is the new volume?

12 Ideal Gas Law Definition: shows the relationship between pressure, volume, temperature and the number of moles within a sample of gas Ideal gases follow KMT. do not take up space. do not have intermolecular forces. follow all gas laws. No gases are truly ideal! But…. Real gases behave like ideal gases under most conditions. but not at very low temp or very high pressure

13 Ideal Gas Law Formula: PV= nRT P – pressure V – volume n - # of moles R – ideal gas constant T – temp in K Ideal gas constant NOT A FORMULA 0.0821Latm molK 8.314L kPa molK 62.4L mmHg molK

14 Ideal Gas Law  Formula can be rearranged to find molar mass or density of a gas.  M= mRT(M - molar massm – mass ) PV  D=MP RT

15 Practice Problems 1. If the pressure exerted by a gas at 25°C in a volume of 0.044L is 3.81 atm, how many moles are present? P=Which constant?? V= T=

16 Practice Problems 2. What is the molar mass of a pure gas that has a density of 1.40 g/L @ STP?


Download ppt "Gas Laws 1.00 atm = 760 torr = 760mmHg = 14.7psi = 101.3kPa Standard Temperature and Pressure (STP) = 1.00 atm at 273K."

Similar presentations


Ads by Google