Unit 13 Gas Laws Gases
The Gas Laws Kinetic Theory Revisited 1. Particles are far apart and have negligible volume. 2. Move in rapid, random, straight-line motion. 3. Collide elastically. 4. No attractive or repulsive forces.
The Gas Laws Variables That Describe A Gas 1. Pressure (P) kpa (atm and mmhg) 2. Volume (V) L (ml) 3. Temperature (T) K T K = 273 + T ºC 4. Amount of gas (n) mol
STP Not Just a Motor Oil STP standard temperature and pressure 0ºC (273 K) 1 atm (101.3 kpa)
The Gas Laws Amount of a Gas As the number of particles increases the pressure increases. n P
The Gas Laws Volume As volume increases the pressure decreases. V P
The Gas Laws Temperature As the temperature increases the pressure increases. T P
The Gas Laws Boyle s Law At constant temperature and mass (moles), the pressure is inversely proportional to the volume. P 1 V 1 = P 2 V 2
Solving a gas law problems 1. Set up a variable table. Are the units consistent and appropriate? Consistent Pressures and volumes must be in the same type of unit. Appropriate Temperature and amounts must be in Kelvin and moles. Pressure and volumes might have to be in a specified unit. 2. Establish which equation you ll be using. 3. Solve the equation for the unknown variable. DO NOT SUBSTITUTE YET!!! 4. Substitute and solve!
Boyle s Law Problem A balloon contains 30.0 L of helium gas at 103 kpa. What is the volume when rises to an altitude where the pressure is only 25.0 kpa? (assume the temp is constant) P 1 = 103 kpa P 2 = V 1 = 30.0 L V 2 = 25.0 kpa? P 1 V 1 = P 2 V 2 P 2 P 2 P 1 V 1 V 2 = = P 2 (103 kpa)(30.0 L) 25.0 kpa = 123.6 L 124 L
The Gas Laws Charles s Law At constant pressure and mass (moles), the volume of a gas is directly proportional to the Kelvin temperature. V 1 V 2 = T 1 T 2
Charles s Law Problem A balloon inflated in a room at 24 C has a volume of 4.00 L. The balloon is then heated to a temperature of 58 C. What is the new volume if pressure is constant? T 1 = 24 C 297 K T 2 = 58 C 331 K V 1 = 4.00 L V 2 =? Temperature MUST be in Kelvin (add 273) T 2 V 1 V 2 = T 2 T 1 T 2 V 2 = T 2 V 1 T 1 = (331 K)(4.00 L) (297 K) = 4.4579 L 4.46 L
The Gas Laws Gay-Lussac s Law At constant volume and mass (moles), the pressure of a gas is directly proportional to the Kelvin temperature. P 1 P 2 = T 1 T 2
Gay-Lussac s Law Problem The gas left in a used aerosol can is at a pressure of 103 kpa at 25 C. If the can is thrown onto a fire and the pressure increases to 415 kpa, what is the temperature? P 1 = 103 kpa P 2 = T 1 = 25 C 298 K T 2 = 415 kpa? P 2 P 1 P 2 = T 1 T 2 P 2 Do we want to solve for? 1 T 2
Gay-Lussac s Law Problem The gas left in a used aerosol can is at a pressure of 103 kpa at 25 C. If the can is thrown onto a fire and the pressure increases to 415 kpa, what is the temperature? P 1 = 103 kpa P 2 = T 1 = 25 C 298 K T 2 = 415 kpa? T 2 = P 2 T 1 P 1 = T 1 T 2 P 1 P 2 T 2 T 1 = P 1 T 1 T 2 P 1 (415 kpa)(298 K) 103 kpa = 1200.6 K 1.20 10 3 K
The Combined Gas Laws Combined Gas Law P 1 V 1 T 1 = P 2 V 2 T 2
Boyle s Law Lab Reminders 1. Watch rounding and units everywhere! 1. Data #1,3,4 and Calculations #1 need units. 2. Rounding and units for Data #3 and 4 are determined by the data used to make your graph. 3. If you have the units correct, you actually should get mmhg/b for Calculation #2 (B is book). 4. The final answer should be in the 1000 s. 5. Does your graph have appropriate labels everywhere (title and axis) and units?
Where do all of these gas laws come from? What happened to moles?
The Ideal Gas Law Ideal Gas Law PV = nrt P = pressure (kpa, atm, or mmhg) V = volume (L) n = moles (mol) T = temperature (K) R = ideal gas constant R = 8.31 L kpa mol K or 0.0820 L atm mol K or 62.4 L mmhg mol K
The Ideal Gas Law ALL OF THE OTHER GAS LAWS CAN BE DERIVED FROM THE IDEAL GAS LAW! PV = nrt Solve for the nt nt constant R! P 1 V 1 n 1 T 1 = R = P 2 V 2 n 2 T 2
Ideal Gas Law Problem What is the volume of 14.2 g of O 2(g) at STP? PV = nrt P P 14.2 g O 2 1 mol O 2 32.0 g O 2 =.444 mol O 2 L kpa V (.444 mol)(8.31 mol K )(273K) = 101.3 kpa = 9.94 L
The Ideal Gas Law Other expressions of the Ideal Gas Law molar mass (M) is: M mass( m) mole(n) Therefore n = m/m, substituting this into the ideal gas law yields PV mrt M
The Ideal Gas Law Solving for molar mass (M) M mrt PV Since density (D) = mass/volume M Solving for density DRT P MP D RT
The Ideal Gas Law Ideal Gases follow the assumptions of kinetic theory. No attractive forces between particles and negligible volume Under most temperatures and pressures most real gases behave like ideal gases When do gases become less ideal and more real? Low temperature High pressure
Going Back particles Avogadro s Principle Equal volumes of gases at the same temperature and pressure contain an equal number of particles.
Molar volume Solving for volume in the Ideal Gas Law for 1 mole at STP we obtain: V (1mole)(8.314 V L kpa mol K 101.3kPa nrt P )(273K) Therefore, at STP 1 mole = 22.4 L This known as the Standard Molar Volume 22.4 L
Remember this problem! What is the volume of 14.2 g of O 2(g) at STP? (Same question we did in Mole Unit) The previous equality gives us shortcut for any question at STP! 14.2 g O 2 1 mol O 2 22.4 L O 2 = 9.94 L O 32.0 g O 2 2 1 mol O 2 This is the same answer as before!
Don t Forget About Dalton!!! Dalton s Law Correction Because most gases are COLLECTED OVER WATER, you will often use this law to correct the pressure. P total = P gas + P water Solving for the gas yields: This is usually the P 1 or P in a gas law problem! P gas = P total P water