Gases Chapter 5

Substances that exist as gases Elements that exist as gases at 25 0 C and 1 atmosphere

5.1

Our Atmosphere: exerts pressure on earth more at sea level less on mountain top The air we breathe: 79% N 2 21% O 2 Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm

Atmospheric pressure 1 atm = 760 mmHg = 760 torr (1 torr = 1mm Hg) 1 atm = 101,325 Pa 1 atm = 101 kPa Pressure of Gas

The Gas Laws Pressure and Temperature Pressure = force/area (we will use torr, mm Hg, Pa & atm) Always use Kelvin temperature (K) K = ° C variables are involved: P = pressure V = volume n = # of moles T = temperature (in Kelvin) R is the gas constant PV = nRT Ideal Gas Equation Ideal gas is a hypothetical gas whose pressure-volume- temperature behavior can be completely accounted for by the ideal gas equation.

What is the pressure of the gas (in atm) when 5.0 moles of CO gas are present in a container of 20.0 L at 27 o C? n= 5.0mole, V=20.0L, T= 27 o C=( )K=300.15K PV=nRT P=nRT / V = 5.0mole*0.082 L atm / (mol K)*300.15K/20.0L = 6.15 atm

The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). R = L atm / (mol K) = J/(K·mol) Experiments show that at STP, 1 mole of an ideal gas occupies L. Molar volume of gas 1 mole of gas at STP = 22.4 Liters 2 moles of gas at STP = 44.8 L

What is the volume (in liters) occupied by 49.8 g of HCl at STP? n = 49.8 g x 1 mol HCl g HCl = 1.37 mol V = 1.37 mol x 22.4 L/mol = 30.6 L 1 mole of gas at STP = 22.4 Liters

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm

Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L 5.4

What is the density of HCl gas in grams per liter at 700 mmHg and 25 o C? d = m V = PMPM RT P=700mmHg=700/760atm=0.92atm T= 25 o C= K=298.15K d = 0.92 atm x g/mol x K Latm molK =1.37g/L

What is the molar mass (g/mol) of 7.10 grams of gas whose volume is 5.40 L at 741 torr and 40 o C? dRT P M = d = m V 7.10 g 5.40 L = = 1.31 g L M = 1.31 g L atm x x K Latm molK M = 34.6 g/mol T=313.15KP= 741torr=741/760atm=0.975atm

Gas Stoichiometry The combustion process for methane is CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l) If 15.0 moles of methane are reacted, what is the volume of carbon dioxide (in L) produced at 23.0 o C and atm? V = nRT P 15mol x x K Latm molK atm = = L x 1CO 2 /1CH 4 15 mole CH  15 mole CO 2

Dalton’s Law of Partial Pressures Partial pressure is the pressure of the individual gas in the mixture. V and T are constant P1P1 P2P2 P total = P 1 + P 2

Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T mole fraction (X i ) = nini nTnT

A sample of natural gas contains 8.24 moles of CH 4, moles of C 2 H 6, and moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = P T = 1.37 atm = P propane = x 1.37 atm= atm

Kinetic Molecular Theory of Gases 1.A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2.Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic. 3.Gas molecules exert neither attractive nor repulsive forces on one another. 4.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy KE = ½ mu 2 u 2 = (u u …+ u N 2 )/N K E  T Mean square speed