1. The document discusses key concepts from general chemistry including: - The van der Waals equation accounting for non-zero particle volumes and interparticle interactions. - Using the ideal gas law to calculate changes in volume with temperature changes. - The value of the gas constant R at standard temperature and pressure. - Conversions between various pressure and volume units. - Calculating moles, pressure, volume, and number of molecules using the ideal gas law.

1. KEY
GENERAL CHEMISTRY-II (1412)
S.I. # 3
1. The van der Waals equation for real gasses recognizes that gas particles have
____________-_____________ volumes and interact with ___________ __________.
NON ZERO EACH OTHER
2. If 30 mL of a certain liquid has an initial temperature of 50°C and is heated to
100°C. What is the new volume of the liquid?
V1 / T1 = V2 / T2
V1 = 30 mL
V2 = ? V2 = (V1T2) / T1  (30 x 373) / 323 = 34.6 mL
T1 = 50°C = 323K
T2 = 100°C = 373K
3. At STP provide the value and units for the gas constant R as used in the Ideal
Gas Law.
R = 0.08206 [(L)(atm)]/[(mole)(K)] or 0.0821 L atm mole-1 K-1
4. Perform the following conversions:
a. 0.850 atm to torr _________________
0.850 atm x (760 torr / 1atm) = 646 torr
b. 785 torr to kilopascals _________________
785 torr x (101.325kPa / 760 torr) = 105kPa
c. 655 mm Hg to atmospheres _______________
655 mm Hg x (1atm / 760 mm Hg) = 0.862 atm
d. 1.323x105 Pa to atmospheres ______________
1.323x105 Pa x (1atm / 1.01325x105Pa) = 1.306 atm
e. 2.50 atm to bars _________________
2.50 atm x (1.01325x105Pa / 1atm)(1bar / 1x105Pa) = 2.53 bar

2. KEY
5. Calculate the volume of a gas, in liters, if 1.75 mol has a pressure of 0.985 atm at
a temperature of -6°C.
PV = nRT  V = nRT / P
n = 1.75 mol V = (1.75)(0.08206)(267) / (0.985) = 38.9L
P = 0.985 atm
T = -6°C = 267K
R = 0.08206 L atm
V=?
6. If the pressure exerted by ozone, O3, in the stratosphere is 3.0x10-3 atm and the
temperature is 250 K, how many ozone molecules are in a liter?
PV = nRT  n = PV / RT
n=?
P = 3.0x10-3 atm n = (3.0x10-3 x 1) / (0.08206 x 250) = 1.46x10-4 mol
V=1L
R = 0.08206 L atm
T = 250K
(1.46x10-4 mol )(6.022x1023 molecules / 1 mol) = 8.8x1019 molecules
7. An aerosol spray can with a volume of 250 mL contains 2.30 g of propane gas
(C3H8) as a propellant. (a) If the can is at 23°C, what is the pressure in the can? (b)
What volume would the propane occupy at STP?
PV = nRT  P = nRT / V & V = nRT / P
a. n = (2.30 g)(1mol / 44.1g) = 0.052154 mol C3H8
R = 0.0821 L atm
T = 23°C = 296K
V = 250 mL = 0.250 L
P = (0.052154 mol C3H8)( 0.0821 L atm)( 296K) / 0.250 L = 5.07 atm
b. T = 273K
P = 1 atm
V = (0.052154 mol C3H8)( 0.0821 L atm)( 273K) / 1atm = 1.1684 L  1.17L
8. Which gas is least dense at 1.00 atm and 298 K? (a) SO2, (b) HBr, (c) CO2.
Explain. c. CO2 (g) is least dense. For gases at the same conditions, density is
directly proportional to molar mass, and CO2 as the smallest molar mass.