Worksheet Chapter 6
1. 14.0 moles of gas has an pressure of 1.57 atm at a temperature of 36°C. What is
the volume?
PV=nRT, so V = nRT/P
V = 14.0 moles x 0.08206 L atm/mole K x 309 K/1.57 atm = 226 L
2. Ammonia (NH3) gas is stored in a 10.0 L tank at 25° C and 14.0 atm. What is the
mass of the gas? (Hint: calculate moles, then use mw to calculate mass).
PV=nRT, n = PV/RT,
mw =mass/mole, mass = mw PV/RT
mass = 17.03 g/mole x 14.0 atm x 10.0 L/ ((0.08206 L atm/mole K)x298K) =
97.5g
3. A sample of gas has a molecular wt of 32.00 grams/mole. The gas has a volume
of 11702 mL under a pressure of 756 torr and a temperature of 4 C. What is the
mass of this sample of gas?
mass = mw x moles = mw x PV/RT
V = 11702 mL x 1 L/1000 mL = 11.702 L
P = 756 torr x 1 atm/760 torr = 0.995 atm
mass = 32.00 g/mole x 0.995 atm x 11.702 L/ ((0.08206 L atm/mole K)x277K) =
16.4 g
4. A 20.0 L sample of gas is initially at 51°C and 10.0 atm. The gas is cooled to 25°
C while the volume is kept constant. What is the new pressure?
(PV/nT)1 = (PV/nT)2
but n and V are same before and after, so P1/T1 = P2/T2, P2 = P1 T2/T1
new pressure = 10.0 atm x 298 K/324 K = 9.20 atm
5. A mixture of gasses contains 16.0 grams of O2, 32.0 g of N2, and 44.0 g of CO2.
The mixture’s total pressure is 777 torr. What is the partial pressure of each gas?
Pressure percent = mole percent, so calculate mole percent
16.0g oxygen x 1 mole/32.0 g = .500 moles oxygen
32.0 g nitrogen x 1 mole/ 28.01 g = 1.14 moles
44.0 g carbon dioxide x 1 mole/44.01 g = 1.00 moles
total moles = 0.500 + 1.14 + 1.00 moles = 2.64 moles
O2 partial pressure = 777 torr x 0.500 moles/2.64 moles = 147 torr
N2 partial pressure = 777 torr x 1.14 moles/2.64 moles = 336 torr
CO2 partial pressure = 777 torr x 1.00 moles/2.64 moles = 294 torr
6. A chamber contains 4.1 moles of N2, 3.3 moles Ar, and 6.7 moles Kr. If the total
pressure in the chamber is 5.5 atm, what are the partial pressures of the individual
gasses?
Pressure percent = mole percent, so calculate mole percent
total moles = 4.1 + 3.3 + 6.7 = 14.1 moles
N2 partial pressure = 5.5 atm x 4.1 moles/14.1 moles = 1.6 atm
Ar partial pressure = 5.5 atm 3.3 moles/14.1 moles = 1.3 atm
Kr partial pressure = 5.5 atm x 6.7 moles/14.1 moles = 2.6 atm
7. The gas in a container is 45% N2, 5% Ar, and 50% O2. If the partial pressures of
N2, Ar, and O2 are 9.0 atm, 1.0 atm and 10 atm respectively, what is the total
pressure in the chamber?
Total pressure is sum of partial pressures = 9.0 + 1.0 +10. = 20. atm
8. The heat of fusion for water is 80 cal/g. The heat of vaporization of water is 540
cal/g.
a. How much heat does it take to turn 43g of H2O into steam from liquid at
100oC?
b. How much heat is released when 43g of H2O freezes at 0oC?
Hmmmm..... we didn’t cover these in class, so I dunno how they managed to find their
way here, but here are the answers. By the way, you won’t be doing these problems, you
will be doing specific heat problems.
a. 43 g x 540 cal/g = 23000 cal
b. 43 g x 80 cal/g = 3000 cal released