Petersen
Chem
(practice) Quiz Chapter 13
Name______________
Pd______
1.000 atm = 760.0 mmHg = 760.0 torr = 14.69 psi = 101.325 kPa = Standard atmospheric pressure
Kelvin = Celsius + 273 STP = 0°C and 1.00 atm
R = 0.08206 L•atm/(mol•K)
SHOW YOUR WORK!!! UNITS!!! Use a periodic table, gas law summary and a molemap.
1.
True or False. At the same temperature and pressure, 3 moles of argon (Ar) occupy more
volume at STP than 3 moles of methane (CH4).
2.
If the pressure on an enclosed gas is increased, and the temperature is not changed, will the
volume increase or decrease? (Circle one)
3.
If the volume of a gas sample does not change, and the temperature is decreased, will the
pressure increase or decrease?
4.
Use Boyle's Law to solve this problem.
A 500. mL bicycle tire is to be filled to a pressure of 3.50 atm (about 50 psi). What volume of
1.00 atm air would be needed?
5.
Use Charles' Law to solve this problem.
A 1.80 L helium balloon at a temperature of 298 K inside a store is carried outside to a car where
the temperature is 265 K. Assuming no pressure change, what is the volume of the balloon in the
car?
6.
Combined Gas Law.
A 3.0 L (leakproof) flexible balloon at STP is allowed to rise in the atmosphere to an altitude
where the temperature is -9.0°C and the pressure is 0.60 atm. What is its volume at this altitude?
7.
Avogadro's Law
10.0 grams of CO2 gas occupy a volume of 5.09 L at STP. What volume is occupied by 6.02E24
molecules of this gas at the same conditions?
8.
Ideal Gas Law
2.0 moles of C2H6 gas occupy what volume at 745 mmHg and 28°C?
9.
The vapor pressure of water is 23.8 mmHg at 25°C. What is the partial pressure of oxygen if it
is collected over water at 25°C if the total pressure is 0.991 atm?
10.
Butane (C4H10(g)) burns in oxygen to form carbon dioxide and water. Write the equation for
this, and then calculate how many liters of CO2 are produced when 1.00 g of butane burns (at
STP)
(practice) Quiz Chapter 13
Petersen
Chem
Name______________
Pd______
1.
False. No matter what the gas, the volume of 3 moles of gas at STP = 67.2 L (Avogadro)
2.
decrease. Pressure and volume of a gas are inversely related. (Boyle’s Law)
3.
decrease. Lowering the temperature slows down the gas molecules and they collide less
frequently with the walls, lowering the pressure (Law of Gay-Lussac)
4.
P1V1=P2V2 (3.50atm)(500.ml) = (1.00atm)V2
V2 = 1750mL
5.
V1/T1 = V2/T2
V2 = 1.60L
6.
P1V1/T1 = P2V2/T2
7.
V1/n1 = V2/n2
8.
PV=nRT
9.
Pt = PO2 + PH2O
Pt = 0.991atm x 760mmHg/1atm = 753mmHg
753mmHg = PO2 + 23.8mmHg
PO2 = 729mmHg
10.
C4H10 + 6.5 O2 ---> 4CO2 + 5H2O (or doubled coefficients)
1.00gC4H10 x 1mol/58.12gC4H10 x 4molCO2/1mol C4H10 x 22.4LCO2/1molCO2 = 1.54L CO2
1.80L/298K = V2/265K
(1.00atm)(3.0L)/(273K) = ((0.60atm)V2/(264K) V2 = 4.8L
n1 = 10.0gCO2 x 1mol/44.01gCO2 = 0.227molCO2
n2 = 6.02E24molecules x 1mol/6.02E23molecules = 10.0molCO2
5.09L/0.227mol = V2/10.0mol
V2 = 224L
(745/760atm)(V)=2.0mol(0.08206L•atm/mol•K)
V = 50.4L