1. a cylinder of argon gas contains 50.0 l of ar at 18.4 ... · pdf filegas law worksheet...
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Gas Law Worksheet Answer Details
1. A cylinder of argon gas contains 50.0 L of Ar at 18.4 atm and 127 °C. How many moles of argon are in the cylinder?
Use Gas Law equationPV = nRTwhere:P = pressureV = volumen = number of moles of gasR = gas constant = 0.08 atm L/mol KT = absolute temperature
Step 1: Convert °C to KelvinT = °C + 273T = 127 + 273T = 400 KStep 2: Solve Gas Law equation for n
n = PVRT
n = 28.75 mol of argon
n = (18.4 atm)(50 L)(0.08 atm L/mol K)(400 K)
Answer: There are 28.75 moles of argon in the cylinder.
2. A 283.3-g sample of X2(g) has a volume of 30 L at 3.2 atm and 27 °C. What is element X?
Step 1: Convert °C to KelvinT = °C + 273T = 27 + 273T = 300 K
Step 2: Solve Gas Law equation for n
n = PVRT
n = 4 mol of X2
n = (3.2 atm)(30 L)(0.08 atm L/mol K)(300 K)
Step 3: Find mass of 1 mol of X24 mol X2 = 283.3 g1 mol X2 = 70.8 gStep 4: Find mass of 1 mol of X1 mol X2 = 70.8 g1 mol X = 35.4 gStep 5: Identify the element with molecular mass 35.4 gChlorine has a molecular mass of 35.4 g
Answer: Element X is Chlorine
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3. An ideal gas sample is confined to 3.0 L and kept at 27 °C. If the temperature is raised to 77 °C and the initial pressure was 1500 mmHg, what is the final pressure?
Step 1: Convert °C to KelvinT = °C + 273T = 27 + 273T = 300 K
Step 2: Use the gas law for constant volume
=PiTi
PfTf
WherePi = initial pressureTi = initial temperaturePf = final pressureTf = final temperature
PiTfTi
Pf =
Solve for Pf
Pf =(1500 mmHg)(350 K)
(300 K)
T = °C + 273T = 77 + 273T = 350 K
Pf = 1750 mmHgAnswer: The final pressure was 1750 mmHg
4. A sample of helium was compressed at 35 °C from a volume of 0.5 L to 0.25 L where the pressure is 500 mmHg. What was the original pressure?
Use Boyle’s Law for a constant temperature case of the ideal gas law:PiVi = PfVfwherePi = initial pressureVi = initial volumePf = final pressureVf = final volumeSolve for Pi
Pi = 250 mmHgAnswer: The original pressure of the helium was 250 mmHg.
Pi =PfVfVi
Pi =(500 mmHg)(0.25 L)
( 0.5 L )
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5. A hot air balloonist puts 125,000 Liters of air into their balloon at 27 °C and atmospheric pressure. When they heat the air to 200 °C at constant pressure, what is the final volume of the air in the balloon?
Step 1: Convert °C to KelvinT = °C + 273T = 27 + 273T = 300 K
T = °C + 273T = 200 + 273T = 473 K
Step 2: Use Charles’ Law for the constant pressure of the ideal gas law
WhereVi = initial volumeTi = initial temperatureVf = final volumeTf = final temperature
ViTfTi
Vf =
Solve for Vf
Vf =(125000 L)(473 K)
(300 K)Vf = 197083 L
=ViTi
VfTf
Answer: The final volume of the balloon is 197,083 L.
Step 3: Using Dalton’s Law to find partial pressures
Px = PTotal ( )nx
nTotalwherePx = partial pressure of gas XPTotal = total pressure of the gasesnx = number of moles of gas XnTotal = total number of moles of gases
Step 2: Using the ideal gas lawPV = nRTSolve for P
6. Air is basically a 80-20 mix of nitrogen and oxygen. A 2 mol sample of air is found to occupy 6.0 L at 27 °C. What is the partial pressure of oxygen in the sample?
P = nRTV
Step 1: Convert °C to KelvinT = °C + 273T = 27 + 273T = 300 K
P = (2 mol)(0.08 atmL/mol K)(300 K)6.0 L
P = 8 atm
nOnTotal
= 20% or 0.2
PO = 8 atm (0.2)PO = 1.6 atm
Answer: The partial pressure of the oxygen is 1.6 atm.About Chemistry http://chemistry.about.com