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