ChemTeam: Ideal Gas Law: Problems #1 - 10

Ideal Gas Law Problems#1 - 10

Fifteen Examples

Problems #11-25

Examples and Problems only

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Problem #0: What is the value of and units on R? What is R called ("A letter" is not the correct answer!)?

R is called the gas constant. It was first discovered, as part of the discovery in the mid-1830's by Emil Clapeyron of what is now called the Ideal Gas Law.

Sometimes it is called the universal constant because it shows up in many non-gas-related situations. However, it is mostly called the gas constant or, sometimes, the universal gas constant.

Depending on the units selected, the "value" for R can take on many different forms. Here is a list. Keep in mind these different "values" represent the same thing.

Problem #1: Determine the volume of occupied by 2.34 grams of carbon dioxide gas at STP.

Solution:

1) Rearrange PV = nRT to this:

V = nRT / P

2) Substitute:

V = [(2.34 g / 44.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (273.0 K)] / 1.00 atm

V = 1.19 L (to three significant figures)

Problem #2: A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon and the mass of argon in the sample.

Solution:

1) Rearrange PV = nRT to this:

n = PV / RT

2) Substitute:

n = [(1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K)]

n = 2.50866 mol (I'll keep a few guard digits)

3) Multiply the moles by the atomic weight of Ar to get the grams:

2.50866 mol times 39.948 g/mol = 100. g (to three sig figs)

Problem #3: At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres?

Solution:

1) Rearrange PV = nRT to this:

T = PV / nR

2) Substitute:

T = [(1.95 atm) (12.30 L)] / [(0.654 mol) (0.08206 L atm mol¯1 K¯1)]

T = 447 K

Problem #4: A 30.6 g sample of gas occupies 22.414 L at STP. What is the molecular weight of this gas?

Solution:

Since one mole of gas occupies 22.414 L at STP, the molecular weight of the gas is 30.6 g mol¯1

Problem #5: A 40.0 g gas sample occupies 11.2 L at STP. Find the molecular weight of this gas.

Solution:

11.2 L at STP is one-half molar volume, so there is 0.500 mol of gas present. Therefore, the molecular weight is 80.0 g mol¯1

Problem #6: A 12.0 g sample of gas occupies 19.2 L at STP. What is the molecular weight of this gas?

Solution:

This problem, as well as the two just above can be solved with PV = nRT. You would solve for n, the number of moles. Then you would divide the grams given by the mole calculated.

1) Use PV = nRT:

(1.00 atm) (19.2 L) = (n) (0.08206) (273 K)

n = 0.8570518 mol (I'll keep a few guard digits)

2) Determine the molecular weight:

12.0 g / 0.8570518 mol = 14.0 g/mol

3) Since it is at STP, we can also use molar volume:

(19.2 L / 12.0 g) = (22.414 L / x )

19.2x = 268.968

x = 14.0 g/mol

Warning: you can only use molar volume when you are at STP.

Problem #7: 96.0 g. of a gas occupies 48.0 L at 700.0 mm Hg and 20.0 °C. What is its molecular weight?

Solution:

1) Solve for the moles using PV = nRT:

n = PV / RT

n = [(700.0 mmHg / 760.0 mmHg atm¯1) (48.0 L)] / [(0.08206 L atm mol¯1 K¯1) (293.0 K)]

n = 1.8388 mol

2) Divide the grams given (96.0) by the moles just calculated above:

96.0 g / 1.8388 mol = 52.2 g/mol

Problem #8: 20.83 g of a gas occupies 4.167 L at 79.97 kPa at 30.0 °C. What is its molecular weight?

Solution:

1) Solve for the moles using PV = nRT:

n = PV / RT

n = [(79.97 kPa / 101.325 kPa atm¯1) (4.167 L)] / [(0.08206 L atm mol¯1 K¯1) (303.0 K)]

n = 0.13227 mol

2) Divide the grams given (20.83) by the moles just calculated above:

20.83 g / 0.13227 mol = 157.5 g/mol

Notice that, in the two problems just above, the I converted the pressure unit given in the problem to atmospheres. I did this to use the value for R that I have memorized. There are many different ways to express R, it's just that L-atm/mol-K is the unit I prefer to use, whenever possible.

Also, you cannot use molar volume since the two problems just above are not at STP.

Problem #9: What is often called the Ideal Gas Constant is 0.0820574 L atm mol¯1 K¯1. What is often called the Universal Gas Constant is 8.31451 J mol¯1 K¯1. Convert the Ideal Gas Constant into the Universal Gas Constant and vice versa.

Solution:

1) To find the conversions, divide one by the other:

8.31451 J mol¯1 K¯1 / 0.0820574 L atm mol¯1 K¯1 = 101.3255 J L¯1 atm¯1

This means that 1 L atm = 101.3255 J

2) The other division:

0.0820574/8.31451 = 0.00986918 (try putting the units in as was done just above)

This means that 1 J = 0.00986918 L atm

You could have also done this:

1 / 101.3255 = 0.00986918

3) Here are the conversions:

(0.0820574 atm L/mol K) (101.3255 J/L atm) = 8.31451 J/mol K

and

(8.31451 J/mol K) (0.00986918 L atm / J) = 0.0820574 L atm / mol K

Problem #10: 5.600 g of solid CO2 is put in an empty sealed 4.00 L container at a temperature of 300 K. When all the solid CO2 becomes gas, what will be the pressure in the container?

Solution:

1) Determine moles of CO2:

5.600 g / 44.009 g/mol = 0.1272467 mol

2) Use PV = nRT

(P) (4.00 L) = (0.1272467 mol) (0.08206 L atm mol¯1 K¯1) (300 K)

P = 0.7831 atm (to four sig figs)

Bonus Problem #1: 2.035 g H2 produces a pressure of 1.015 atm in a 5.00 L container at −211.76 °C. What will the temperature (in °C) have to be if an additional 2.099 g H2 are added to the container and the pressure increases to 3.015 atm.

Solution:

1) What gas law should be used to solve this problem?

Notice that we have pressure, volume and temperature explicitly mentioned. In addition, mass and molecular weight will give us moles. It appears that the ideal gas law is called for.

However, there is a problem. We are being asked to change the conditions to a new amount of moles and pressure. So, it seems like the ideal gas law needs to be used twice.

2) Let's set up two ideal gas law equations:

P1V1 = n1RT1

This equation will use the 2.035 g amount of H2 as well as the 1.015 atm, 5.00 L, and the −211.76 °C (converted to Kelvin, which I will do in a moment).

P2V2 = n2RT2

This second equation will use the data in the second sentence and T2 will be the unknown.

What I need to do is set the two equations equal to each other. First, I rearrange a bit.

3) Like this:

P1V1 = n1RT1 leads to:

		P1V1
R	=	–––––
		n1T1

and

P2V2 = n2RT2 leads to:

		P2V2
R	=	–––––
		n2T2

4) I will use the fact that R is the same value in each equation:

R = R, therefore:

P1V1		P2V2
–––––	=	–––––
n1T1		n2T2

5) I'm going to isolate T2 on one side of the equals sign:

Since the volume never changes, we can eliminate it from the equation:

P1		P2
–––––	=	–––––
n1T1		n2T2

Now, cross-multiply:

P1n2T2 = P2n1T1

Isolate T2:

		P2n1T1
T2	=	–––––
		P1n2

Another way to write it is this:

T2 = P2n1T1 / P1n2

6) One more comment: it's about the moles:

Each of the mole amounts would be arrived at by dividing the grams by the molar mass (in this case, H2). However, notice the molar masses will cancel, being the same numerical value and one in the nominator and one in the denominator.

After cancelling, this is what we wind up with:

		P2mass1T1
T2	=	–––––––––
		P1mass2

7) We are now ready to solve:

		(3.015 atm) (2.035 g) (61.24 K)
T2	=	–––––––––––––––––––––––––
		(1.015 atm) (4.134 g)

T2 = 89.546867 K

8) Convert Kelvin to Celsius:

°C = 89.546867 K − 273.15 (I decided to use 273.15 rather than 273.)

Using four sig figs gives −183.6 °C for the final answer

Bonus Problem #2: 1.00 mole of gas occupies 22.414 L at STP. Calculate the temperature and pressure conditions needed to fit 2.00 moles of a gas into a volume of 22.414 L.

Solution:

1) Notice that the problem asks for two conditions: one of temperature and one of pressure. The answer we arrive at will not be a value of T and one of P, but a ratio between the two. Start here:

PV = nRT

2) Insert our known values:

(P) (22.414 L) = (2.00 mol) (0.08206 L atm / mol K) (T)

3) Since the question mentions T first, let's determine a T/P ratio:

T/P = 22.414 L / [(2.00 mol) (0.08206 L atm / mol K)]

T/P = 136.57 K/atm

Any T/P combination that gives 136.57 will be an answer.

4) If you wanted a P/T ratio, it would be 0.007322.

Fifteen Examples

Problems #11-25

Examples and Problems only

Return to KMT & Gas Laws Menu