Humboldt State University ® Department of Chemistry

Richard A. Paselk

General Chemistry

Discussion Modules

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Gas Law Problems - Answers

In these exercises we will be using the Ideal Gas Law or Perfect Gas Law equation:

PV = nRT

where R = the gas constant with units appropriate to the various measurements. We will use atm, L, K, and moles, so that R = 0.0821 L*atm/mole*K.

I will base all of my examples on this equation because that requires a minimum of memorization. However you may find it easier to memorize a series of equations such as the "combined gas law equation"

1. Find the molar volume of a gas under standard conditions of temperature and pressure (STP).

The first thing we need is to determine what STP is. Looking in the text we find that STP are defined as: P = 1 atm and T = 0° C (both defined so exact values). Thus we need to solve the gas equation for one mole of gas at 273.15 K (= 0°C+ 273.15) and 1 atm.

Next we can look up molar volume and find that it is the volume of one mole of gas.

To solve the problem, first write out the equation for the ideal gas law:

PV = nRT

Since we want to find molar volume we want volume/mole = V/n. Dividing both side of the gas law equation by n and then by P we get:

PV/n = RT and then

V/n = RT/P

Listing the values for the terms:

R = 0.0821 L*atm/mole*K

T = 273.15 K

P = 1.000 atm (express to one significant figure more than limiting term, R)

n = 1.000 mole (since we are finding volume/mole)

Substituting these values into our equation we get:

V/(1.000 mole) = (0.0821 (L*atm)/(mole*K))(273.15 K)/1.000 atm

Checking the units: atm cancels atm, K cancels K and mole cancels mole giving units of L, so we are confident we set our work up correctly.

Solving:

V = 22.426 L = 22.4 L

and the molar volume of an ideal gas = 22.4 L/mol

2. A 1.00 L sample of gas weighed 1.25 g at a temperature of 0 °C and a pressure of 1.00 atm. What is the MW of this gas?

The first thing to do in a problem like this is to ask what you need to know and what you do know.

Since we are asked to find MW we need to know grams and moles.
In the problem we are given grams, so half the problem is solved. All we need is moles. Recall that moles appears in the gas law equation as n: PV=nRT, or rearranging,

n=PV/RT

At this point let's list what we know:

P = 1.00 atm

V = 1.00 L

R = 0.0821 L*atm/mole*K

T = 0 °C But our value of R is in K, so the temperature MUST be converted to K! So,
T = 0 + 273.15 = 273.15 K (note that addition gave us three significant figures instead of one!)

Plugging these values into our formula we get:

n = (1.00 atm)(1.00L)/(0.0821 L*atm/mole*K)(273.15 K)

Looking at the equation we see that atm cancels atm, L cancels L and K cancels K leaving moles, which is what we want, so it looks good! Solving:

n = 0.04462 mole

So we have 0.04462 moles in a 1.00 L sample which weighs 1.25 g.

Finally, MW = g/mole = 1.25 g/ 0.04462 mole

MW = 28.02 g/mole = 28.0 g/mole

3. A rigid, sealed gas cylinder with a volume of 2.50 L is filled with hydrogen at a pressure of 820 mmHg and a temperature of 24 °C.

a. What will the pressure in the cylinder become if the temperature is raised to 450°C?

The first step in this problem is to write out the ideal gas law:

PV = nRT

We next note that not all of the terms are provided in our data - we don't have n. However, on reading the problem we can see that n won't change since we have a sealed container, so moles = moles at both temperatures. And of course R is also constant.

With this in mind, we can now rearrange our equation to put the constants (n, R and V {since the container is rigid}) on one side and the variables (P and T) on the other:

P/T = nR/V, and since nR/V = constant we can write :

P_o/T_o = nR/V = P_f/T_f or

P_o/T_o = P_f/T_f

comparing the initial and final conditions.

Listing the values of the variables:

P_o = 820 mmHg

T_o = 24 °C

P_f = ?

T_f = 450 °C

Before going further we MUST convert our temperatures into K since gas volumes are proportional to Absolute temperature. The pressure can remain as is unless different units are requested.

T_o = 24 + 273.15 = 297.1 K (keeping one extra significant figure for calculations)

T_f = 450 + 273.15 = 723 K (keeping one extra sf for calculations)

Solving for P_f

P_f= P_oT_f/T_o

inserting values

P_f= (820 mmHg)(723 K)/(297.1 K)

canceling units, checking and solving

P_f= 1.996x10³mmHg = 2.0x10³mmHg

b. How many moles of hydrogen are there in this container?

To solve this problem we again start with the gas law equation:

PV = nRT

rearranging to solve for n:

n = PV/RT

Listing values we get:

P_o = 820 mmHg

V = 2.50 L

R = 0.0821 L*atm/mole*K

T_o = 24 + 273.15 = 297.1 K

In this case we see that P has the wrong units, since they won't cancel with atm in R. Recalling that one atm = 760 mmHG exactly

P_o = (820 mmHg)(1 atm/760 mmHg) = 1.079 atm

Inserting values

n = (1.079 atm)(2.50 L)/(0.0821 L*atm/mole*K)(297.1 K)

canceling units, checking and solving

n = 0.11058 mole = 0.11 mol (sf fixed by 820 mmHg)

4. A student ignores the warning labels and throws an empty (no liquid left, no spray or hissing) can of hair spray into his campfire.

a. Assuming an ambient temperature of 25 °C an atmospheric pressure of 7.20 x 10²mmHg, and a temperature in the coals of 600 °C (assume two significant figures), find the pressure in the can in the fire, assuming it doesn't burst or expand (it is rigid).

First write out the Ideal Gas Law equation

PV = nRT

Then let's ask what we know:

If the can is sealed, then n is constant.
If it doesn't expand, then V is constant.
If the can is empty and wouldn't spray or hiss, then P = P_atmosphere = 7.20 x 10²mmHg
T_o = 25 °C = 25 + 273.15 = 298.1 K
T_f = 600 °C = 600 + 273.15 = 870 K

Since n, R and V are all constant we can rearrange to put the variables on one side:

P/T = nR/V, and

P_o/T_o = nR/V = P_f/T_f or

P_o/T_o = P_f/T_f

Solving for P_f

P_f= P_oT_f/T_o

P_f = (720 mmHg)(870 K)/(298.1 K)

P_f = 2109 mmHg = 2.1 x 10³mmHg

or in atm

P_f = (2109 mmHg)/(760 mmHg/atm) = 2.8 atm.

b. If the volume of the can is 420 mL, how many moles of gas are in the can?

Again, start with gas law equation to show we are educated people,

PV = nRT

Rearranging,

n = PV/RT

Listing values we get,

P = 720 mmHg/760 mmHg/atm = 0.9474 atm

V = 420 mL/1000 mL/L = 0.420 L

n = ?

R = 0.0821 L*atm/mol*K

T = 25° + 273.15° = 298 K

Putting in values and solving:

n = (0.9474 atm)(0.420 L)/(0.0821 L*atm/mol*K)(298 K)

n = 1.626 x 10^-2

n = 1.6 x 10^-2 moles

5. One of the student's colleagues on this ill fated trip tossed an "empty" 0.500 L propane cylinder into the fire. Unfortunately, 2.20 g of propane remained in the cylinder. What pressure would be reached in the cylinder assuming no deformation and no bursting at 600 °C (assume 3 sig figs for the temperature this time).

Again, start with gas law equation to show we are knowledgeable folk,

PV = nRT

Rearranging to find pressure,

P = nRT/V

P = ?

V = 0.500 L

R = 0.0821 L*atm/mol*K

T = 6.00 x 10² + 273.15 = 873 K

n = ?

Need to find moles from mass and formula. Propane = C₃H₈, MW = 3 (12.01) + 8 (1.008) = 44.09 g/mol, so

n = 2.20 g/44.09 g/mol = 4.989 x 10^-2 mol

Putting in values and solving,

P = (4.989 x 10^-2 mol)(0.0821 L*atm/mol*K)(873 K)/(0.500 L)

P = 7.152 atm

P = 7.15 atm

6. 2.40 L of ethene gas (C₂H₄) is combined with 7.35 L of oxygen and ignited to give carbon dioxide as the only carbon containing product. If all volumes of reactants and products are measured at the same temperature and pressure (above 100 °C - so water is a vapor), calculate the volume of each substance after the reaction is complete.

First let's write the equation for the reaction

	C₂H₄	+	O₂		CO₂	+	H₂O
Balancing the reaction then gives:
	C₂H₄	+	3 O₂		2 CO₂	+	2 H₂O
Because V is proportional to n in PV = nRT wew can express the stoiciometry of this reaction in either L or moles and get the same ratios shown below.
	1	:	3	:	2	:	2
Now we can look at the starting volumes and the ratios to determine the final amounts of stuff.
Before reaction:	2.40 L		7.35 L		0		0
From the stoichiometry need 3 volumes of oxygen for every volume of ethene, or 3*2.4 = 7.2 L. We have 7.35, so 7.35 - 7.2 = 0.15 L is left over. Each of the products has a 2:1 ratio, so each will be 4.8 L
After reaction:	0 L		0.15 L		4.8 L		4.8 L

7. Assume you have an engine with a 500 mL cylinder with a 10:1 compression ratio. If 0.200 L of methane and 0.300 L of oxygen are introduced into the cylinder at 765 mmHg and 25 °C, what will be the pressure "at the top of the stroke" if ignition gives a temperature of 557 °C?

First we need to write a balanced equation, determine the stoichiometry, and the before and after situation.

The easiest way to do this is to assume no temperature or pressure change during the reaction - we'll take care of that later:

Note that the total volume (0.500 L) doesn't change in this reaction.
Equation:	CH₄	+	O₂		CO₂	+	H₂O
Balancing:	CH₄	+	2 O₂		CO₂	+	2 H₂O
Stoichiometry (n or V):	1	:	2	:	1	:	2
Before reaction:	0.200 L		0.300 L		0		0
From the stoichiometry can see that oxygen is limiting - some methane will be left over.
After reaction:	0.050 L		0 L		0.150 L		0.300 L

Now use Gas Laws to solve:

PV = nRT,

putting constants together,

PV/T = nR, or

P₁V₁/T₁ = P₂V₂/T₂

Rearranging:

P₂ = (P₁)(V₁/V₂)(T₂/T₁)

Listing values:

P₁ = 765 mmHg

V₁ = 0.500 L

V₂ = 0.050L

T₁ = 25° + 273 ° = 298 K

T₂ = 557 ° + 273 ° = 830 K

Substituting values and solving

P₂ = (765 mmHg)(0.500/0.050)(830 K/298 K)

P₂ = 2.13 x 10⁴ mmHg = 28.23 atm

P₂ = 2.1 x 10⁴ mmHg = 28 atm

In addition to these exercises you should familiarize yourself with the text materials.

Gas Law Problems

© R A Paselk

Last modified 4 March 2014