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General Chemistry
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Fall 2009 |
| Exercise: Gas Laws |
© R. Paselk 2008
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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/L
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 :
Po/To = nR/V = Pf/Tf or
Po/To = Pf/Tf
comparing the initial and final conditions.
Listing the values of the variables:
- Po = 820 mmHg
- To = 24 °C
- Pf = ?
- Tf = 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.
- To = 24 + 273.15 = 297.1 K (keeping one extra significant figure for calculations)
- Tf = 450 + 273.15 = 723 K (keeping one extra sf for calculations)
Solving for Pf
Pf = PoTf/To
inserting values
Pf = (820 mmHg)(723 K)/(297.1 K)
canceling units, checking and solving
Pf = 1.996x103 mmHg = 2.0x103 mmHg
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:
- Po = 820 mmHg
- V = 2.50 L
- R = 0.0821 L*atm/mole*K
- To = 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
- Po = (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, 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:
P/T = nR/V, and
Po/To = nR/V = Pf/Tf or
Po/To = Pf/Tf
Solving for Pf
Pf = PoTf/To
Pf = (720 mmHg)(870 K)/(298.1 K)
Pf = 2109 mmHg = 2.1 x 103mmHg
or in atm
Pf = (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 = 1.6 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 = 7.15 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 |
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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 |
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0.15 L |
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4.8 L |
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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 |
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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.
© R A Paselk
Last modified 6 November 2008