Scientific Understanding

**What is the ideal gas law?**

The ideal gas law is an equation. What the equation describes is how you can manipulate different factors to control how a gas behaves. There are 4 factors in the ideal gas law; pressure, volume, moles, and temperature. I have already explained pressure in the previous section so here I will give a brief description of volume, moles, and temperature. Volume is how much space something takes up. Moles are how many or how much stuff you have (gas molecules or particles in this case). Temperature is how fast the molecules or particles of gas are moving. The ideal gas law equation is below (it is very important equation throughout the rest of your chemistry learning):

PV = nRT

The letter P represents Pressure, the letter V represents Volume, the letter n represents Moles, the letter R represents a constant (a certain number to make the equation work together), the letter T represents temperature. The most common units you will see for the ideal gas law are pressure in atmospheres (atm), volume in Liters (L), moles in moles (mol), and temperature in (K). If the equation has all the units stated in the previous sentence then the constant R is 0.08206. It also has the units of (atm * L / mol * K). However, if the units of the equation change the constant number will also change. They usually call R the gas constant in chemistry books and classes.

The most important thing to know about all gas law equations is that they require you to use the Kelvin temperature scale, which is why I taught you how to convert to the Kelvin temperature scale before I started this section. All other units of pressure, volume, and moles can possibly have different units and the gas law equation will still work (although other units tend to make it more difficult because you have to recalculate what the gas constant R will be). However, YOU CAN NEVER USE ANY OTHER UNITS BESIDES KELVIN FOR THE TEMPERATURE SCALE. Another pain in the ideal gas law problems is that most books and some teachers do not tend to do a good job of telling you that YOU SHOULD ALWAYS HAVE ACCESS TO THE GAS CONSTANT R (0.08206) whenever you are doing homework problems or test. I will now go on to teach you how to solve equations using the ideal gas law. In my demonstrations and example I will not emphasize the units of the gas constant R in the equations because it is too long and complex for this text format. However, if you are more interested in the units of the gas constant then you can click on this link. First let us look at some examples to get an idea of what we will be asked of us.

**Examples**: Solve for the following ideal gas law problems.

If the volume of a gas container is 3L and the amount of moles is 2.2 moles and the temperature is 273K. What is the Pressure in atmospheres given that the gas constant ( R ) is 0.08206?

Answer: 16.4 atm

A container of gas is 530ml at 600 torr, and 23 C. If the gas constant is 0.08206 how many moles are in the container?

Answer: 0.0172 mol….(don’t forget to convert all the units before you use the ideal gas law)

**VIDEO Ideal Gas Law Demonstrated Example 1**: If the pressure of a gas is 1.5 atm at 13 mol and 300 K with as gas constant of 0.08206. Then what is the volume?

**Step 1:**

What information are we given?

Answer:

P = 1.5atm

n = 13 mol

R = 0.08206

T = 300 K

**Step 2:**

What does the question ask for?

Answer: V = ?

**Step 3:**

How do we set up the problem?

Answer: Start with the equation

PV = | nRT |

1 | 1 |

**Step 4:**

What can we fill in for the equation?

Answer: The information we are given (red).

1.5 atm * V = | 13 mol * 0.08206 * 300 K |

1 | 1 |

**Step 5:**

How do we rearrange the equation?

Answer: Divide both sides by pressure (red)

1.5 atm * V = | 13 mol * 0.08206 * 300 K |

1.5 atm | 1.5 atm |

**Step 6:**

Cross off 1.5 atm from the left side.

1.5 atm * V = | 13 mol * 0.08206 * 300 K |

1.5 atm | 1.5 atm |

**Step 7:**

Simplify

V = | 13 mol * 0.08206 * 300 K |

1.5 atm |

**Step 8:**

How do I do the calculations?

Answer: (13 * 0.08206 * 300) / 1.5 = 213

213 L = | 13 mol * 0.08206 * 300 K |

1.5 atm |

**Step 9:**

What is the complete answer?

COMPLETE ANSWER: 213 L

**VIDEO Ideal Gas Law Demonstrated Example 2**: If a gas is 250 mL and 600 mmHg at 230 C, how many moles will that gas be if the gas constant is 0.08206?

**Step 1:**

What information are we given?

Answer:

V = 250 mL

P = 600 mmHg

T = 230 C

R = 0.08206

**Step 2:**

What conversions are needed?

Answer:

V = 250 mL ——> 0.25 L

P = 600 mmHg ——> 0.789 atm

T = 230 C ——> 503 K

R = 0.08206

**Step 3:**

What does the question ask for?

Answer: n = ?

**Step 4:**

How do we set up the problem?

Answer: Start with the equation

PV = | nRT |

1 | 1 |

**Step 5:**

What can we fill in for the equation?

Answer: The information we are given (red).

0.789 atm * 0.25 L = | (n) * 0.08206 * 503 K |

1 | 1 |

**Step 6:**

How do we rearrange the equation?

Answer: Divide both sides by the gas constant and the temperature (red).

0.789 atm * 0.25 L = | (n) * 0.08206 * 503 K |

0.08206 * 503 K | 0.08206 * 503 K |

**Step 7:**

Cross off 0.08206 and 503 K from the right side

0.789 atm * 0.25 L = | (n) * 0.08206 * 503 K |

0.08206 * 503 K | 0.08206 * 503 K |

**Step 8:**

Simplify

0.789 atm * 0.25 L = | (n) |

0.08206 * 503 K |

**Step 9:**

How do I do the calculations?

Answer: (0.789 * 0.25) / (0.08206 * 503) = 0.004779

0.789 atm * 0.25 L = | 0.004779 mol |

0.08206 * 503 K |

**Step 10:**

What is the complete answer?

COMPLETE ANSWER: 0.00478 mol…or…4.78 * 10^{-3} mol

**PRACTICE PROBLEMS**: Solve the Ideal gas law equations below. Keep in mind the gas constant is always 0.08206 if you are in atm, L, mol, and K and that you can use that number with any of these problems.

If the pressure of a gas is 4 atm at 3 mol and 250 K, then what is the volume?

Answer: 15.4 L

If the volume of a gas is 5 L at 42 C and 2.5 mol, then what is the pressure?

Answer: 12.9 atm….(convert Celsius to Kelvin)

How many moles are in a container of gas that is 940 torr at 25 C and 0.2 L?

Answer: 0.01014 mol…..(convert torr and Celsius)

A gas that is 1.05 * 10^{6} Pa, 710 mL, and 9 mol is at what temperature in Celcius?

Answer: -263 C….(Convert everything before you put it into the equations then convert the temperature after you are done with the equation).