Scientific Tutor
Chem – STP (Standard Temperature and Pressure)
Author: Andrew
What is STP?
STP stands for the Standard Temperature and Pressure. It should be more correctly named the standard temperature and pressure of human inhabitants on earth. That is to say it is approximately the average of the pressure and temperature that most humans live at during most of their lives. The standard temperature is 0 C or 273 K and the standard pressure is 1 atm or 1.013 * 105 Pa. To put that into perspective, the standard pressure is the pressure at approximately sea level. Since about 80% of people live within about 100km (60 miles) of a coastline, that actually makes a lot of sense. But what about the temperature, doesn’t that seem a little low? Well if you think about all the places people live on the earth, the temperatures actually range from about 40 C (a hot dessert) to about -30 C (in the tundra). So to say that 0 C is average is not a stretch. It also makes it easy to remember with only one digit. Although most humans prefer a temperature of about 25 C. All that STP is meant to do is give a quick reference for a person who is trying to calculate a gas under the average conditions where that container might be needed. In other words, if you were to design a container to hold gas that could go anywhere on the surface of the earth at a moments notice these are pretty good guides to the conditions you want to assume. How can STP be used in a problem? Most commonly it is used in an ideal gas law problem (PV = nRT) or in a dynamic gas law problems like (P1V1 / T1 = P2V2 / T2). I will give an example of each below and then a demonstrated example.
Examples: Solve for the following STP gas law problems. Remember STP is 0 C and 1 atm. The gas constant R is 0.08206.
What is the volume of a container of 1.4 mols of H2 gas at STP?
Answer: 31 L….the type of gas H2 has nothing to do with this problem.
If a gas at 5 atm 300K and 0.13L is changed to STP conditions what is the new volume?
Answer: 0.59 L….use combined gas law (P1V1 / T1 = P2V2 / T2).
VIDEO STP Gas Law Demonstrated Example 1 (Ideal Gas Law): How many moles is a gas at 4.9L under STP(1 atm and 273 K) conditions? The gas constant R is 0.08206 if you are in atm, L, mol, K.
Step 1:
What information are we given?
Answer:
V = 4.9 L
P = 1 atm
T = 0 C
R = 0.08206
Step 2:
What conversions are needed?
Answer:
V = 4.9 L
P = 1 atm
T = 0 C ——> 273 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).
| 1 atm * 4.9 L = | n * 0.08206 * 273 K |
| 1 | 1 |
Step 6:
How we do rearrange the equation?
Answer: Divide both sides by the gas constant and the temperature (red).
| 1 atm * 4.9 L = | n * 0.08206 * 273 K |
| 0.08206 * 273 K | 0.08206 * 273 K |
Step 7:
Cross out the 0.08206 and the 273 K on the right side.
| 1 atm * 4.9 L = | n * 0.08206 * 273 K |
| 0.08206 * 273 K | 0.08206 * 273 K |
Step 8:
Simplify
| 1 atm * 4.9 L = | n |
| 0.08206 * 273 K |
Step 9:
How do I do the calculations?
Answer: (1 * 4.9) / (0.08206 * 273) = 0.219
| 1 atm * 4.9 L = | 0.219 moles |
| 0.08206 * 273 K |
Step 10:
What is the complete answer?
COMPLETE ANSWER: 0.219 mols
VIDEO STP Gas Law Demonstrated Example 2 (Combined Gas Law): At STP (1 atm and 273 K) the volume of a gas is 6.3 L. What is the pressure of the same gas at 7.8 L and 340 K?
Step 1:
What information are we given and what do they ask for?
Answer: Group them into two categories (those that came first in the sentence versus those that came next in the sentence).
Category 1
P1 = 1 atm
V1 = 6.3 L
T1 = 273 K
Category 2
P2 = ?
V2 = 7.8 L
T2 = 340 K
Step 2:
What gas law equation does this fall into?
Answer: Set up the dynamic gas law equation to start.
| P1V1 = | P2V2 |
| n1T1 | n2T2 |
Step 3:
Which terms or variables are missing from the information of the question?
Answer: n, so cross that off of the equation
| P1V1 = | P2V2 |
| n1T1 | n2T2 |
Step 4:
Simplify
| P1V1 = | P2V2 |
| T1 | T2 |
Step 5:
What can we fill in for the equation?
Answer: The information we are given (red).
| 1 * 6.3 = | P2 * 7.8 |
| 273 | 340 |
Step 6:
How do we rearrange the equation?
Answer: Multiply both sides by 340 and divide both sides by 7.8 (red).
| 1 * 6.3 * 340 = | P2 * 7.8 * 340 |
| 273 * 7.8 | 340 * 7.8 |
Step 7:
Cross out the 340 and the 7.8 on the right side
| 1 * 6.3 * 340 = | P2 * 7.8 * 340 |
| 273 * 7.8 | 340 * 7.8 |
Step 8:
Simplify
| 1 * 6.3 * 340 = | P2 |
| 273 * 7.8 |
Step 9:
How do I do the calculations?
Answer: (1 * 6.3 * 340) / (273 * 7.8) = 1
| 1 * 6.3 * 340 = | 1 atm |
| 273 * 7.8 |
Step 10:
What is the complete answer?
COMPLETE ANSWER: 1 atm
PRACTICE PROBLEMS: Solve these STP gas law problems below. Remember STP means (1 atm and 273 K) and the gas constant R is 0.08206 if you are in atm, L, mol, K.
How many liters is a gas with 43 mol under STP conditions?
Answer: 964 L
If a gas is 9 L under conditions of STP. How many moles of gas do you have?
Answer: 24.9 mol
If you take a gas from STP conditions and then increase the pressure to 7.5 atm. What will be the new temperature assuming the moles stay constant?
Answer: 2048 K
At STP the volume of a gas is 2.1 L. What is the volume of the same gas at 3.4 atm and 700 K?
Answer: 1.58 L
Posted in Chemistry
