Chemistry Gas Laws: Combined Gas Law - Video Tutorials & Practice Problems

Chemistry Gas Laws: Combined Gas Law

now if we have the combined gas law. The combined gas law was created from combining Boyle's law, Charles Law and Gay Loose Ex Law. We're gonna say it highlights the relationship between the variables off pressure, volume and temperature. Remember, Boyle's Law says that pressure is inversely proportional to volume Charles losses. That volume is directly proportional. It's a temperature and gay loose X losses. That pressure is directly proportional to temperature, and we take a look down here. We could figure out how the combined gas law was derived. So it comes from combining Boyle's long Charles law and gay loose ax law. If we look, we have pressure and volume as numerator. So volume here numerator pressure here. Numerator. So they share those variables in common with one another so that B p v as enumerators, Charles law and gay loose sacks law have as their denominators temperature. So that comes down here. So here we say that the combined gas law is PV over T equals some constant. So okay, here would be a constant. This year would represent our combined gas law. And if we're dealing with two sets of pressures, volumes and temperature, then it can go further and say p one v one over t one equals p two V two over t to So this is how the combined gas law can be derived from these earlier chemistry gas laws.

Chemistry Gas Laws: Combined Gas Law Example 1

here in this example question. It says a sample of gas initially has a volume of 900 mL at 520 Kelvin and 1.85 atmospheres. What is the pressure of the gas at? The volume decreases to 330 mL while the temperature increases to 770. Kelvin. Alright, So in this question, they're giving me a volume. We're gonna say it's V one because later on they give me a second new volume V two. They give me a temperature t one and then later give me a second temperature. T two, they give me this pressure in atmosphere. So this is P one because later they asked me what is the new pressure? So they're asking us to find P two. Now we have our ideal gas law. PV equals NRT. If you have watched my videos on ideal law derivations, you know that we can manipulate the ideal gas law in order to get our combined gas law. Now, if you haven't watched those videos, I highly suggest you go back and take a look. So look under ideal gas derivation videos. Now here. We're talking about two pressures. We're talking about two volumes. We're talking about two temperatures. Moles aren't being discussed because they're being held. Constant are is a constant. We divide out the tea so that everything is on the left side. And we see that since we're dealing with two different values for these variables, it becomes p one V one equal over t one equals p two V two over t to. So here we just showed how we derived the combined gas law. All right, so now we're gonna plug in the values that we have. Our pressure is 1.85 atmospheres. Initially, our volume when we change milliliters toe leaders, it gives us 0.900 leaders. Remember, temperature must always be in Kelvin's when doing calculations. It's already in Kelvin, so we don't have to worry about that. Equals we don't know what p two is. V two is 20.330 leaders and then temperature to a 770 Kelvin. We're just looking toe isolate r p two so you can cross multiply these and then you can cross multiply these so that you can isolate P two. So if I come over here, I'm going to see that p two equals P. Two times 330 liters times 5 20 kelvin equals 1. atmospheres. Times 0.9900 leaders times the 770 kelvin. So divide out the volume and the temperature that we have here on both sides here. So we divide them both, both sides. By that we'll have isolated at the end r p two So leaders cancel out Kelvin's cancel out and we'll see here that p two equals 7.47 atmospheres so we can see that by changing our volume and changing our temperature. It's caused this change in our pressure to which comes out again to 7. atmospheres.

A 4.30 L gas has a pressure of 7.0 atm when the temperature is 60.0 ºC. What will be the temperature of the gas mixture if the volume and pressure are decreased to 2.45 L and 403.0 kPa respectively?

• A. 108 K
• B. 181 K
• C. 206 K
• D. 310 K
• E. 381 K

A sealed container with a movable piston contains a gas with a pressure of 1380 torr, a volume of 820 mL and a temperature of 31°C. What would the volume be if the new pressure is now 2.83 atm, while the temperature decreased to 25°C?

• A. 0.0253 L
• B. 0.167 L
• C. 0.326 L
• D. 0.516 L
• E. 1.46 L