Natural gas is a mixture of hydrocarbons, primarily methane (CH₄) and ethane (C₂H₆). A typical mixture might have X_methane = 0.915 and X_ethane = 0.085. Let's assume that we have a 15.50 g sample of natural gas in a volume of 15.00 L at a temperature of 20.00 °C. (b) What is the pressure of the sample in atmospheres?

Video transcript

Hello everyone today, we are being given the following problem. An L. P. G. Or a liquefied petroleum gas is made up of a mixture of propane and butane assume that the mole fractions of propane and butane in a certain L. P. G mixture are 0.8 22 and 220.17 8 respectively. A 36.5 g sample of an L. P. G mixture has a volume of 10 liters at 15 degrees Celsius, calculate the pressure in atmospheres of our mixture. So the first thing we want to do is make note of our mole fractions. So we have our mole fractions here and they were given to us in the question for propane, It was 0.822 and for butane It was 0. secondly, we want to calculate the molar mass for both propane and beauty. And so our molar mass of our propane is going to be 12. g per mole given on the periodic table. The molar mass for carbon is 12.1 and then we have three of them. So you multiply that by three and then we add that by the molar mass of hydrogen which is 30.1 point 008 g per mole. And there's eight of them in propane. We multiply that by eight and we get 44 point oh nine g per mole. We'll save that for later. Next for our molar mass for our butane, It's going to be 12.01 because that's the molar mass for one carbon. But the routine has four carbons. So that's going to be times four plus the molar mass of hydrogen. And in this situation there's 10 of them. So we're gonna hold it by 10 and we will get 58. g thermal. We will hang onto those two numbers and so now we have to find the average molar mass and we do that by using our mole fraction. So to find the average molar mass of this mixture, we're going to take our mole fraction of propane which is zero 8-2. And we will multiply that by our molar mass of propane which would be 44. g per mole. And then we're gonna add that to the mole fraction of butane, which is 0.178 times the molar mass of butane, which is 58.12 g per mole. And through that we're going to get 46.5906 g per mole for our average molar mass. And so from that average molar mass, we need to find the total moles that we have. And so to do that, we take our mass that was given to us in our sample, our 36.5 g sample and we multiply that by the molar mass of this mixture, which we said was for every one mole we had 46.5906 g units of grams will disappear. And canceled out. We'll be left with 0.78 moles. We're going to save that for later For our 5th step we're going to recall our ideal gas law and rearrange it to solve for P. So we're gonna solve for our pressure. Our ideal gas law says that our pressure times are volume is equal to the number of moles times of gas constant, times our temperature. And so we arranging this, we get our pressure as you go to our most times gas constant times temperature over our volume. and so moving on up here to step six, We need to make sure that all of our units are in order and we noticed that we have our temperature here which is in 15°C. However, that must be converted to Kelvin. So we're gonna do is we're gonna add 273.15 and we're gonna get 288.15 Kelvin. And now that our units are in check, we can go ahead and solve for our pressure. So we have the number of moles that we solved for which is 0.7834 moles. R gas constant is 0.8206 Leaders Times atmospheres in the numerator moles, times kelvin in the denominator And they were gonna multiply that by our temperature which we just saw it for was to 88.15 Kelvin and all of that's gonna go over our volume of 10, L Giving us a final pressure of 1.85 atmospheres of the mixture. And with that we have solved the problem overall, I hope that this helped, and until next time.