Propane, C3H8, liquefies under modest pressure, allowing a
large ...

Question

Propane, C3H8, liquefies under modest pressure, allowing a
large amount to be stored in a container.
(a) Calculate the number of moles of propane gas in a 20-L container at 709.3 kPa and 25 C. 
(b) Calculate the number of moles of liquid propane that can be stored in the same volume if the density of the liquid is 0.590 g/mL. 
(c) Calculate the ratio of the number of moles of liquid to moles of gas. Discuss this ratio in light of the kinetic-molecular theory of gases.

Accepted Answer

Hey everyone in this example, we're told that ammonia has hydrogen bonding among its molecules due to the high election negativity of nitrogen. We're told that due to the attractive forces, ammonia can be easily liquefied for part A. We need to determine how many moles of ammonia can, a 15 liter cylinder hold at 1.25 A. T M. S, and 30 degrees Celsius. So we're going to begin with solving out part A. And for part A. We want to recall our ideal gas equation where we have pressure times volume equal to the moles of our gas times the gas constant R times temperature. So what we're going to do since part A is asking for moles of ammonia is cell four moles of NH three by reorganizing our formula so that we have pressure times volume divided by the gas constant R times temperature. And so what that's going to look like is in our numerator, we're going to have our pressure given as 1.25 A t M's in the prompt for part A Multiplied by the given volume. In the prompt for part a as 15 L. And then this is going to be divided by r gas constant R which we should recall is equal to 0.8206 leaders times A T M's divided by moles, times kelvin And then multiplied by our temperature, which in this equation should be in Kelvin. So in the prompt were given the temperature 30°C and to convert that to Kelvin, we're adding to 73.15. And so what we're going to get for part A for the moles of our ammonia gas is a value equal to 0.754. And our units should be moles. But to show how that cancels out, we would be able to cancel out our units of leaders as well as a T. M's and then also kelvin from our edition of 73.15 here. And that leaves us with moles, which is what we want for moles of our ammonia. So this would be our answer for part A. And now we can move on to Part B of the prompt which states that if the liquid ammonia has the following density, how many moles of it could the same cylinder hold. So we're going to move on and sulfur part B. To solve for moles of our liquid. And this is specifically of our liquid ammonia. So we're going to begin our setup by using the volume that they give us from part a of our prompt which is 15 liters of ammonia. And we're going to go ahead and convert this from leaders into middle leaders. And that's due to the fact that as you can see in prompt B were given density in grams per middle leaders. So we want our volume to be in middle leaders so we can cancel out middle leaders out. So we're going to recall that our prefix milli gives us for one millimeter 10 to the negative third power leaders. And just so it's clear this is a -3. So I'm just going to fix this. So now we're able to cancel out our units of leaders were left with the middle leaders and now we can go ahead and plug in that density by multiplying by our density as a conversion factor where according to the part B of our prompt we have for one middle leader A density of 0.682 g per meal leader. And so just to make more room to solve for part B, we're going to scoot this over because we're going to now that we have milliliters canceled out. We're left with grams. But again we want moles of our liquid ammonia. So we're going to multiply by another conversion factor. Where we're going to use the molar mass of ammonia to end up with moles of ammonia. And so we should recall that molar masses in grams per mole. And from our periodic tables, we would see that for one mole of ammonia we have a molar mass of 17 point oh four g of ammonia. And so this allows us to cancel our units of grams from our density As well, leaving us with moles of Ammonia as our final unit, which is what we want to answer part b. And what we're going to get here is a value equal to moles of our liquid ammonia. So this is our answer for part B of the prompt. And now we can go ahead and move on to PART C of the prompt. So going to Part C, it says to determine the ratio of the number of moles of liquid to the number of moles of gas contained in the same cylinder. So we're going to be utilizing our answers from parts A and B. To answer part C. Because we need to give the ratio between the molds of our liquid To the moles of our gas. So we're going to be dividing these two quantities from one another and that's going to be in our numerator. We have our moles of our liquid ammonia as 600 moles of ammonia. And we're dividing this by the molds of our gas and our denominator, which according to part A of our problem, we determined that our moles of ammonia gas is 0.754 moles of ammonia gas. So now we have our ratio of ammonia liquid to ammonia gas. And this quotient is going to give us a result equal to 796 because our units of moles are going to cancel out here. So this would be our final answer for part C. And we should recognize that this is such a high value for the ratio of liquid ammonia to gaseous ammonia because in the liquid phase we should recall that our molecules are going to be much closer to one another. And so our container with our ammonia liquid molecules can hold a lot more ammonia molecules in comparison to the gaseous phase, where our gaseous phase molecules are farther apart. And so our container would hold less ammonia. So everything boxed in blue represents our final answers for part A through C of this question. I hope that everything I explained was clear. But if you have any questions, please leave them down below and I will see everyone in the next practice video.

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