Ethanol has a heat of vaporization of 38.56 kJmol and a normal
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Question

Ethanol has a heat of vaporization of 38.56 kJ>mol and a normal 
boiling point of 78.4 °C. What is the vapor pressure of ethanol 
at 15 °C?

Accepted Answer

Welcome back, everyone. We need to calculate the vapor pressure of bromine BR two at 27.5 degrees Celsius given its heat of vaporization of 30. kg per mole. And it's normal boiling point of 58.8 degrees Celsius. Let's begin by recalling that we can use the following formula where we take the natural log of our final pressure P two over our initial pressure P one which is set equal to negative one times the entropy of vaporization delta H Vape which is divided by our constant are for the gas. And then this quotient is multiplied by the following brackets where we have the difference between one over T two minus one over T one, our initial temperature. So now with this formula outlined, let's make note of some of our variables. So for our recall that in this formula, our value for the gas constant R is equal to 8.314 with units of joules per mole times Kelvin. And now focusing on our variable for entropy of vaporization. We're told in the prompt that we have a value and sorry, this is a delta symbol we're told that we have a value of 30.91 killer jules Permal. Now note that R gas constant R is again jules, Permal times Kelvin. So we want our unit in the numerator for the entropy of vaporization to also be jewels, meaning that we're going to do a conversion to go from killer joules per mole to jules per mole. And so actually, what we want is we're going to multiply by kilo joules in the denominator so that it can cancel out and jewels in the numerator. And so recall that our prefix kilo tells us that we have To the third power of our base unit jewels. This allows us to cancel out killer jewels leaving us with jewels per mole as we wanted. And we will result in an entropy of vaporization of 30,910 joules per mole. Now, with that outlined, let's focus on our initial temperatures and pressures. So for temperature, the initial temperature were given a value of the normal boiling point being 58.8 degrees Celsius. But we're going to want to convert this to Kelvin because we see that we have units of Kelvin in R gas constant. So we're going to add to 73.15 And this is going to give us an initial temperature equal to 331.95. Kelvin. Now focusing on the final temperature T2, we are given the vapor pressure, we're sorry, not the vapor pressure, but we're given that we need to calculate the vapor pressure of remain at 25 degrees Celsius. And so 25 degrees Celsius would be the final temperature when we calculate the vapor pressure, which we're told is 27.5 degrees Celsius. So we also want this in Kelvin. So we're going to add to 73.15 and this results in a final temperature equal to 300.65 Kelvin. And so we just need to outline our pressure values, the final pressure, the vapor pressure, we don't know that's what we're solving for. But for the initial pressure, we're going to recognize that because the normal boiling point is given to us for our bro mean, then we can understand that initially, we likely have standard conditions. And with that said, we can say that our pressure is equal to 1 80 M, which is our standard pressure at S T P. So now with all of our variables outlined, let's get into our formula. What we have is the natural log of our quotient P two where we're solving for the numerator and dividing by our initial pressure of 1 80 M. This is set equal to negative one times our entropy of vaporization in the numerator which we converted to joules per mole as 30,910 joules per mole, which is divided by R gas constant R which we recall is always 8.314 joules divided by moles times Kelvin. Now, with this quotient, we want to and sorry, let's make this a parenthesis. With this quotient, we're multiplying now by our brackets where we have our inverse final temperature which again, T2 We said is 300.65 Kelvin subtracted from our inverse initial temperature which we converted to 331.95. Kelvin. Now let's focus on units before we type everything into our calculator. Recognize that we can cancel out our units of Kelvin here with our units of Kelvin being multiplied here or subtracted rather, then we can also cancel out our units of jewels as well as most. And so we're left with A T M as our final unit, which is what we want. Since we're solving for P two R vapor pressure, now we can go ahead and simplify. And so we're going to begin by simplifying the entire right hand side in our calculators so carefully type this in and make sure you have your parentheses and brackets for each part on the right hand side. And what we should result in when we also multiply by this negative one here is a value of negative 1.166. Again, we have no units left on the right hand side and our left hand side is the same, we just have the same thing. So I'll just copy and paste that. So we can just now focus on getting rid of our Ln term. Now, so we're called that we can cancel out Ln by taking both sides as exponents to Ulis number E. And so this cancels out Ln on the left hand side. And what we'll find is that now we have P 2/1 80 M equal to the right hand side, which results in a value of 0.3116. And so now we just need to get rid of that denominator 1 80 M. So we're going to multiply both sides by 1 80 M. It will cancel out on the left hand side and we're left with exactly P two equal to 0. ATMs which we can round to exactly our minimum number of sig figs being three sig fix from our problem text which would give us a final answer of 0. to 80 M as our vapor pressure of romaine At 27.5°C. And our final answer highlighted in yellow corresponds to choice a in the multiple choice. Thanks for watching. If you have any questions, leave them down below and I'll see everyone in the next video.

Ethanol has a heat of vaporization of 38.56 kJ>mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 15 °C?

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