Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1 °C. At what temperature does benzene boil when the external pressure is 445 torr?

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Hey everyone in this example, we need to calculate the temperature at which cyclo octane will boil at an external pressure of 565 Tour. We're told that the heat of vaporization is 43.3 kg per mole and the normal boiling point for cyclo octane is 149 degrees Celsius. So in giving us the boiling point for cyclo octane, the prompt tells us our initial temperature T one which we can convert to um from Celsius to kelvin. So we're going to take the Celsius temperature 1 49 degrees Celsius and add to 73.15 to give us our kelvin temperature which is going to be equal to 422.15 kelvin. Next we want to recognize that the prompt is asking us to find the final temperature. So this is what we're solving for. We also want to recognize that because we're given the boiling point for Cyclo octane. We're told that at this boiling point the external pressure is 565 tour. So we're gonna go ahead and recognize that this is going to be our final pressure P two 565 Tour. And so that means that we can assume that our initial pressure is going to be standard pressure which should be 760 tour Or 1 80 M. But we're going to use the unit tour because we want to keep our units consistent. So because we have temperature and values for pressure. We need to go ahead and recall the following formula where we would take the natural log of the final pressure divided by our initial pressure. We're setting this equal to the negative value of our heat or entropy of vaporization divided by r gas constant R. Where we will also multiply in brackets by our inverse final temperature subtracted from our inverse initial temperature. And because we're going to utilize this formula, we should recall that r gas constant R. Is equal to 8.314 jewels divided by most times kelvin as the units. And because we use these units for our gas constant R. When we look at our heat of vaporization, Which according to the prompt is equal to 43.3 kg jewels Permal. We want to convert from kilograms per mole so that we have units of jewels promote to cancel out with our gas constant R. Here. And so we're going to just apply a conversion factor to go from kilo jewels. Two jewels. By recalling that are prefix kilo, tells us that we have 10 to the third power of our base unit. Jewels for one kg jule. And so now we're able to cancel out kayla jewels were left with jewels Permal for our um entropy heat of vaporization sorry. And this is going to give us a new value for heat of vaporization equal to 43,300 jules Permal. And so now that we have all of our info properly laid out for our equation, we're going to go ahead and solve for our final temperature. So what we should have is that our natural log of our final pressure which again we set is 565 tour given from the prompt divided by our initial pressure, which we assume is standard pressure being 760 tour. Set equal to negative one times our heat of vaporization in the numerator, which again be converted to 43,300 joules per mole divided by our gas constant. R which were called above is 8.314 joules per mole times kelvin. This is then multiplied in brackets to our inverse final temperature, which we are solving for here subtracted from one over R Our initial temperature which we converted to Kelvin as 422. Kelvin. So our first step is to simplify this left hand side as well as this portion of the right hand side of our equation. So we're going to go ahead and do the math there. And what we should get in. Our next line is -0.2965 for the left hand side. And then on the right hand side when we divide our heat of vaporization by our gas constant R we're going to get a value of 0.8. And as far as the units will be able to cancel out jewels as well as moles leaving us with kelvin. So we would have the unit kelvin here and this would be inverse kelvin since it's in the denominator actually, it would remain as just kelvin now because we've we're now in just the numerator here, we don't have a fraction anymore and we're going to make this value negative because we also need to recall that we're going to multiply by negative one here. So for the right hand side we have that and then we also will still be multiplying by our brackets here where we have one over our final temperature subtracted from one over our initial temperature 422.15 Kelvin. So the next step to simplify here, since we're solving for final temperature is to divide both sides by negative 5002 08.08 Kelvin. And what that's going to give us when we simplify is on the left hand side, will now have our temperature term since it's isolated even more. So we have one over T2 -1 over initial temperature set equal to And we'll keep this in brackets here. So on the right hand side, when we divide these two values on the left here, we're going to get a result equal to 5.69, two Times 10 to the negative 5th power. And because we divided by Kelvin here, we're going to have units of inverse kelvin. So I'm just going to scroll down for more room and again, we want to continue to isolate for t to our final temperature. So what we want to do is get rid of this term here. So we're going to get rid of it by adding one over our initial temperature to both sides. And so now what we would have is that one over our final temperature is equal to on the right hand side. We're going to have 5.693 times 10 to the negative fifth power, inverse kelvin added to 1/4 22. kelvin. And so actually, since we technically wrote that before, we can just write the value of what that would equal. So when we take this value and add it to 1/4 22.15 kelvin, we're going to get a result equal to 2.4 to 5 times 10 to the negative third power inverse kelvin. And so we should recall that when we have diagonals in algebra we can just switch the places of those terms. So what we can have is now that our final temperature is equal to 1/2 0.4 to five times 10 to the negative third power inverse kelvin. And when we take the result of this fraction in our calculators, we should get that our final temperature is equal to a value of 412.24 Kelvin, Which we can say is just about 412 Kelvin. And this would be our final answer to complete this example. As our temperature that cycle, octane is going to boil at At the given pressure tour. So what's highlighted in yellow is our final answer. If you have any questions, leave them down below. Otherwise, I'll see everyone in the next practice video.

Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1 °C. At what temperature does benzene boil when the external pressure is 445 torr?

Verified Solution

Key Concepts

Heat of Vaporization

Boiling Point and External Pressure

Clausius-Clapeyron Equation