This table displays the vapor pressure of ammonia at several diff...

Question

This table displays the vapor pressure of ammonia at several different
temperatures. Use the data to determine the heat of
vaporization and normal boiling point of ammonia.
Temperature (K) Pressure (torr)
200 65.3
210 134.3
220 255.7
230 456.0
235 597.0

Accepted Answer

hey everyone in this example, we need to calculate heat of vaporization and the normal boiling point of digital ether. Using the table below that shows the vapor pressure of digital ether at different temperatures. So in our calculators, we're going to plot a graph where we're going to take the natural log of our pressure values versus our inverse temperature values. And this is going to generate an equation of a line where we have our line equal to -2642.8 as our slope X plus our Y intercept 16.463. And so this is our slope value. And we should recall that to get our heat of vaporization. We would take negative one times our slope value. And we're going to multiply this by our gas constant R. And so what we can say is that our heat of vaporization is equal to negative one times our slope. Which from our calculators we set is negative 2642.8. And then we're going to multiply this by a gas constant. R. Which we recall is 8.314 in units of joules divided by moles times kelvin. We also want to make sure our heat of vaporization is in units of kilograms per mole. So we're going to multiply by the conversion factor to go from jules, tequila jewels by recalling that our prefix kilo tells us we have 10 to the third power jewels for one kg jule. And so now we're able to cancel out jewels were left with kayla jewels per mole times kelvin. And we can actually interpret our slope value in units of kelvin, since it's based off of our temperatures of digital ether. So we can also cancel out our units of kelvin here. And so now we're just left with kilograms for mole officially. And sorry about that. So we don't want to scroll down yet. So what we're going to get is an official value for heat of vaporization equal to 21.97 kg joules per mole, sorry, that should say kilo jewels. So this would be kayla jules Permal for heat of vaporization. So going back to our equation of our line, we can reinterpret it so that we can understand why as now the Ln of our pressure Equal to our y intercept, which we said is negative 2642.8 Kelvin multiplied by our inverse temperature as X. And then we're going to add our y intercept 16.463. We're going to assume that our pressure at normal boiling point Is equal to 760 of Mercury. And so we can re simplify this so that now we have the natural log of 300 or sorry, millimeters of mercury Equal to our negative value of our slope negative 2642.8 kelvin multiplied by our inverse temperature Added to our y intercept 16.463 Because we want to solve for the normal boiling point, we're going to move the temperature term to the left hand side so that we can now have on the left hand side, 2642.8 times are inverse temperature equal to on the right hand side. We should now have 16. Being subtracted from our natural log of of Mercury. And so now to isolate for temperature, we're going to divide both sides by 2000, sorry, by 642.8 on both sides. And so this allows it to cancel it on the left hand side. So now what we would have is our inverse temperature is equal to and actually to simplify even further. So that temperature is no longer in the inverse, we're just going to isolate temperature completely by having Our right hand side be in an inverse fraction so that we have in the denominator 16.463 Subtracted from the Ln of 760. And then this is going to be divided by 2642.8, which should still have the units of kelvin here. So sorry for not writing that in. And so we should get for our final temperature A value equal to 268. Kelvin. And this would be our second final answer, which is our normal boiling point of our di ethyl ether. And then our first answer we found above was our heat of vaporization and kilograms per mole of our devil ether. So everything highlighted in yellow represents our final answers. I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

This table displays the vapor pressure of ammonia at several different temperatures. Use the data to determine the heat of vaporization and normal boiling point of ammonia. Temperature (K) Pressure (torr) 200 65.3 210 134.3 220 255.7 230 456.0 235 597.0

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