Dichlorodifluoromethane, CCl2F2, one of the chlorofluo- rocarbon refrigerants responsible for destroying part of the Earth's ozone layer, has Pvap = 40.0 mm Hg at -81.6 °C and Pvap = 400 mm Hg at -43.9 °C. What is the normal boiling point of CCl2F2 in °C?

Video transcript

Hi everyone for this problem. It reads talia line is an organic solvent used in paint thinners and markers. It's favor pressure at 31.8°C and 40.3°C are 40 of mercury and 60 of mercury, respectively. Using this information estimate the normal boiling point of toggling and degrees Celsius. So the answer that we want to find here is the normal boiling point of the line and degrees Celsius. And in this problem, okay, we're given two pressure's all right. So we're given 40 millimeters of mercury and 60 millimeters of mercury and were also given to temperatures 31.8 degrees Celsius and 40.3 degrees Celsius. The equation that we're going to want to use to solve this problem to calculate the heat of vaporization is because that's the first thing we're going to need to do is calculate the heat of vaporization and the problem that we're going to use or the equation that we're going to use is Ln of P two over P one where these are the pressures is equal to the negative entropy of vaporization over gas constant. R times one over T two. That's our second temperature minus one over T one. Okay, so for our temperatures we're going to need these temperatures in units kelvin. Alright, So T one Okay, is equal to 31.8°C. And we need to convert this to Kelvin. So we're going to add 273.15. And so the temperature we're going to get is 304.95 Kelvin And T two is 40.3°C. So we need to also convert this to Calvin. So the temperature we're going to get for T two is 313.45 kelvin we can go ahead and also write out our P one and P two. So P one is Our first pressure 40 millimeters of mercury and P two is 60 millimeters of mercury and gas constant R is 8.314 jewels per mole kelvin. Alright, so these are the this is all the information given aside from our which is a constant we should know. So we have everything that we need to solve for our entropy of vaporization. So let's go ahead and plug in these values. Alright, so when we plug in these values we're going to get L. N. Of P two over P one. So P two is 60 millimeters of mercury over P one is 40 of mercury Is equal to the negative entropy of vaporization over gas constant R which is 8.314 jewels Permal Calvin. This is going to be times one over T two which is 313.45 kelvin minus one over T one is 304.95 kelvin. Okay, so we've written everything out so now we can go ahead and simplify. Alright, so for our left side we're going to get Ln of 1.5. Okay, We're gonna take 60 and divided by 40. So that is going to be 1.5 and this is going to equal the negative entropy of vaporization over gas constant R. And we're gonna go ahead and simplify our one over T two minus one over T one. Okay, And when we simplify that, we get negative 8.58 68287 Times 10 to the -5 Calvin inverse. All right, So, let's go ahead and simplify More. Alright, and so are Ln of 1.5 is going to become zero .405 four 651081. And this is going to equal the same thing that is on the right side. Alright, So we just simplified RLN of one 0.5. All right. So now what we're going to do is we're going to we're going to bring we're gonna divide both sides of our equation by negative 8.5. We're gonna divide it by this by this number here. All right, so, we're gonna divide both sides by negative 8.5868-2 87 times 10 to the -5 Kelvin's inverse. Okay. Alright, So when we do that, what we're going to get then is we're gonna bring are negative Entropy of vaporization over 8.314 joules per mole Calvin is equal to now. We're gonna do this calculation right here And it's going to equal -4721.84 243 Calvin. Okay, so now we're going to multiply both sides by r gas constant. R Alright, so we're gonna multiply by 8.314 joules per mole kelvin. We're going to multiply by 8.314 joules per mole kelvin. All right. So this is going to cancel. And what we're going to get is our negative entropy of vaporization is going to equal 39,000 257.398 Jules per mole. And we can divide both of both sides by negative ones so that our entropy of vaporization becomes positive. So we'll divide both sides by -1. And so our and there'll be a vaporization. Once we do this is going to become positive or this number should be negative here. This should be a negative 39,000. Okay, So we get positive. 39,000 257.398 jules Permal Okay, so that is our entropy of vaporization but we're not done. Okay. We need to note that at normal boiling point. It's alright. Bp at normal boiling point. The pressure of vaporization is equal to the atmospheric pressure. Okay. And That is equal to 760 millimeters of mercury. Alright. So using that information we're going to rewrite the equation that we wrote up above. All right. so we have L. N. of P two over P one is equal to our negative entropy of vaporization over gas constant R times one over T two minus one over T. One. And in this case T one is equal to 358.15 Kelvin and T two is we're going to say this is equal to our boiling point. Okay, so tv So we're going to plug everything in again and essentially we're solving for T. Two because when we saw for T. Two that's gonna let us know what the boiling point temperature is. And this temperature we're going to get it in kelvin. Okay. And so we're going to need to convert that temperature two Celsius because the question asked us for it in Celsius. So let's go ahead and plug in our values. So we're going to get Ellen of 760 of mercury over 40 of Mercury is equal to the negative and there'll be a vaporization Over 8.314 jewels Permal kelvin. This is gonna be multiplied by or we can plug in our heat of vaporization because we solved for it. Okay, so it's gonna equal negative 39,000 257.398 jewels, her mole times one Over T two. Which is what we're trying to solve for -1 over T. one Which is 304.95 Kelvin. Okay, So let's go ahead and simplify the left side Ln of 7, 60/40 Is equal to 2.94444. And this is going to equal Our heat of vaporization over R. is negative 4721.842 43 Kelvin Times one over T two -1 over T one. R. T one over T one becomes minus 3.279 226103 times 10 To the negative three Calvin's inverse. All right, so let's isolate our one over T two. Okay, so when we isolate our one over T two this becomes one over T two is equal to negative 6.23578 times 10 to the negative four Calvin's inverse plus 3.279-2 6103 times 10 to the negative three kelvin's inverse. So when we saw for T. two. Okay, that's what we're solving for. When we saw for T. Two we're going to get T two or what? Let's simplify first. Okay, when we simplify this We get one over T two Is equal to 2.65556 2.6556 times 10 to the negative three kelvin's inverse. Alright, so we're solving for T. Two here. So we're gonna get T two is going to equal 376.56 kelvin and we want this in degrees Celsius. Alright, so when we convert this two degrees Celsius we're going to subtract 273.15, so T two is going to equal 103.4 degrees Celsius. And that's going to be our final answer, degrees Celsius. So our answer to this problem is the normal, the estimation for the normal boiling point of towline and degrees Celsius is 103.4. That is it for this problem? I hope this was helpful.