What is the vapor pressure of SiCl4 in mm Hg at 30.0 °C? (See Pro...

Question

What is the vapor pressure of SiCl4 in mm Hg at 30.0 °C? (See Problem 11.27.)

Accepted Answer

Hey everyone, we're asked to determine the vapor pressure in millimeters of mercury for tetra brahma, methane at 178 degrees Celsius. Using the following information with our normal boiling point being at 189.5 degrees Celsius and our change of entropy of vaporization being at 48.2 kg joules per mole for this question. We do have to use the following formula. The natural log of our pressure to is equivalent to the natural log of our pressure one plus the change of entropy of vaporization, divided by our gas constant R times one, T one, which is our temperature -1 over our second temperature. So let's go ahead and write out our values for pressure one. This will be the pressure at our normal point which is one atmospheric pressure Which is also 760 of Mercury. Next looking at our temperature one We were told that this is 189.5°C. But we need to convert this into Kelvin and we can do so by adding 273.15. And this will get us to a temperature of 462.65 Kelvin. Now looking at our second temperature, This is going to be 178°C Plus 273.15 which will get us to 451.15 Kelvin. Now looking at our change of entropy of vaporization, we were told that this was 48.2 kg joules per one mole. Now, since our gas constant R. is in Jules, let's go ahead and convert this into jewels. Now we know that one killer jewel contains 10 to the third jewels. So when we calculate this out, we end up with a value of 48,200 joules per mole. And as we've learned, our gas constant is going to be 8.314 joules per mole times Kelvin. Now that we have our values, let's go ahead and plug these in. So we have the natural log of our pressure to is going to be equal to the natural log of our pressure one, which is 760 of mercury Plus the change of entropy of vaporization, which is 48,200 joules per mole divided by our gas constant R which is 8.314 joules per mole times Calvin. Next we're going to multiply this by one Over our temperature one Which is 462.65 Kelvin. And we're going to subtract one over our temperature to which is 451.15 Kelvin. Now, when we calculate this out and cancel out all of our units, We end up with a natural log of pressure to being 6.31389, Solving for pressure 552.1 we end up with 552.1 a millimeters of mercury And we can further simplify this into 552 mm of mercury, which is going to be our final answer. Now, I hope that made sense and let us know if you have any questions.

What is the vapor pressure of SiCl4 in mm Hg at 30.0 °C? (See Problem 11.27.)

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