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### Topic: Calculation of the temperature and pressure of triple point  (Read 12943 times)

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#### Christine1

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##### Calculation of the temperature and pressure of triple point
« on: June 20, 2011, 11:29:02 AM »
Hi, could anyone help me out with this question?

The temperature dependence of the vapor pressure of solid sulfur dioxide can be approximately represented by the relation log (p/Torr)=10.5916-1871.2/(T/K) and that of liquid sulfur dioxide by log (p/Torr)=8.3186-1425.7/(T/K). Estimate the temperature and pressure of the triple point of sulfur dioxide.

Thank you so much

#### enahs

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##### Re: Calculation of the temperature and pressure of triple point
« Reply #1 on: June 20, 2011, 11:30:50 AM »
Just use excel and make some data points and plot it and take a look at the graph.

#### BluePill

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##### Re: Calculation of the temperature and pressure of triple point
« Reply #2 on: June 21, 2011, 02:22:00 AM »
If I remember correctly, triple point is the pressure and temperature where three phases coexist. So, all three phases would have the same pressure. Then, you can make the assumption that logP(solid SO2) = logP(liquid SO2). Solve for T.

Once you get T, you could probably substitute that T to either one of the equations to get the temperature.

I'm not that entirely sure though. Just a thought.

#### rober2et

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##### Re: Calculation of the temperature and pressure of triple point
« Reply #3 on: March 15, 2015, 09:55:33 PM »
I can't for the life of me get the right pressure.
I set the equations equal to each other and got T=195.99K
Then substituting back into the second equation:

log(p)=8.316 - 1425.7(195.99K)
p=10^(8.316 - 1425.7(195.99K))

I either get 0, 1, or an error due to overflow when I'm supposed to be getting ~11 Torr.
What on earth am I doing wrong?

#### sjb

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##### Re: Calculation of the temperature and pressure of triple point
« Reply #4 on: March 16, 2015, 02:44:56 AM »
I can't for the life of me get the right pressure.
I set the equations equal to each other and got T=195.99K
Then substituting back into the second equation:

log(p)=8.316 - 1425.7(195.99K)
p=10^(8.316 - 1425.7(195.99K))

I either get 0, 1, or an error due to overflow when I'm supposed to be getting ~11 Torr.
What on earth am I doing wrong?

Where has the 8.316 come from? Also, it's 1425.7รทT, not multiplied by.