Wednesday

July 30, 2014

July 30, 2014

Posted by **Anonymous** on Tuesday, July 19, 2011 at 2:38pm.

- Chemistry -
**DrBob222**, Tuesday, July 19, 2011 at 2:52pmUse the Clausius-Clapeyron equation. Remember the definition of boiling point; i.e., the temperature at which the vapor pressure of the liquid equals atmospheric pressure. So you know p1 and p2 and T1. T2 is the only unknown.

- Chemistry -
**Anonymous**, Tuesday, July 19, 2011 at 4:38pmI know im using that equation but would you set it up like this

1/t2= (760mm hg/480torr)/(30720j/8.314)+(1/353.25degrees celsius) when i do it out i get a number that does not make sense.

- Chemistry -
**DrBob222**, Tuesday, July 19, 2011 at 6:39pmI didn't work it out but I suspect the problem is that you are trying to solve for 1/t2 from the get go. It is far simpler if you plug the numbers into the equation and solve that way.

ln(P2/p1) = DHvap/R (1/T1 - 1/T2)

ln(480/760) = (30720/8.314)(1/353.25 - 1/T2)

First term = ln(480/760) = -0.4595

30720/8.314 = 3694.97

1/353.25 = 0.002831

1/t2 = 1/t2

-0.4595 = 3694.97(0.002831-1/T2)

-0.4595 = (10.4599 - 3694.97/T2)

Check my work. I'll let you finish.

- Chemistry -
**Anonymous**, Tuesday, July 19, 2011 at 9:17pmI have -.4595= -3684.51/T2. Would you then just divide -3684.51/-.4595? and how did you het 10.4599 in the last equation?

- Chemistry -
**DrBob222**, Tuesday, July 19, 2011 at 10:20pmNo. I think your algebra is cockeyed.

Assuming my work is ok to where I stopped, then multiply through by T2 to get rid of the T2 term in the denominator.

-0.4595 = (10.4599 - 3694.97/T2)

-0.4595T2 = 10.4599T2 - 3694.97

-10.919T2 = -3694.97

T2 = 338.4 or thereabouts which is approximately 65 C. Check my work but that sounds reasonable to me. Lower pressure means lower boiling point. I can see why 8,000 K didn't sound right to you.

- Chemistry -
**1994**, Wednesday, November 28, 2012 at 5:53amIsn't the formula suppose to be ln(p2/p1)=Dvap/R(1/T2 -1/T1) ? That is what my textbook says.

**Related Questions**

chemistry - Benzene has a heat of vaporization of 30.72kJ/mol and a normal ...

Chem. - Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling...

Chemistry - Benzene has a heat of vaporization of 30.72kJ/mol and a normal ...

problem - Benzene has a heat of vaporization of 30.72 kj/mol and a normal ...

chemistry - Benzene has a heat of vaporization of 30.72kJ/mol and a normal ...

science - benzene has a heat of vaporization of 30.72kj/mol and a normal boiling...

chemistry - Benzene has a heat of vaporization of 30.72kJ/mol and a normal ...

chemistry - Benzene has a heat of vaporization of 30.72kj/mol and a normal ...

chemistry - Benzene has a heat of vaporization of 30.72kj/mol and a normal ...

Chem. - Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling ...