What is the normal boiling point of the substance?

Critical point: 117.9 atm and 548°C

Triple point: 0.107 atm and 114°C

Normal melting point: 1.00 atm and 124°C

What is the normal boiling point of the substance?

I don't understand how to get the boiling point. I drew a phase curve and plotted the points given. I know the answer should be somewhere between 124 and 548 *C, wherever the 1 atm meets the liquid to gas curve. However, since the only information I am given are those few points, I don't know how to get the actual value from there.

Can someone please explain how to do this. Thank you! I'll pick a best answer. :-)

Update:

I now know that the answer is NOT 253*C. Still don't understand how to do this. 

Update 2:

I ran out of attempts, so I won't actually know if any of those answers are correct or not, but thank you so much for answering anyway! :-)

2 Answers

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  • Dr W
    Lv 7
    3 weeks ago
    Favorite Answer

    so you have two data points with T and vapor pressure info.  and you want T for a 3rd point.  That should immediately lead you to the clausius clapeyron equation.

    let's start there

    .. ln(P1/P2) = dHvap/R) * (1/T2 - 1/T1)

    rearranging

    .. (dHvap/R) = ln(P1/P2) / (1/T2 - 1/T1)

    now you want a third data point.. .so let's use point 1 and set up another CC

    type equation to solve for T3

    .. ln(P1/P3) = (dHvap/R) * (1/T3 - 1/T1)

    rearranging

    .. 1/T3 = ln(P1/P3) / (dHvap/R) + 1/T1

    subbing in that previous equation for dHvap/R

    .. 1/T3 = ln(P1/P3) / (ln(P1/P2) / (1/T2-1/T1)) + 1/T1

    or better yet.. 

    .. 1/T3 = ln(P1/P3) * (1/T2 - 1/T1) / ln(P1/P2) + 1/T1

    now.. the normal mp gives use the P3 = 1.00atm

    (that was the point of throwing in the normal melting point fyi)

    and we can sub in that and the rest to solve this

    .. .. . .. . .ln(0.107atm / 1.00atm) * (1/(273+548) + 1/(273+114)).. . ... .1

    .. 1/T3 = ----- ---- ----- ---- ---- ---- ---- ---- ---- ---- ---- ---- ----- ----- + ----- -----

    .. .. .. . .. .. .. ... .. .. .. . . . .ln(0.107atm / 117.9atm) .. ... ... ... .. .. . (273+114)

    .. 1/T3 = 0.002148 / K

    .. T3 = 465K = 192°C

  • 3 weeks ago

    Use the  Clausius-Clapeyron equation and the temperatures and pressures that you have for the triple and critical points to calculate Delta Hvap. Then, using that value and one of the points, calculate the Temperature at 1 atm pressure:

    ln (P2/P1) = (Delta Hvap / R) (1/T1- 1/T2)

    ln (117.9 / 0.107) = Delta Hvap / 8.314 J/molK (1/387 K - 1/823 K)

    7.0048= Delta Hvap (1.6565X10^-4)

    Delta Hvap = 4.23X10^4 J/mol

    Then,

    ln (117.9 / 1 ) = (4.23X10^-4 / 8.314) (1/T1 - 1/823)

    T1 = 464 K = 191 C

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