Recent content by DiffusConfuse

yes, it is 155 + 0.086T - 1.8\times106 T-2 (J/mol K)

Yes this is the standard reference state of the material. Heat of vaporization is not constant. The heat of fusion is unknown (no measured data). The entropy of vaporization is from liquid to vapour at 460 K. There is also vapour pressure data for the solid-gas; Log10 (mmHg) = -3571/T...

at 298K it is a solid

I am presented a review of data which gives: vapour pressures of a liquid have been measured and fit to the following equation: Log10 (mmHg) = -3571/T + 8.999 The melting point has been determined to be 392.7 K. A Cp value given for the liquid is 250 J/mol K and the ΔSvap is 117 J/mol K...

I would use L= TΔS = 68361 J/mol = 229.29 J/mol K x 392.7 K = 392.7(S2-S1), At this point I am unsure what to use for S2 to solve for S1 or vice versa.

Homework Statement Vapour pressures of a liquid have been measured and fit to the following equation: Log10 P (mmHg) = -3571/T + 8.999 The melting point has been determined to be 392.7 K. A Cp value given for the liquid is 250 J/mol K and the ΔSvap is 117.20 J/mol K Homework Equations...

I rechecked it, I get 8222.53 instead of what is listed above, which doesn't change my final answer drastically

for some reason I cannot edit the formula above, it should be: Log10 P (mmHg) = -3571/T + 8.999

Just noticed an error in the formula I gave. The +6.124 is for the pressure in bar, when I gave it in mmHg. I will correct it to the given data in mmHg (+8.999)

Yes, thank you haruspex, it is Log of the Pressure in mmHG and assuming Kelvin for temperature units

Homework Statement vapour pressures of a liquid have been measured and fit to the following equation: Log10 (mmHg) = -3571/T + 6.124 The melting point has been determined to be 392.7 K. A Cp value given for the liquid is 250 J/mol K and theΔSvap is 117 J/mol K Homework Equations...

My attempt was as follows: T1= 392.7 K P1 = 0.8045 mmHg (from the equation, giving the vapour pressure of the LIQUID at the melting point). T2= 298.15 K P2= 1.052E-3 mmHg From the Clausius Clapeyron: ln(0.8045/1.052E-3)= ΔH/R*(1/298.15 K - 1/392.7 K) ΔH/R = 8221.89 ΔH = 68361...

Homework Statement The vapour pressures of a liquid have been measured and fit to the following equation: Log10 (mmHg) = -3571/T + 6.124 The melting point has been determined to be 392.7 K. Calculate the standard entropy of the liquid at the melting point. Homework Equations...

I did not actually plot the concentration. My diffusion profile consists of an initial amount of substance a in moles n(a) and plots it as a function of distance in the sample. so I am trying to figure out what information I get when I integrate this profile. moles*meters would be the units

Hello- I have a diffusion profile, in which I plot the decrease in concentration versus distance of my sample. I am trying to find the increase in mass of the overall sample. How would I do this? I have integrated the function and am wondering what type of information that would supply...