The vapor pressure, P, of a certain liquid was measured at...

The vapor pressure, P, of a certain liquid was measured at two temperatures, T. The data is shown in the table. T (K) | P (kPa) 225 | 3.55 875 | 8.54 Keep the pressure units in kilopascals. If you were going to graphically determine the enthalpy of vaporizaton, ?Hvap, for this liquid, what points would you plot? To avoid rounding errors, use three significant figures in the x-values and four significant figures in the y-values. point 1: x = .004 point 1: y = point 2: x = point 2: y = Determine the rise, run, and slope of the line formed by these points. rise = run = slope = What is the enthalpy of vaporization of this liquid? ?Hvap = J/mol

Transcript

00:01 All right, so we have some data points of the temperature in kelvin versus the pressure in kilopascals.

00:06 And we want to determine the delta h of vaporization.

00:12 So an easy way to do this is to take the linear form of the relationship between pressure and temperature, which is given by the closest clapper in equation, which is that the natural log of the ratio of the pressures, p2 over p1, is equal to negative.

00:31 Delta h vaporization over r times one over t2 minus one over t1 the one and the two represent different points on the graph and so this is a linear equation where your slope is going to be the delta h vaporization negative over the ideal gas constant so you plot all the points and then you can calculate the slope and from there deduce delta h all right so what would we graph well these would be the y values and these would be the x values.

01:05 So for y, we want to graph the natural logarithms of the points.

01:10 So one point that we could have, if this is data set one and data point one and this is data point two, one xy point would be the ln of, well, actually that's the y value.

01:24 This one over the temperature in calvin and then the y value would be the natural log of 3 .55.

01:37 And then the other point, xy, for the corresponding to the second data point, would be one over the 875 kelvin plotted against natural log of 8 .54, which we can go ahead and simplify both of these so that they're in numbers.

01:56 1 divided by 225 is 0 .00444 for repeating.

02:05 And then the ln of 3 .55 is 1 .265.

02:18 And then we have 1 over 875 is 0 .0014, and then the ln of 8 .54 is 2 .14.

02:45 What is the slope of the line formed by these two points? so we take the change in the y values.

02:51 I'll do the second minus the first, and then divided by the change in the x values.

03:00 All right, so 2 .145 minus 1 .267 comes up to 0 .878, and then 0 .0014 minus 0 .0044 comes out to negative 0 .0033...

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