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I'm given a container of gas that is kept at constant temperature, and whose volume is decreased by applying pressure, from A to E, as the graph shows. I am given the temperature, the total moles in the container, and the following data: At B the molar volume is 1.153, and the pressure is 17.76. At C the molar volume is .8167, and the pressure is 17.76. At D the molar volume is .032, and the pressure is 17.76.

Given this data, and I emphasize, only this data (no critical P, T, or Vmolar), find the van der Waals constants, a and b, for the gas.

I've been working on it for 3 hours and so far all I can come up with is a single equation in terms of a and b at point B: 17.76=RT/(Vmolar-b) - a/Vmolar^2 At point C, the equation would be the same as what I came up with for B, since the temperature and pressure remains the same as B, and we're in the region of co-existence, so Vmolar is at its maximum for the gaseous phase and therefore also is the same as in B (any further increase in Vmolar causes precipitation).

Any and all help is greatly appreciated.

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    $\begingroup$ Point C is nothing but smoke and mirrors. It contains no new information. Of course the pressure in it is the same as in B and D; we could have guessed that much ourselves. Use B and D. Write down the equation, plug in all the known values... $\endgroup$ – Ivan Neretin Nov 14 at 13:53
  • $\begingroup$ I don't quite understand why this question is getting closed as homework because OP clearly demonstrated their attempt and explained the problem in details. And if the question were unclear, there probably weren't two answers already… $\endgroup$ – andselisk Nov 15 at 6:16
  • $\begingroup$ @MishaelHibshoosh It would be nice if you could add some information regarding the source of the image (e.g. a textbook reference/website URL). $\endgroup$ – andselisk Nov 15 at 6:18
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I would approach this a little differently. I would integrate the pressure between $V_D$ and $V_B$, such that the average pressure is 17.76 bars: $$\frac{\int_{V_D}^{V_B}{PdV}}{(V_B-V_D)}=17.76$$or, $$17.76(V_B-V_D)=RT\ln{\frac{(V_B-b)}{(V_D-b)}}-a\left[\frac{(V_B-V_D)}{(V_B-b)(V_D-b)}\right]$$ This would provide an equation for expressing a in terms of b. So the pressure could then be expressed exclusively in terms of b. Then, to determine the value of the constant b, I would set $P(V_B)$ equal to $P(V_D)$ and solve the resulting equation for b.

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    $\begingroup$ Took me a second, but this is very elegant and clever. Me gusta. $\endgroup$ – Michael Green Nov 15 at 19:42
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What this ultimately boils down to is that this is a system of equations which we're trying to solve. Since we have two unknowns, we need two equations. Luckily, we have 3.

$$17.76 atm = \frac{RT}{V_m - b} - \frac{a}{V_m^2}$$

where our three equations come from having three different known values of the molar volume.

As is typical with systems of equations, easiest way to deal with them is to reduce a first equation to a single unknown variable, and then substitute that variable into a second equation to explicitly solve for the other variable. Then, turn around and use that known second variable to explicitly calculate the first.

Van der Waals equation is kinda ugly to work with, but we can do that first step by solving for a with an equation 1 (I'm going to use subscripts on our molar volumes to allow us to keep track easily)

$$a = -V_{m,1}^2\Bigg[P-\frac{RT}{V_{m,1}^2 - b}\Bigg]$$

we'll take this piece and stick it into this equation where a is

$$17.76 atm = \frac{RT}{V_{m,2} - b} - \frac{a}{V_{m,2}^2}$$

and now since we have a single equation with a single unknown, just rearrange all the variables to solve for b. Use b back in equation 1 to solve for a.

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