# Chemical Forums

## Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Suze on April 08, 2005, 08:32:47 AM

Title: Second virial coefficient
Post by: Suze on April 08, 2005, 08:32:47 AM
I don't seem to be getting anywhere with this problem:

The second virial coefficient B' can be obtained from measurements of the density (d) of a gas at a series of pressures. Show that the graph of p/d against p should be a straight line with slope proportional to B'. Find the values of B and B'at 298K... and I've been given six pressure values with corresponding densities.

I've plotted p/d vs p but don't see how that helps. I've plotted Z vs p and don't see how that helps. I've tried using pVm=RT(1+B/Vm) but don't get the correct value for B. I've tried using pVm=RT(1+B'p) but don't get the correct value for B'. I know B=B'RT but that's not useful until I've found one of the second virial coefficients.

Please could somebody help me with this??  ???
Title: Re:Second virial coefficient
Post by: Donaldson Tan on April 08, 2005, 03:19:09 PM
pv = RT + Bp where B is a function of temperature and v is molar volume

the above relationship holds sufficiently well for p < 5bar

d = M/v where M is the molar mass
pv - Bp = RT
p(v - B) = RT
v - B = RT/p
v = B + RT/p
d = M/v where M is the molar mass
d = M/(B + RT/p) = Mp/(Bp + RT)
d/p = M/(Bp + RT)
p/d = (Bp + RT)/M = Bp/M + RT/M
p/d = (B/M)p + RT/M

from the above equation, it can be seen the gradient of the graph for p/d against p is a function of B. note: density is varied with pressure at constant temeprature.
Title: Re:Second virial coefficient
Post by: Suze on April 11, 2005, 08:25:11 AM
I understand the manipulation of 'pV=RT + Bp' and 'd=M/V' to get an equation in the form of 'y = mx + c', and after plotting the graph of p/d vs p, I multiply the gradient by M (molar mass) but still don't get the answer printed in my textbook.
Here are the given pressure and density values:
p /Torr         91.74     188.98     277.3     452.8     639.3     760
d /(g L-1)     0.225      0.456      0.664     1.062     1.468    1.734

I get: B = 2.119 L mol-1, B' = 1.140E-4 Torr-1 = 1.5003E-7 atm-1
The printed answer is: B = -4.4 L mol-1, B = -0.18 atm-1

Please could someone check if my answer is correct or if I have a problem with my method??  :-\
Title: Re:Second virial coefficient
Post by: Donaldson Tan on April 11, 2005, 08:46:05 PM
what is the molar mass of the gas? i used the data to directly calculate the gradient, which is 439.93 Tor L / g
Title: Re:Second virial coefficient
Post by: Suze on April 12, 2005, 02:49:36 AM
Sorry, I forgot to say the gas is dimethyl ether so the molar mass is 46.0682 g mol-1.

If the gradient is the y-value divided by the x-value, shouldn't the units of the gradient be Torr g-1 L / Torr = g-1 L, and then multiplied by 46.0682 g mol-1 to give 'x' L mol-1 ?

So at least neither of us get a minus value which the printed answer is....
Title: Re:Second virial coefficient
Post by: Donaldson Tan on April 12, 2005, 02:05:59 PM
at what temperature was the measurement taken?
Title: Re:Second virial coefficient
Post by: Suze on April 13, 2005, 09:15:33 AM
I'm really sorry - I keep leaving vital information out!!!  :P

The measurements were taken at 25 deg C.