Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started 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?? ???

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.
i hope this proves helpful

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 L1) 0.225 0.456 0.664 1.062 1.468 1.734
I get: B = 2.119 L mol1, B' = 1.140E4 Torr1 = 1.5003E7 atm1
The printed answer is: B = 4.4 L mol1, B = 0.18 atm1
Please could someone check if my answer is correct or if I have a problem with my method?? :\

what is the molar mass of the gas? i used the data to directly calculate the gradient, which is 439.93 Tor L / g

Sorry, I forgot to say the gas is dimethyl ether so the molar mass is 46.0682 g mol1.
If the gradient is the yvalue divided by the xvalue, shouldn't the units of the gradient be Torr g1 L / Torr = g1 L, and then multiplied by 46.0682 g mol1 to give 'x' L mol1 ?
So at least neither of us get a minus value which the printed answer is....

at what temperature was the measurement taken?

I'm really sorry  I keep leaving vital information out!!! :P
The measurements were taken at 25 deg C.