Calculate the pressure using the Van der Waal equation on reduced variables

[il postino]

Summary:: Find the pressure using the vdw equation in reduced variables

Hi everyone!
I have a doubt when I try to solve this exercise. The result was very high pressure.
Calculate the pressure using the reduced variable vdW equation for a sample of 74.8 grams of ethane in a $200 cm^3$ container and at $310.5 K$. The $P_c = 48.2 atm$ and $T_c = 305.4K$

I calculated the correction factors a and b as a function of $T_c$ and $P_c$
$a=(27R^2T_c^2)/(64P_c)$
$b = (RT_c)/(8P_c)$

$a=5,489 L^2 atm/mol^2$
$b=0,0649 L/mol$

Then I calculated the moles and then calculated the molar volume:
74.8 gramos = 2,493 mol

$V_m = 0,2L/2,493mol$
$V_m = 0,08022 L/mol$

and the van der Waals equation of state:

$(P + \frac{a}{V_m^2})(V_m -b) = RT$

and clearing $P$ and replacing:

$(P + \frac{5.489}{(0,08022)^2})(0.08022 -0.0649) = (0.082)(310.5)$

$P = 808,5 atm$

This is possible?
Is the procedure correct?
Thank you very much for the help

 

[hutchphd]

First of all carry the units and be sure to check each step of the calculation.
Always. Habitually. Never Forget.
I believe the first calculation (for a ) my be wrong. Now check all of them.

 

[il postino]

First of all carry the units and be sure to check each step of the calculation.
Always. Habitually. Never Forget.
I believe the first calculation (for a ) my be wrong. Now check all of them.

hi hutchphd! :)
I checked and they give me well the factor "a".
I did the calculation again and it gives $a=5,489 L^2 atm/mol^2$
thanks for the advice!

 

[hutchphd]

I'll take a look again later if you still don't find the problem. Does seem wrong to me...shouldn't the molar volume at 50 atm be ~half a liter (22.4 l/mole at stp)?
sorry i revised

 

[il postino]

I'll take a look again later if you still don't find the problem. Does seem wrong to me...shouldn't the molar volume at 50 atm be ~half a liter (22.4 l/mole at stp)?
sorry i revised

I can't find the error, if there is one
:(

 

[hutchphd]

The values for a and b seem correct (and match published values!). The "occluded space" "b" for 2.5 mole is therefore ~.165 liter and so the pressure is going to be pretty damned high! I think your answer is correct (sorry for my lousy intuition here) Good luck.

 

[il postino]

The values for a and b seem correct (and match published values!). The "occluded space" "b" for 2.5 mole is therefore ~.165 liter and so the pressure is going to be pretty damned high! I think your answer is correct (sorry for my lousy intuition here) Good luck.

You have been very helpful! thank you very much!