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

• il postino

#### 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

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.

• anorlunda
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 ##

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
:(

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!

• hutchphd