Find Van der Waals constants and what is the pressure for C2H6?
Find van der waals constants for C2H6 if critical temperature=32,1 deegres, critical pressure is 494, 76kPa.
Find pressure for 5g C2H6 in the container of volume=1L and on 15 degrees.
NOTE: it seems that Pc should be closer to 4947.6 kPa .
- Truong-Son
Find van der waals constants for C2H6 if critical temperature=32,1 deegres, critical pressure is 494, 76kPa.
Find pressure for 5g C2H6 in the container of volume=1L and on 15 degrees.
NOTE: it seems that
- Truong-Son
1 Answer
Using the result derived in full here that:
a=27R2T2c64Pc
b=RTc8Pc
and the van der Waals equation of state:
P=RT¯¯¯V−b−a¯¯¯V2 ,
First, I'd convert
Pc=4947.6 kPa×0.01 bar1 kPa=49.476 bar
(compared to49 bar from the NIST database)
Tc=32.1∘C=305.25 K
Then:
a=(27)(0.083145 L⋅bar/mol⋅K)2(305.25 K)264(49.476 bar)
= 5.493 L2⋅bar/mol2
(compared to5.570 L2⋅bar/mol2 from here.)
The
b=(0.083145 L⋅bar/mol⋅K)(305.25 K)8(49.476 bar)
= 0.06412 L/mol
(compared to0.06499 L/mol here.)
Using this information, we can find the pressure of
5 g ethane×1 mol ethane30.0694 g=0.1663 mols
giving us a molar volume of (exactly):
¯¯¯V=1 L0.1663 mols=6.01388 L/mol
Finally, we can solve for the pressure to get:
P=(0.083145 L⋅bar/mol⋅K)(288.15 K)6.01388 L/mol−0.06412 L/mol−5.493 L2⋅bar/mol2(6.01388 L/mol)2
= 3.875 bar
Using the ideal gas law, we can check our answer:
P=nRTV=RT¯¯¯V=(0.083145 L⋅bar/mol⋅K)(288.15 K)6.01388 L/mol
= 3.984 bar
which is indeed close! The real gas actually takes up less volume (its attractive forces dominate), meaning that there is more empty space in the container than we expected, and thus, a slightly smaller pressure is exerted.
Note: some textbooks ask you to go through and derive
