# Find Van der Waals constants and what is the pressure for C2H6?

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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* #P_c# *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* *should be closer to*

*- Truong-Son*

Using the result derived in full here that:

and the van der Waals equation of state:

Then:

giving us a molar volume of (exactly):

Finally, we can solve for the pressure to get:

Using the ideal gas law, we can check our answer:

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.

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The Van der Waals equation for C2H6 is (P + a(n^2/V^2))(V - nb) = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. The Van der Waals constants for C2H6 are a = 2.250 atm L^2/mol^2, b = 0.0427 L/mol.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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