Question

According to the ideal gas law, a **10.59** mol
sample of **argon** gas in a **0.8229** L
container at **495.4** K should exert a pressure of
**523.2** atm. By what percent does the pressure
calculated using the van der Waals' equation differ from the ideal
pressure? For **Ar** gas, a = **1.345**
L^{2}atm/mol^{2} and b =
**3.219×10 ^{-2}** L/mol.

??? %

*Hint:* % difference = 100 × (P _{ideal} - P_{van
der Waals}) / P _{ideal}

Answer #1

According to the ideal gas law, a 9.939 mol
sample of argon gas in a 0.8276 L
container at 500.6 K should exert a pressure of
493.3 atm. What is the percent difference between
the pressure calculated using the van der Waals' equation and the
ideal pressure? For Ar gas, a =
1.345 L2atm/mol2 and b =
3.219×10-2 L/mol.

According to the ideal gas law, a 10.08 mol
sample of krypton gas in a 0.8488
L container at 496.7 K should exert a pressure of
484.0 atm. By what percent does the pressure
calculated using the van der Waals' equation differ from the ideal
pressure? For Kr gas, a = 2.318
L2atm/mol2 and b =
3.978×10-2 L/mol.
----------------%
Hint: % difference = 100 × (P ideal - Pvan
der Waals) / P ideal

According to the ideal gas law, a 10.38 mol sample of krypton
gas in a 0.8420 L container at 502.0 K should exert a pressure of
507.8 atm. What is the percent difference between the pressure
calculated using the van der Waals' equation and the ideal
pressure? For Kr gas, a = 2.318 L2atm/mol2 and b = 3.978×10-2
L/mol.

A 10.09 mol sample of argon
gas is maintained in a 0.8203 L container at
301.8 K. What is the pressure in atm calculated
using the van der Waals' equation for Ar gas under
these conditions? For Ar, a =
1.345 L2atm/mol2 and b =
3.219×10-2 L/mol.
_______atm

If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘C
, what is the difference between the ideal pressure (as predicted
by the ideal gas law) and the real pressure (as predicted by the
van der Waals equation)? For argon, a=1.345(L2⋅atm)/mol2 and
b=0.03219L/mol.

If 1.00 mol of argon is placed in a 0.500-L container at 29.0 ∘C
, what is the difference between the ideal pressure (as predicted
by the ideal gas law) and the real pressure (as predicted by the
van der Waals equation)?
For argon, a=1.345(L2⋅atm)/mol2 and
b=0.03219L/mol.

Use the van der Waals equation and the ideal gas equation to
calculate the pressure for 2.00 mol He gas in a 1.00 L container at
300.0 K. 1st attempt
Part 1 (5 points)
Ideal gas law pressure_____ atm
Part 2 (5 points)
Van der Waals pressure_____ atm

1. Part A
A 3.00-L flask is filled with gaseous ammonia, NH3. The gas
pressure measured at 24.0 ∘C is 1.75 atm . Assuming ideal gas
behavior, how many grams of ammonia are in the flask?
Express your answer to three significant figures and include the
appropriate units.
2. Part B
If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘C
, what is the difference between the ideal pressure (as predicted
by the ideal gas...

Use the ideal gas equation and the Van der Waals equation to
calculate the pressure exerted by 1.00 mole of Argon at a volume of
1.31 L at 426 K. The van der Waals parameters a and
b for Argon are 1.355 bar*dm6*mol-2
and 0.0320 dm3*mol-1, respectively. Is the
attractive or repulsive portion of the potential dominant under
these conditions?

Use the van der Waals equation of state to calculate the
pressure of 2.90 mol of CH4 at 457 K in a 4.50 L vessel. Van der
Waals constants can be found here.
P= ________ atm
Use the ideal gas equation to calculate the pressure under the
same conditions.
P= ______ atm

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