Question

#25

A **10.86** mol sample of **krypton**
gas is maintained in a **0.7529** L container at
**296.3** K. What is the pressure in atm calculated
using the van der Waals' equation for **Kr** gas under
these conditions? For **Kr**, a =
**2.318**L^{2}atm/mol^{2} and b =
**3.978×10 ^{-2}** L/mol.

____atm

According to the ideal gas law, a **1.077** mol
sample of **methane** gas in a **1.670**
L container at **265.4** K should exert a pressure of
**14.05** atm. By what percent does the pressure
calculated using the van der Waals' equation differ from the ideal
pressure? For **CH _{4}** gas, a =

______%

According to the ideal gas law, a **9.308** mol
sample of **nitrogen** gas in a
**0.8235** L container at **502.7** K
should exert a pressure of **466.3** atm. By what
percent does the pressure calculated using the van der Waals'
equation differ from the ideal pressure? For
**N _{2}** gas, a =

%

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

.26

A **9.65×10 ^{-2}** mol sample of an unknown
gas contained in a

A **7.01×10 ^{-2}** mol sample of

_______g/L

What is the molar volume of **Xe** gas under the
conditions of temperature and pressure where its density is
**2.84** g/L?

_____ L/mol

Answer #1

**25:**

**krypton gas
:**

We know the van der Waals' gas equation = [P + a(n/V)^2] [V/n-b] = RT

From the given data a = **2.318**
L^{2}atm/mol^{2} and b =
**3.978×10 ^{-2}** L/mol

Volume V= 0.7529 L, T = 296.3 K, n = 10.86 moles

if plugin all the above values in the equation

[P+ 2.318 L^{2}atm/mol^{2} (10.86 moles/0.7529
L)^2] [0.7529 L/10.86 moles -
**3.978×10 ^{-2}** L/mol] = 0.0821 L.atm
/mole.K x 296.3 K

[P + 482.2791 atm] [2.9547 x 10-2 L/mol] = 24.32623 L.atm/mole

[P + 482.2791 atm] = 8.23306 x 10^{2} atm

P = **341.0269** atm

**Methane gas
:**

Similarly if we calculated for CH4 the pressure calculated using the van der Waals' equation

[P+ 2.253 L^{2}atm/mol^{2} (1.077 moles/1.670
L)^2] [1.670 L/1.077 moles - **4.278×10 ^{-2}**
L/mol] = 0.0821 L.atm /mole.K x 265.4 K

P = 13.514 atm

Ideal pressure Pi = 14.05 atm

% difference = 100×(P _{ideal} - P_{van der
Waals}) / P _{ideal} = 100 (14.05-13.514)/14.05 =
**3.815 %**

**Nitrogen gas
:**

[P+ 1.390 L^{2}atm/mol^{2} (9.308 moles/0.8235
L)^2] [0.8235 L/9.308 moles -
**3.910×10 ^{-2}** L/mol] = 0.0821 L.atm
/mole.K x 502.7 K

177.5825 x 0.04937 =

P = 658.3840 atm

Ideal pressure Pi = 466.3 atm

% difference = 100×(P _{ideal} - P_{van der
Waals}) / P _{ideal} = 100 (14.05-13.514)/14.05 = -
**41.19** %

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

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
L2atm/mol2 and b =
3.219×10-2 L/mol.
??? %
Hint: % difference = 100 × (P ideal - Pvan
der Waals) / P ideal

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.

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 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

15.0 moles of gas are in a 4.00 L tank at 21.2 ∘C . Calculate
the difference in pressure between methane and an ideal gas under
these conditions. The van der Waals constants for methane are
a=2.300L2⋅atm/mol2 and b=0.0430 L/mol.

12.0 moles of gas are in a 4.00 L tank at 24.4 ∘C . Calculate
the difference in pressure between methane and an ideal gas under
these conditions. The van der Waals constants for methane are
a=2.300L2⋅atm/mol2 and b=0.0430 L/mol.

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