Question

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

Answer #1

Given,

Number of moles of He = **2.00 mol**

Volume of container(V) = **1.00 L**

Temperature(T) = **300.0 K**

**Part 1)**

We know, the ideal gas equation,

PV = nRT

Here, R =0.08206 L atm /mol.K

Rearranging the equation,

P = nRT /V

Substituting the known values,

P = (2.00 mol x 0.08206 L.atm /mol.K x 300.0 K) / 1.00 L

**P = 49.2 atm**

**Thus, ideal gas law pressure is 49.2 atm**

**Part 2)**

We know, Van der Waals equation,

**(P - n ^{2}a/V^{2}) (V- nb) =
nRT**

**Here,**

a (for He) = 0.0346 L^{2} atm / mol^{2}

b (for He) =0.0238 L/mol

Rearranging the van der Waals equation,

P = (nRT /V-nb) - (n^{2}a / V^{2})

Substituting the known values,

P = [( 2.00 mol x 0.08206 L. atm/mol.K x 300.0 K) / (1.00 L -
2.00 mol x 0.0238 L/mol)] - [(2.00 mol)^{2} x 0.0346
L^{2} atm / mol^{2}/(1.00 L)^{2}]

P = 51.6968 - 0.1384

**P = 51.6 atm**

**Thus, Van der Waals pressure is 51.6 atm**

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