14.0 moles of gas are in a 3.00 LL tank at 23.3 ∘C∘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/mol2a=2.300L2⋅atm/mol 2 and b=0.0430 L/molb=0.0430 L/mol. Express your answer with the appropriate units.

The difference in pressure = 21.59 atm

Explanation:

Step 1: Data given

Moles of gas = 14.0 moles

Volume = 3.00 L

Temperature = 23.3 °C

The van der Waals constants for methane are a=2.300L²*atm/mol² and b=0.0430 L/mol

Step 2: Calculate pressure

p*V = n*R*T

⇒ p = the pressure = TO BE DETERMINED

⇒ V = the volume of the tank = 3.00 L

⇒ R = the gas constant = 0.08206 L*atm/mol*K

⇒ T = the temperature = 23.3 °C = 296.45 K

⇒ n = the number of moles gas = 14.0 moles

p = (n*R*T)/V

p = (14.0 *0.08206 * 296.45)/3.00

p = 113.52 atm

Step 3: Calculate the pressure via the van der waals equation

The Van der waals equation is (p + n²a/V²)*(V-nb) = nRT 

⇒p = To be determined

⇒ n =the number of moles = 14.0 moles

⇒a=2.300 L²*atm/mol²

⇒b=0.0430 L/mol

(p + 14.0²*2.300 / 3.00²) * (3.00 - 14.0*0.0430) = 14.0*0.08206*296.45

(p + 14.0²*2.300 / 3.00²) * (3.00 - 14.0*0.0430) = 340.57

(p + 14.0²*2.300 / 3.00²) * 2.398 = 340.57

(p + 14.0²*2.300 / 3.00²) = 142.02

(p + 50.09) = 142.02

p = 91.93 atm

The difference in pressure = 113.52 atm - 91.93 atm = 21.59 atm