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According to the ideal gas law, a 9.843 mol sample of argon gas in a 0.8425 L container at 502.0 K should exert a pressure of 481.3 atm

Hint: % difference = 100×(P ideal - Pvan der Waals) / P ideal

According to the ideal gas law, a 9.843 mol sample of argon gas in a 0.8425 L container at 502.0 K should exert a pressure of 481.3 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For Ar gas, a =1.345L2 atm/mol2 and b = 3.219×10-2 L/mol.


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Answer #1


van der Waals' equation


(P+an^2/V^2)(V-nb) = nRT

(p+(1.345*9.843^2/0.8425^2))(0.8425-9.843*3.219*10^(-2)) = (9.843*0.0821*502.0)

p = 588.16 atm

according to ideal gas equation

PV = nRT

(P*0.8425) = 9.843*0.0821*502

P = 481.51 atm

let me take ideal pressure = 481.51 atm = 100%

vandewaals pressure = 588.16 atm

difference = 588.16-481.51 = 106.65

percent difference = 106.65/481.51*100 = 22.15% extra

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