Ask Question
26 September, 21:27

A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 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.

+4
Answers (1)
  1. 26 September, 21:34
    0
    The pressure in atm calculated using the van der Waals' equation, is 337.2atm

    Explanation:

    This is the Van der Waals equation for real gases:

    (P + a/v²) (v-b) = R. T

    where P is pressure

    v is Volume/mol

    R is the gas constant and T, T° in K

    a y b are constant for each gas, so those values are data, from the statement.

    [P + 1.345 L²atm/mol² / (0.7564L/10.21mol) ² ] (0.7564L/10.21mol - 3.219*10-2 L/mol) = 0.082 L. atm/mol. K. 296.9K

    [P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219*10-2 L/mol) = 0.082 L. atm/mol. K. 296.9K

    (P + 245.05 atm) (0.04181L/mol) = 0.082 L. atm/mol. K. 296.9K

    (P + 245.05 atm) (0.04181L/mol) = 24.34 L. atm/mol

    0.04181L/mol. P + 10.24 L. atm/mol = 24.34 L. atm/mol

    0.04181L/mol. P = 24.34 L. atm/mol - 10.24 L. atm/mol

    0.04181L/mol. P = 14.1 L. atm/mol

    P = 14.1 L. atm/mol / 0.04181 mol/L

    P = 337.2 atm
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using the van der ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers