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Firlakuza [10]
3 years ago
10

Help please. 15.0 moles of gas are in a 3.00L tank at 23.4∘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..
Chemistry
1 answer:
CaHeK987 [17]3 years ago
3 0
Using PV=nRT or the ideal gas equation, we substitute n= 15.0 moles of gas, V= 3.00L, R equal to 0.0821 L atm/ mol K and T= 296.55 K and get P equal to 121.73 atm. The Van der waals equation is (P + n^2a/V^2)*(V-nb) = nRT.  Substituting  a=2.300L2⋅atm/mol2 and b=0.0430 L/mol, P is equal to 97.57 atm. The difference is <span>121.73 atm- 97.57 atm equal to 24.16 atm.</span>
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