Use the van der Waals equation of state to calculate the pressure of 2.90 mol of

Question

Use the van der Waals equation of state to calculate the pressure of 2.90 mol of SO_2 at 491 K in a 3.70 L vessel. Van der Waals constants can be found in this table. P = Use the ideal gas equation to calculate the pressure under the same conditions. P = In a 16.3 L vessel, the pressure of 2.90 mol of SO_2 at 491 K is 7.17 atm when calculated using the ideal gas equation and 7.03 atm when calculated using the van der Waals equation of state. Why is the percent difference in the pressures calculated using the two different equations greater when the gas is in the 3.70 L vessel compared to the 16.3 L vessel? The attractive forces between molecules become less of a factor at the higher pressure in the 3.70 L vessel. The molecular volume is a smaller part of the total volume of the 3.70 L vessel. The attractive forces between molecules become a greater factor at the higher pressure in the 3.70 L vessel. The molecular volume is a larger part of the total volume of the 3.70 L vessel.

Explanation / Answer

for SO2, a= 6.803 L2 bar/mol2 = 6.803*0.9869 atmL2/mole2 =6.71 atmL2/mole2

b= 0.05636 L/mol

Vanderwaal equation (P+an2/V2)*(V-nb)= nRT

P =nRT/ (V-nb)- an2/V2

= 2.90*0.0821*491/(3.7 - 2.9*0.05636)- 6,71*2.9*2.9/ 3.7*3.7 =28.93 atm

2. From ideal gas law, P= nRT/V= 2.9*-0.0821*491/3.7 =31.59 atm

3. when pressure is higher, the attractive forces become significant ( C is correct)

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