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why is the difference between the result for an ideal gas and that calculated using the van der waals equation greater when the gas is confined to 5.00 L compared to 22.4 L

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The question is incomplete, the complete question is;

In Sample Exercise 10.16, we found that one mole of Cl2 confined to 22.41 L at 0 °C deviated slightly from ideal behavior. Calculate the pressure exerted by 1.00 mol Cl2 confined to a smaller volume, 5.00 L, at 25 °C. (a) First use the ideal-gas equation and (b) then use the van der Waals equation for your calculation. (Values for the van der Waals constants are given in Table 10.3.) (c) Why is the difference between the result for an ideal gas and that calculated using the van der Waals equation greater when the gas is confined to 5.00 L compared to 22.4 L?

Answer:

See explanation

Explanation:

Since we have

For 5.00 L using Ideal gas law;

PV= nRT

P = nRT/V

P = 1 * 0.082 * 298 K/5.00

P= 4.887 atm

Using Van der walls equation;

[P + an^2/v^2] *  [V - nb] = nRT

P = (nRT/ [V - nb]) - (an^2/v^2)

P= 1 * 0.082 * 298/5 - (1 * 0.0562) - (6.49(1^2)/5^2)

P = (24.436/4.9438) - 0.2596

P= 4.683 atm

For 22.4 L

P = nRT/V

P = 1 * 0.082 * 298 K/22.4

P= 1.0909 atm

Using Van der Walls equation;

[P + an^2/v^2] *  [V - nb] = nRT

P = (nRT/ [V - nb]) - (an^2/v^2)

P= 1 * 0.082 * 298/22.4 - (1 * 0.0562) - (6.49(1^2)/22.4^2)

P= (24.436/ 22.3438) - 0.0129

P= 1.081 atm

According to Boyle's law, pressure is inversely proportional to volume. Increase in volume leads to decrease in pressure. Therefore the pressure calculated both by the ideal gas equation and Van der Waals equation at 5.00 L is greater than that calculated using both methods at 22.4 L

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