Question

Assuming non-ideal behavior, a 2.0 mol sample of CO₂ in a 7.30 L container at 200.0 K has a pressure of 4.50 atm. If a = 3.59 L²･atm/mol² and b = 0.0427 L/mol for CO₂, according to the van der Waals equation what is the difference in pressure (in atm) between ideal and nonideal conditions for CO₂?

Answer 1

The difference between the pressure in ideal and nonideal conditions for the CO₂ gas is 0.22 atm.

The given parameters;

• number of moles of the gas, n =  2 mol
• volume of the gas, V = 7.3 L
• temperature of the gas, T = 200 K
• pressure of the ideal gas, P₁ = 4.5 atm
• constant, a = 3.59 L²atm/mol²
• constant, b = 0.0427 L/mol

The van der Waals equation is calculated as follows;

$(P + \frac{an^2}{V^2} )(V-nb) = nRT\\\\P = \frac{nRT}{V-nb} - \frac{an^2}{V^2} \\\\P = \frac{2\times 0.082057 \times 200}{7.3 \ -\ 2\times 0.0427} - \frac{3.59\times 2^2}{7.3^2} \\\\P = 4.281 \ atm$

The difference between the pressure in ideal and nonideal conditions for the CO₂ gas is calculated as;

$\Delta P = P_{ideal} - P_{nonideal}\\\\\Delta P = 4.5 \ atm \ - \ 4.281 \ atm\\\\\Delta P = 0.22 \ atm$

Thus, the difference between the pressure in ideal and nonideal conditions for the CO₂ gas is 0.22 atm.

Learn more here:https://brainly.com/question/17355868