Question

If 1.00 mol of argon is placed in a 0.500-L container at 27.0 degree C , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? For argon, a=1.345(L2⋅atm)/mol2 and b=0.03219L/mol. Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

2.0 atm is the difference between the ideal pressure and  the real pressure.

Explanation:

If 1.00 mole of argon is placed in a 0.500-L container at 27.0 °C

Moles of argon = n = 1.00 mol

Volume of the container,V  = 0.500 L

Ideal pressure of the gas = P

Temperature of the gas,T = 27 °C = 300.15 K[/tex]

Using ideal gas equation:

$PV=nRT$

$P=\frac{1.00 mol\times 0.0821 L atm/mol K\times 300.15 K}{0.500 L}=49.28 atm$

Vander wall's of equation of gases:

The real pressure of the gas= $p_v$

For argon:

$a=1.345 L^2 atm/mol^2$

b=0.03219 L/mol.

$(p_v+(\frac{an^2}{V^2})(V-nb)=nRT$

$(p_v+(\frac{(1.345 L^2 atm/mol^2)\times (1.00 mol)^2}{(0.500 L)^2})(0.500 L-1.00 mol\times 0.03219L/mol)=1.00 mol\times 0.0821 L atm/mol K\times 300.15 K$

$p_v = 47.29 atm$

Difference :$p - p_v= 49.28 atm - 47.29 atm = 1.99 atm\approx 2.0 atm$

2.0 atm is the difference between the ideal pressure and  the real pressure.