Respuesta :
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:
[tex]PV=nRT[/tex]
[tex]P=\frac{1.00 mol\times 0.0821 L atm/mol K\times 300.15 K}{0.500 L}=49.28 atm[/tex]
Vander wall's of equation of gases:
The real pressure of the gas= [tex]p_v[/tex]
For argon:
[tex]a=1.345 L^2 atm/mol^2[/tex]
b=0.03219 L/mol.
[tex](p_v+(\frac{an^2}{V^2})(V-nb)=nRT[/tex]
[tex](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[/tex]
[tex]p_v = 47.29 atm[/tex]
Difference :[tex] p - p_v= 49.28 atm - 47.29 atm = 1.99 atm\approx 2.0 atm[/tex]
2.0 atm is the difference between the ideal pressure and the real pressure.
