Thank you!!! Calculate the pressures exerted at 250 K and 500 K by = 1.42 L/mol

Question

Thank you!!!

Calculate the pressures exerted at 250 K and 500 K by = 1.42 L/mol of a gas that obeys the van der Waals equation of state with parameters a = 1.355 bar dm6 mol -2 and b = 0.0320 dm3 mol-1. Use these results to deduce whether the attractive or repulsive parts of the intermolecular interaction potential are dominant at these two temperatures? At sufficiently high temperatures the van der Waals equation can be reasonably approximated by P RT/( - b), suggesting that the attractive part of the intermolecular interaction potential can be neglected. Justify this approximation qualitatively by referring to the potential energy diagram we discussed in class (shown also in EDR Ch. 1, Fig. 7).

Explanation / Answer

a) P
T =250 K , 500K

R= 0.0821
V' =1.42 L/mol
a = 1.355
b= 0.032
(P+(a/v'^2)) (v'-b) =RT
P = RT/(v'-b) - (a/v'^2)
T =250 K
P = 0.059*T - 0.67

P =14.11 bar

Z = PV /nRT = 0.976

T=500K

P =28.9 bar

Z = PV /nRT = 0.99

If Z < 1

* Vreal < Videal

* The attractive forces are more significant than the repulsive forces.

* The gas can be liquefied easily.

* Usually the Z < 1 for gases like NH3, CO2, SO2.

b)

At high temperatures: In this case, V is very large and attractions are negligible. Hence

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