If 1.00 mol of argon is placed in a 0.500-L container at 22.0 ∘C , what is the difference between the ideal pressure (as predict ed 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/mo
1 answer:
Answer:
The answer to your question is ΔP = 1.57 atm
Explanation:
Data
moles = 1
volume = 0.5 L
Temperature = 22°C or 295°K
Pressure = ?
a) Ideal Gas
PV = nRT R = 0.082 atm L / mol°K
-Solve for P
P = nRT/V
P = (1)(0.082)(295)/ 0.5
P = 23.944/ 0.5
P = 47.9 atm
b) Van der Waals
(P + a/v²)(v - b) = RT
- Substitution
(P + 1.345/0.5²)(0.5 - 0.03219) = (0.082)(295)
- Simplification
(P + 5.38)(0.46781) = 24.19
P + 5.38 = 24.19/0.46781
P + 5.38 = 51.71
-Solve for P
P = 51.71 - 5.38
P = 46.33 atm
c) The difference between both Pressures is
ΔP = 47.9 - 46.33
= 1.57 atm
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