F 1.00 \(\rm mol\) of argon is placed in a 0.500-\(\rm L\) container at 21.0 \({\rm \, ^{\circ}C}\) , 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, \(a = 1.345 \;\rm (L^2\cdot atm)/mol^2\) and \(b = 0.03219\; \rm L/mol\).

2.05 atm

Explanation:

From ideal gas equation;

PV=nRT

Making P the subject of the formula;

P= nRT/V

Substituting values;

P= 1 × 0.0821 × 294/ 0.5

P= 48.27 atm

Real gas pressure;

(P + an^2/v^2) . (V - nb) = nRT

(P + 1.345 × 1^2/0.5^2) . (0.5 - 1 × 0.03219) = 1 × 0.0821 × 294

(P + 5.38) . (0.46781) = 24.1374

(P + 5.38) = 24.1374/0.46781

(P + 5.38) = 51.5966

P= 51.5966 - 5.38

P= 46.22 atm

Difference = Pideal - Preal

Difference = 48.27 - 46.22

Difference = 2.05 atm

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

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