Answer to Question #261870 in Inorganic Chemistry for Smiles

Question

Using the appropriate gas laws, calculate the quantities denoted by the blanks below.

Conditions before Conditions after

n₁, mol V₁, L P₁, atm T₁, K n₂, mol V₂, L P₂, atm T₂, K

1.25 13.2 1.27 298 1.25 (a)____ 0.987 298

1.25 13.2 1.27 298 1.25 (b)____ 1.27 407

1.25 13.2 1.27 298 4.00 (c)____ 1.27 298

1.25 13.2 1.27 298 1.36 (d)____ 1.27 372

1.25 13.2 1.27 298 1.25 30.5 (e)____ 298

Answer 1

a) We need V2

n and T are constants hence we use Boyles Law

P1V1 = P2V2

So V2$=\dfrac{1.27 x 13.2 }{0.987} = 16.98L$

b) We also need V2

n and P are constant hence we use Charles Law

$V2 = \dfrac{V1T2}{T1} = \dfrac{13.2 x 407}{298} = 18.03L$

c) Here since n is varied we can use Ideal Gas Law PV = nRT to calculate for V2 in conditions after; In this case, R = 0.082 LatmK^-1mol^-1

Thus V $=\dfrac{nRT}{P} = \dfrac{4.00 x 0.082 x 298}{1.27} = 76.96L$

d) Like in (c) above we use PV = nRT

Thus, V $=\dfrac{1.36x0.082x372}{1.06}$ $=39.14L$

e) We are solving for P2 hence we can use P1V1=P2V2 (Boyles Law) since n and T are constants.

Thus, P2$=\dfrac{P1V1}{V2}= \dfrac{1.27 x 13.2}{30.5}=0.55 atm$