# Charles' Law/Boyle's Law Gas Problems?

Can anyone help me with these? I got my answer but I don't know if I did it right

An aerosol can with a volume of 325mL contains hair spray at 445kPa and 12 degrees C. What volume would the gas occupy if it was allowed to escape at 101kPa and at 21 degrees C?

Update:

I forgot the other one:

For the combustion of pentane:

a) Complete the balanced equation.

b) If 250L of pentane at STP was combusted, what volume of carbon dioxide would be produced at 25 degrees C and 99.9kPa?

Relevance

(V1*P1)/T1 = (V2 *P2)/T2

(325ml * 445kPa)/(273 + 12) = (V * 101kPa)/(273 + 21)

507.46 = (V * 101)/294

101V = 149192.11

V = 1477.15ml

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here is to your combustion of pentane:

a. C5H12 + 8 02---------->5 CO2 + 6 H20

b. (V1*P1)/T1 = (V2 *P2)/T2

(250L*101.3kPa)/273 = (V * 99.9kPa)/(273 + 25)

92.77 = (99.9V)/298

27,644.14 = 99.9V

V = 276.72L

pV/T is constant, so determine the constant value first, then reverse the equation and determine V (V = (T/p) × constant)

(445 × 325)/(12 + 273) = 144625/285 kPa mL K^-1.

Now plug in the new values.

V = [(273 + 21)/101] × (144625/285) = 1477.150 mL

a) C5H12 + 8O2 -> 5CO2 + 6H2O

b) Use pV/T = constant again. I'm not sure about the STP definition you use, apparently the IUPAC definition has changed since I needed it.

• Erika
Lv 4
4 years ago

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• Anonymous