If a balloon filled with air has a volume of 15.0 L at 25.0°C, what is its volume (in L) at -100.0°C?

Question

 

If a balloon filled with air has a volume of 15.0 L at 25.0°C, what is its volume (in L) at -100.0°C?

 

Answer 1

At constant pressure, we are dealing with Charles' Law:

V1/T1 = V2/T2

V1 = 15.0 L; T1 = 25 + 273 = 298K; V2 = ?; T2 = -100 + 273 = 173K

15.0/298 = V2/173

298V2 = (173)(15)

V2 = (173)(15) / 298

V2 = 8.71 L

Answer 2

The ideal gas law PV = nRT should be easy enough to remember, so let's use it here.

PV₁ = nRT₁ and PV₂ = nRT₂

so (PV₁)/(PV₂) = (nRT₁)/(nRT₂) which means that V₁/V₂ = T₁/T₂ and

V₂ = V₁(T₂/T₁)

Plug in

V₁ = 15.0 L

T₁ = 25.0°C = (25.0 + 273)°K = 298°K

T₂ = –100.0°C = (–100.0 + 273)°K = 173°K

So V₂ = (15)(173/298) =

I'll leave the calculator work to you...

Hope this helps. Add a comment if you have any questions.