I need help solving equation before midnight! Help, please and thank you very much!

Question

A sample of 568 mL of chlorine gas originally at 25° C was cooled to -25° C. What is the

volume of the cooled gas?

Answer 1

Here you would use Charles' Law: V₁ V₂

---- = ----

T₁ T₂

Here:

V₁= 568 mL V₂ = unknown (or what you're trying to find)

T₁ = 25°C T₂ = -25°C

Both T₁ and T₂ must be converted to Kelvin. To do that, you just add 273.15 to each of the temperatures making T₁ and T₂:

T₁ = 298.15 K and T₂ = 248.15 K

Then just plug in everything to the formula and you get:

568 mL V₂ ----------->1.905 mL/K = (0.00403K)·V₂ Making V₂ = 472.7 mL or 473 mL

------------- = ---------------

298.15 K 248.15 K

This calculation makes sense because as

temperature decreases, so does volume