gas law tutorial boyle

Question

alculate the decrease in temperature when 6.00 L at 20.0 °C is compressed to 4.00 L

Answer 1

The question asks about Boyles law and then asks to calculate the change in temperature. This makes no sense because....

Boyles law is used when the temperature is constant. It can be written simply as...

P1V1 = P2V2

P1 = initial pressure

V1 = initial volume = 6.00 L

P2 = final pressure

V2 = final volume

(P1)(6.00) = (P2)(4.00)

P2 = 6P1 / 4

P2 = 1.5 P1 so the final pressure will be 1.5 x the initial pressure.

NOW...if you want to find the final temperature when the pressure is constant, then we use Charles Law...

V1/T1 = V2/T2

V1 = initial volume = 6.00 L

T1 = initial temperature = 20 + 273 = 293

V2 = final volume = 4.00 L

T2 = final temperature = ?

6.00 / 293 = 4.00 / T2

T2 = (293)(4.00)/6.00

T2 = 195K

T2 = 195 - 273 = -78K

Answer 2

ideal gas law is PV = nRT

rearranged we have PV/nT = R

therefore for any gas P₁V₁/n₁T₁ = R

we can set two conditions equal to each other as the following equation:

P₁V₁/n₁T₁ = P₂V₂/n₂T₂

In your question only pressure and temperature are needed so pressure and n can be removed as they are the same:

V₁/T₁ = V₂/T₂

plug in numbers but use kelvin

(6L)/(20+273 K ) = 4/(273+x)

The answer from here is trivial.