Question

An air bubble has a volume of 0.510 L at 19 ∘C. What is the volume, in liters, at 422 K , if n and P do not change? What is the volume, in liters, at -13 ∘C , if n and P do not change? What is the volume, in liters, at 574 K , if n and P do not change?

Answer 1

According to the explanation given in the previous session about Charles's law, now we have different temperatures:

We have in situation 1:

V1 = 0.510 L

T1 = 292 K

V2 = ?

T2 = 422 K

In situation 2:

V1 = 0.510 L

T1 = 292 K

V2 = ?

T2 = 260 K

In situation 3:

V1 = 0.510 L

T1 = 292 K

V2 = ?

T2 = 574 K

Now solving each situation:

1.

0.510/292 = V2/422

0.00175 = V2/422

V2 = 0.738 Liters

2.

0.510/292 = V2/260

0.00175 = V2/260

V2 = 0.455 Liters

3.

0.510/292 = V2/574

0.00175 = V2/574

V2 = 1.00 Liters