Question

A gas occupies 0.67 L at 350 K. What temperature is required to reduce
the volume to 0.21 L *

Answer 1

109.7K

Step-by-step explanation:

Using Charles law equation as follows;

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

V1 = 0.67L

V2 = 0.21L

T1 = 350K

T2 = ?

Using the formula above;

0.67/350 = 0.21/T2

0.001914 = 0.21/T2

0.001914T2 = 0.21

T2 = 0.21/0.001914

T2 = 109.7K