What is the volume of a gas a t320 K, if the gas occupies 50.0 mL at a temperature of 273 K?

58.61mL

Explanation:

V1 (initial volume) = ?

T1 (initial temperature) = 320 K

V2 (final volume) = 50mL

T2 (final temperature) = 273 K

Using the Charles' law equation V1/T1 = V2/T2

The initial volume of the gas can obtained as follow:

V1/320 = 50/273

Cross multiply to express in linear form

V1 x 273 = 320 x 50

Divide both side by 273

V1 = (320 x 50) / 273

V1 = 58.61mL

Therefore, the initial volume of the gas is 58.61mL