Question

A gas sample is held at constant pressure. The gas occupies 3.62 L of volume when the temperature is 21.6°C. Determine the temperature at which the volume of the gas is 3.45 L.

Answer 1

Answer:

281 K

Explanation:

Charles's Law. V1/T1 = V2/T2.

The temperature must be in K = 21.6°C + 273 = 294.6K.

V1T2 = V2T1.

3.62L x T2 = 3.45L x 294.6K

T2 = (3.45 x 294.6) / 3.62 = 1016.4 / 3.62 = B): 281K.

(By direct proportion of volume change: (3.45L / 3.62L) x 294.6K = 281K).