A hot air balloon is filled with 1.35 × 106 l of an ideal gas on a cool morning (11 °c). the air is heated to 127 °c. what is the volume of the air in the balloon after it is heated? assume that none of the gas escapes from the balloon.

Respuesta :

Louli
Answer:
Volume = 1.901 * 10^6 liters

Explanation:
We will assume that the pressure of the gas is kept constant.
Charles's law states that: "At constant pressure, the volume of a given mass of gas is directly related to its temperature in kelvin."
This means that:
V1 / T1 = V2 / T2
where:
V1 is the initial volume = 1.35 × 10^6 liters
T1 is the initial temperature = 11 + 273 = 284 degrees kelvin
V2 is the final volume that we want to find
T2 is the final temperature = 127 + 273 = 400 degrees kelvin

Substitute with the givens in the above equation to get the final volume as follows:

V1 / T1 = V2 / T2
(
1.35 × 10^6) / (284) = V2 / (400)
V2 = 400 * [
(1.35 × 10^6) / (284)]
V2 = 1.901 * 10^6 liters

Hope this helps :)

From the Charle's law V1/T1=V2/T2
V1= 1.35 ×10^6 L
T1= 11 +273 = 284 K
T2 = 127+273 = 400 k 
Therefore, V2 = V1T2/T1
                       = (1.35 ×10^6 × 400)/284

                       = 1.901 ×10^6 Liters