A hot air balloon holds 7.40 x 107 liters of gas. In order for the hot air balloon to rise into the sky, it must be heated to 120°C. Once the balloon is up in the air, the air needs to be cooled in order for the balloon to land. If the temperature of the balloon is decreased to 40.0°C, what will the new volume of the gas be? Remember to pay close attention to the units of temperature before beginning your calculations.

2.467×10^7 litres

Explanation:

Using the formula V1/T1 = V2/T2

V1 and V2 are the initial and final volume respectively

T1 and T2 are their respective temperatures

Given

V1 = 7.4×10^7litres

T1 = 120°C

T2 = 40°C

Required

V2

From the formula

V2 = V1T2/T1

V2 = 7.4×10^7(40)/120

V2 = (7.4×10^7)/3

V2 = 2.467×10^7 litres

Hence the new volume of the gas will be 2.467×10^7 litres

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat