A toy balloon has an internal pressure of 1.05 atm and a volume of 5.0 L. If the temperature where the balloon is released is 20° C, what will happen to the volume when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is –15° C

Then volume of the balloon will increase to 7.11 L

Explanation:

Step 1:

Data obtained from the question. This includes:

Initial pressure (P1) = 1.05 atm

Initial volume (V1) = 5.0 L

Initial temperature (T1) = 20°C

Final pressure (P2) = 0.65 atm

Final temperature (T2) = –15°C

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below

K = °C + 273

T1 = 20°C + 273 = 293K

T2 = –15°C + 273 = 258K

Step 3:

Determination of the final volume.

Applying the general gas equation P1V1/T1 = P2V2/T2, the final volume of the balloon can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 5/293 = 0.65 x V2/258

Cross multiply to express in linear form

293 x 0.65 x V2 = 1.05 x 5 x 258

Divide both side by 293 x 0.65

V2 = (1.05 x 5 x 258) / (293 x 0.65)

V2 = 7.11 L

Therefore, the volume of the balloon will increase to 7.11 L when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is –15° C

the device is called “ rocket engine”

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