A child has a toy balloon with a volume of 1.80 L. The temperature of the balloon when it was filled was 20.ºC and the pressure was 1.00 atm. If the child were to let go of the balloon and it rose 3 km into the sky where the pressure is 0.667 atm, the balloon’s volume becomes 1.55 L. What is the new temperature, in ºC?

Explanation:

Step one:

given data

V1=1.80L

T1=20°C

P1=1atm

V2=1.55L

T2=?    new temperature

P2=0.667atm

Step two:

The combined gas equation is given as

P1V1/T1=P2V2/T2

make T2 which is the new temperature subject of formula we have

T2=(P2*V2*T1)/P1

T2=1.8*1.55*20/0.667

T2=55.8/0.667

T2=83.65°C

Therefore the new temperature is 83.65°C when the balloon rises 3km

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