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A hot air balloon is filled with 1.19 x 106 L of an ideal gas on a cool morning (11 °C). The air is heated to 113 °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.

Answer :

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Answer:

using the Charles law

v1/t1= v2/t2

Explanation:

1.19* 10^6/11+273=v2/113+273

1.19*10^6/284=v2/386

v2= 1.19*10^6/284/386

v2= 10.89L

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