Question

A hot air balloon is filled with 1.27 x 10^6 L of an ideal gas on a cool morning (11C). The air is heated to 131C What is the volume of the balloon after it is heated? Assume that none of the gas escaped from the balloon.

Answer #1

As temperature increases , volume also increases , Therefore, According to Charls law, volume & temperature related as,

V_{1} / T_{1} = V_{2} /
T_{2}

Here, V_{1} = Initial volume = 1.27 x
10^{6} L

V_{2} = Final volume = ? = x

T_{1} = Initial temp = 11^{o}c = 11 + 273 = 284
K

T2 = filan temp = 131^{o}c = 131 + 273 = 404 K

1.27 x 10^{6} / 284 = X / 404

By doing cross multiplication,

1.27 x 10^{6} x 404 = X x 284

513.08 x 10^{6} = X x 284

Therefore X = 513.08 x 10^{6} / 284

= 1.806 x 10^{6} L of Volume.

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