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Using Charles's Law, if the balloon increased in size by 0.5 liters when the temperature increased from 280 Kelvin to 300 Kelvin, how many liters of air did the balloon contain originally at the Fair?

a) 11 liters
b) 10 liters
c) 5 liters
d) 8 liters

1 Answer

2 votes

Final answer:

Based on Charles's Law, the volume of a gas is directly proportional to temperature. Setting up a proportion reveals that the original volume should be 7 liters when considering the temperature increase and volume increase provided, which is not one of the provided options.

Step-by-step explanation:

According to Charles's Law, the volume of a gas is directly proportional to its temperature when pressure is held constant. In the scenario where the balloon increased in size by 0.5 liters upon a temperature increase from 280 Kelvin to 300 Kelvin, we can set up a proportion to find the original volume of the balloon at the fair.

Let V1 be the original volume of the balloon at 280 K and V2 be the increased volume at 300 K. Since V2 = V1 + 0.5 L and Charles's Law is given by V1/T1 = V2/T2, we have V1/280 K = (V1 + 0.5 L)/300 K.

By solving for V1, we get V1 × 300 K = V1 × 280 K + 140 L. Simplifying, we find that 20 K × V1 = 140 L, and thus V1 = 140 L / 20 K, which results in V1 = 7 L.

However, this is not one of the options provided, which indicates a possible mistake in the given choices or in the calculation. Re-examining the calculations shows that the correct calculation should result in V1 = 7 liters, which suggests an error in the question since none of the answer options match. A proper answer based on the calculation is 'None of the above'.

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