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1 February, 09:15

Boyle's law balloon was filled to a volume of 2.50 l when the temperature was 30.0∘

c. what would the volume become if the temperature dropped to 11.0∘

c.

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  1. 1 February, 13:02
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    To solve this we assume that the gas inside the balloon is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

    T1 / V1 = T2 / V2

    V2 = T2 x V1 / T1

    V2 = 284.15 x 2.50 / 303.15

    V2 = 2.34 L
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