A boy on the beach holds a spherical balloon filled with air. at 10:00am, the temperature on the beach is 20°c and the balloon h as a diameter of 30 cm. two hours later, the balloon diameter is 30.5 cm. assume that the air is an ideal gas and that no air was lost or added, what is the temperature on the beach at noon?
1 answer:
For idea gases, volume is directly proportional to temperature. That is, an increase in temperature leads to increase in volume and vice versa. Therefore, V1/T1 = V2/T2 => T2 = (V2*T1)/V1 Assuming that the balloon is spherical in shape, V= 4/3*pi*R^3.... In the formula for calculating T2, 4/3*pi cancels out. R1 = 30/2 15 cm; R2 = 30.5/2 = 15.25 cm; T1 = 20+273.15 =293.15 K Therefore, T2 = (R2^3*T1)/R1^3 = (15.25^3*293.15)/15^3 = 308.05 K = 34.9 °C
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t = 444.125 sec
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