A weather balloon is inflated to a volume of 28.5 L at a pressure...

Question

A weather balloon is inflated to a volume of 28.5 L at a pressure
of 748 mmHg and a temperature of 28.0 °C. The balloon rises in
the atmosphere to an altitude of approximately 25,000 ft, where
the pressure is 385 mmHg and the temperature is -15.0 °C. Assuming
the balloon can freely expand, calculate the volume of
the balloon at this altitude.

Accepted Answer

Hey everyone. So here we have 30 degrees Celsius and 1.1 atmosphere and a balloon has a volume of three liters. Then the balloon rises where the pressure, 0.6 atmosphere and the temperature is 18 degrees Celsius. Or ask catholic the volume of the balloon in the sky. But here we need to use our combined gas law which is P one times B one about about T. One, it was P two V two by T. Two you want is 1. atmosphere in one. It's three liters T one, it's 30 degrees Celsius. Me too. 0.6 atmosphere. Me too. What we're looking for is he too, it's 18 degrees Celsius. So now we need to convert the temperatures into kelvin R. T. One, you have 30 degrees Celsius. If we add 273 15, get 303 15 Elvin for T. Two, we have 18 degrees Celsius. And if we add 273.15, get 291 15 kelvin. So now we can play the bodies into the equation and solve for V two. We're gonna get 1.1 atmosphere times three Leaders about about 303 kelvin, 0.6 atmosphere times V two, divided by 291 15 kelvin. So now we can cross multiply to solve for B two. Gonna get 1.1 atmosphere times three Leaders, 191 15 kelvin equal to 0.6 atmosphere comes V two times 303 15 kelvin divide both sides by 0.6 atmosphere times 303 15 kelvin. We're gonna get B two equals 5. leaders. Thanks for watching my video, and I hope it was helpful.

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