Question

Answer all parts including a picture showing all variables please! Thank you

1. While I was in the Serengeti I saw people watching wildlife froma hot air balloon. Seemed great. Imagine this: You fly down there and on a wonderful day at 20°C, you are supposed to take off You are skeptical about the hot air balloon carrying you, your two friends, the operator, the operator's dog, the weight of the balloon itself and all the sandwiches and beverages. From the balloon data sheet you estimate that the balloon needs to lift 600kg (1300lbs). That means that the hot air in the balloon has to somehow be 600kg lighter than the same balloon would be filled with 20°C air. The balloon is a perfect sphere with a diameter of 15m. From chemistry class you remember that one mole of air at 20°C has a mass of 29g. You know that the air pressure during your short flight will be constant and the volume of the balloon does not change. From studying the ideal gas law you know that an increasing temperature results in a change of the gas, and you intuitively understand that hot air balloons are lighter than the air around (that's why the fly). You calculate the air temperature in the balloon for successful lift off and more importantly to keep you safely in the air. A quick search on the internet tells you, that the gas burners used for hot air balloons heat the air in the balloon consistently to 100°C After some quick calculations you discuss with your friends how you will spend the day a) Draw a picture including all the properties given. b) Why is the hot air balloon lighter than the 20°C air balloon? Use the ideal gas law for your explanation c) How many moles of air have a mass of 600kg? d) How many moles of air fit into the balloon at 20°C? e) What temperature do you need in the balloon for take-off? f) Would you take the flight based on the information given, or would you consider taking another balloon operator for your adventure? Can the balloon safely carry 600kg?

Answer 1

a. b. we know Pv = nRT T = Pv/nR or n = Pv/RT no of moles (n) alpha 1/T so it T is higher n is less so mass is also less (lighter) c. 29 gm gas = perpendicular mole 600 times 1000 gm = 1/29 times 600 times 1000 moles = 20689.655 moles \r\n d. using Pv = nRT V = 4/3 pi R^3 = 4/3 times pi times (15/2)^3 = 1767.146 m^3 P = P_atmosphere = 1.01 times 10^5 Pa n = Pv/RT = 1.01 times 10^5 times 1767.146/8.314 times (273.15 + 20) (at 20 degree c) n = 73230.81 moles e. Let the ballon is of temp T no of moles inside ballon should be n� = (73230.81 -0 20689.655) to lift 600 kg weight so n� = Pv/RT = 1.01 times 10^5 times 1767.146/8.314 (T) T = 1.01 times 10^5 times 1767.146/8.314 times (73230.81 - 20689.655) T = 408.59 K = 135.44 degree f. hence T = 135.44 <