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Answer the following questions:

(a) Is the pressure of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?(b) Is the density of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?(c) At a pressure of 1 atm and a temperature of 20 °C, dry air has a density of 1.2256 g/L. What is the (average) molar mass of dry air?(d) The average temperature of the gas in a hot air balloon is $1.30 \times 10^{2} \circ F$ Calculate its density, assuming the molar mass equals that of dry air.(e) The lifting capacity of a hot air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?(f) An average balloon has a diameter of 60 feet and a volume of $1.1 \times 10^{5} \mathrm{ft}^{3}.$ What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?(g) A balloon carries 40.0 gallons of liquid propane(density 0.5005 $\mathrm{g} / \mathrm{L} )$ . What volume of $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$ gas is produced by the combustion of this propane?(h) A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ/min) from the hot air in the bag during the flight?

a) The pressure of the gas in the hot air balloon shown at the opening of this chapter is equal to the pressure of the atmosphere outside the balloon because the balloon is free to expand until the pressures are equalized.b) The density of the gas in the hot air balloon shown at the opening of this chapter is less than the density of the atmosphere outside the balloon. Due to less density of gas present inside the balloon than the surrounding air, the hot air balloon rises.c) 29.48 $\mathrm{g} \cdot \mathrm{mol}^{-1}$d) 1.096 $\mathrm{g} \cdot \mathrm{L}$e) 0.12894 $\mathrm{g}$f) 384 $\mathrm{lb}$g) 269.6 $\mathrm{L}$h) 39.1 $\mathrm{kJ} \cdot \mathrm{min}^{-1}$

05:18

Aadit S.

Chemistry 101

Chapter 9

Gases

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podcast. We're taking a look at density. So against is often denoted. Is this strange little P that is the mass of the substance w divided by the volume feet. It can also be expressed as P is equal to p M. Divided by R T. You can rearrange for em. That is density R T overpay. LaPierre's pressure are is 8.31 40 is temperature. And so in the fast part. Well, starting page the pressure of the gas in the hot air balloon show at the opening off. The entire chapter is equal to the pressure of the atmosphere outside of the balloon because the balloon is free to expand until the pressures are equalized. So moving on to the next partner, the density of the gas in the hot air balloon is less than the density of the atmosphere. So due to the last density of gas present inside the balloon than the surrounding air, that the hot air balloon will rise as a result of this. So in part seal, start fresh page because you've got a small calculation. So but the molecular mass of the dryer was calculated, so is equal to 1.2256 grams per liter, minus one multiplied by 9.98 to 1 atmosphere per liter per MOL minus one Calvin's the minus warm that is multiplied by 293 Calvin. And then that is all divided by one atmosphere, so M equal to 29.48 grams. Put mall to the minus one in the next part. The density of the gas inside the hot air balloon is calculated as follows we have. The density is equal to 29.48 g per mole, minus one multiplied by one atmosphere. Let me divide that by 9.98 to 1 atmosphere. Elita multi minus one. Calvin's minus one multiplied by 1.30 times 10 square part two degrees Fahrenheit. So what we need to be doing is changing up some units is so we need to come vote this Calvin. So we have 3 to 7.59 Calvin, So density is even 1.96 g per liter. So what we've got in the next part because the lifting capacity of the higher balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. So are lifting capacity equal to 1.2256 g. Take away 1.9 666 g. That's equal to not 0.12894 g, so the lifting capacity is equal to not put 1 to 894 g. Next ramped apart F. So we're calculating the volume in liters and the multiplying the volume by the density difference to find the lifting capacity of the balloon. Then we subtract the way of the balloon after converting £2 so volume in liters is equal to 3.11 times 10 to the six liters. Then the mass of the balloon is calculated. That is 4.1 times 10 to the 5 g, so that stands to multiplied by volume than in pounds. That is 884 pounds lifting capacity people to 884 pounds. Subtract £500. Lifting capacity is £384 Female parts to go in this podcast. So G, we're calculating the volume of gas produced from the combustion off propane. So fast we calculate the mass of propane. So that is 75 point 78 g. The moles of propane people to 1.72 miles. We take a look at our reaction. We look at the total volume standard temperature pressure, that is 269 0.6 liters. One last part h. So we have our reaction. Delta reach not a combustion is equal to negative 2043.96 kg joules per mole. And so that we multiply this value Our 2043.96 by 1.72 models to get 3.52 times 10 to the 3 kg jewels. So the heat released, divided by 90 minutes. So the heat lost from the hot air bag people to study 9.1 killer jewels per minute.

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