Planetary Atmospheres. (a) Calculate the density of the atmospher...

Question

Planetary Atmospheres. (a) Calculate the density of the atmosphere at the surface of Mars (where the pressure is 650 Pa and the temperature is typically 253 K, with a CO2 atmosphere), Venus (with an average temperature of 730 K and pressure of 92 atm, with a CO2 atmosphere), and Saturn’s moon Titan (where the pressure is 1.5 atm and the temperature is -178°C, with a N2 atmosphere).

Accepted Answer

Hey everyone welcome back in this problem. We are asked to find the density of the air. Okay, inside a truck tire, we're told that the pressure is 7.58 times 10 to the five pascal's the temperature is 25 degrees Celsius. We're told to assume that the average molar mass of air is 28.97 g per mole. Okay, Alright, so we're seeing something with pressure with temperature with Mueller mass. So this makes us think of our ideal gas law, ideal gas equation. Okay, so let's go ahead and write that. Recall that the ideal gas equation P V is equal to N. R. T. Anyway, you'll notice is that we don't have density in this equation. Ok. We have pressure volume and the number of moles are are ideal constant or the gas constant K N. T. Temperature. We don't have density. Well, let's think about density. We know that we can write density rho as mass per volume mass divided by volume. Okay, so we do have volume in this equation but we don't have mass. So let's try to figure out how we can get mass into this equation. Well, recall that N The number of moles can be written as M. The mass divided by big M. The molar mass. Okay, substituting this into our equation, we have PV is equal to little M over big M. R. T. Okay, now we have M and we have V. In our equation. Okay, And we can rearrange to write rho is equal to M. V. Okay, so we know that rho is equal to M divided by V. Okay, now, in order to rearrange our equation to have M divided by V. We need to multiply big M. So we get p big M. Okay? And then divide by bigger big T. And now we have an equation for rho the density that we're looking for. And all of the quantities in here we know we can find, okay, we have p the pressure which we know and the molar mass which we know are the gas constant or ideal gas constant which we know the value of. And we have t the temperature. Okay, so now it's just a matter of substituting in the values that we know to find our density. Okay, so let's just go ahead on the side here and write out the quantities we were given. Okay, so the pressure 7.58 Times 10 to the five Pascal's. Mhm. The molar mass M is 28.97 grams per more. When we start dividing by temperature and we're multiplying by pressure. Okay, we're gonna want this in kg per more. So this is gonna be equal to 0. kilograms thermal. We have are the ideal gas constant which is eight .314 jules Permal Calvin. And finally, T the temperature we're told is 25 degrees Celsius. Okay. And converting into Calvin. 298.15. Okay, we just have to add 273.15 Calvin. Alright so we have all our values. Let's go ahead and substitute them in. So we have rho is equal to p 7.58 times 10 to the five Pascal's Many times the Mueller mass which is 0.02 kg / Mole. Okay, divided by 8.314 jewels. Her more Calvin Times 298. Calvin. Okay what you'll see here is that we have some units that start to cancel. The unit of mole is going to cancel. The unit of Calvin is going to cancel. Now let's recall to the unit of pascal. Okay, pascal is equal to a jewel. Her meter cubed. Okay so when we think of that if we have jewel per meter cube then the unit of jewel is also going to cancel. Okay so let's just write this out. So in the numerator we're gonna have 21, 260. And again our unit is pascal which is going to be jewel per meter cubed times kilogram because we're gonna jewel per meter cubed times kilograms in the numerator. And in the denominator we have 2478.8191. In the unit we have left here is jewel. Okay so the unit of jewel will cancel when we divide. We're gonna be left with rho is equal to 8.85 kg per meter cubed. And so that is the density on the air inside the truck tire that's going to correspond with answer B. Thanks everyone for watching. I hope this video helped see you in the next one.

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