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If the gauge pressure of air in a enclosed 12 m3 tank is measured to be 34 kPa ad the air is at 40 %C determine the mass of air in the tank Complete your calculatio...

Question

If the gauge pressure of air in a enclosed 12 m3 tank is measured to be 34 kPa ad the air is at 40 %C determine the mass of air in the tank Complete your calculation and enter the value below:Assume 'atm 101 kPa ANSWER:The mass of air in the tank is

If the gauge pressure of air in a enclosed 12 m3 tank is measured to be 34 kPa ad the air is at 40 %C determine the mass of air in the tank Complete your calculation and enter the value below: Assume 'atm 101 kPa ANSWER: The mass of air in the tank is



Answers

A $1-m^{3}$ tank containing air at $10^{\circ} \mathrm{C}$ and 350 kPa is connected through a valve to another tank containing $3 \mathrm{kg}$ of air at $35^{\circ} \mathrm{C}$ and 200 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at $20^{\circ} \mathrm{C}$. Determine the volume of the second tank and the final equilibrium pressure of air. Answers: $1.33 \mathrm{m}^{3}, 264 \mathrm{kPa}.$

Speaking first heard the continuity equation massive two months MP's of one. This would be equal to the excess mass that is let out some of the air escaping essentially the mass of the air escaping. I'm so to multiplied by the specific energy said to minus M civil one use of one. This would be equaling, then to the negative mass of the escaped air times the entropy of that air plus heat transfer minus the work. Um, we can say that for the entropy, M sub two s a to minus amsa born s sub one. This would be equal to the negative mass of the air. Escaped times the entropy of that air, plus the entropy associated with he, uh, she transfer, Plus the entropy generated. Ah, here we have an 80 Batic, So 80 about a process meaning there's no heat transfer. Um, there's no work also, So this becomes zero, and this becomes there s no work and needy back. So given this, we can we don't know state to We don't know any values for state too. So we can say that I m sub two being the mass that remains in the tank We can then say that I am so too multiplied by use of to minus use of one, this would be equaling. Then two cues up to minus works up to I'm some to as to minus s of one. This is equal to the square root of CQ over t plus sm two he entropy generated. So again, um, we have that This is zero. This process is reversible. So the entropy generated is also zero. And we can say that the nests of two equals as someone. So this is a ice and tropic process. We can then use the equation teeth of two equaling teeth of one multiplied by the ratio of the pressures raised to the K minus one over K power. So this would be equaling them to 400 multiplied by 200 over 300 raised to the 0.2857 power. This is equaling them to 356.25 Calvin and you can find the mass that is left over in the tank. So this would be 200 times two, divided by 0.287 multiplied by three. 56.25 Calvin and we find the masses equaling 3.912 kilograms. This is the mass that is left inside the tank. That is the end of the solution. Thank you for watching.

Mhm. This question covered the concept of absolute pressure and the absolute pressure is Cuban into it some of the gauge pressure but the atmospheric pressure from this we can write the cage pressure equals absolute pressure minus the atmospheric pressure. Okay. And the absolute pressure is 1275 kill a basket. And that most very pressure Equals 101 point created Philip. Ask. Therefore we can write the gauge pressure equals 1275 Philip asking -101.32. You know pascal are the gauge pressure in the tank is 1174 Philip. Ask

Discretion. We have to find the rate of shamed of the density of air in the tank or d roti by D. T. We can use this basic equation to find that And we'll have that we tank tom girl. Uh well tank by DT plus um go exit tom be time a CO ceo. So we have that the rotating by DT is minus grow exit time. We time. Eh Good bye. We tank and we have that there is a CO P exit. We took a but by are a tom ti exit. So we tank. So if we substitute these datas, we'll have that. It is When I was 310 to the power of three time to 50 time. 100 Dwight by 1000 square um one over 236.9 Tom my next 20 plus 273 down 24. And that would be my next point 258 for um the read option in density of our intent as a unit of kilo camp where meter cubic meter time second. So that is the answer.

Okay. A compressed air tank holds 1/2 a cubic meter of air and then it's released into the atmosphere and we'll assume that the same amount of error of the same number of moles of air are released in the atmosphere. So it asks what volume it will need to occupy in the atmosphere. So what do we know? You know, the initial volume of the tank is 1/2 a cubic meter says you're a point. Fine. 00 cubic meters. All right. It has a temperature initially of 295 Kelvin. Just a good unit for comforter. And the pressure is thea 820. Kill it. Past scales will always want to get into units. Abascal sweet 120,000. Haskell's great 20 times, 10 to the third. And then, uh so we can actually use R p B perfect gas ball PV equals and Artie to find out how much air we're dealing with. All right, So to find the number of rules, we're gonna take the PV and divide by Artie. Always isolate your unknown symbolically, and then we do our plug in shop. So we plug in the pressure one, the volume one in there. So we have 820 1000 and that's pass skills. You plug in for peace in the middle of a fight with volume, which is you're a 0.500 Cuban meters. These are all standard unit. So if you work with these units, you should be able to get the upper balls. And 8.31 is our gas constant universal guests constant, which is in units of jewels from all Killman. And then we multiply by our temperature initially just to 95. So what we find out is the amount of gas we're dealing with is that that gas has stayed the same and I get used with 36 667 moments of gas. Okay, well, each mole contains six times we know to two times that of the 23rd particles. So we're talking about quite a few particles of gas. All right, so what's our new volume? That's what we're looking for, William, too. We know that its temperature is given to be 303 Kelvin. You know, pressure in the second state is equal to 101 1000 skills. So to calculate the volume, we rearranged the perfect gas formula by taking the number moles. Times are times t all divided by the pressure cats this is going to and this this end is going to stay that same 167 malt civil to plug that right into here. We'll plug in the are playing in temperature to and pressure too, into this relationship. So we've got everything we need and we end up getting a volume of 4.17 cubic meters with three sick fix. And there you have it. The amount of gas stayed the same. That's why we wanted to find the amount of moles of gas. And we just needed to find out what volume expanded, so went from 1/2 a cubic meter all the way to well over eight times as much that expanded out into the atmosphere. So that is thing solution


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