5

Q16. What is the net work Ws,net Ws,12 + Ws,23 + Ws,34 + Ws,4I done by the gas during the entire cycle?Ws,net 1550 J Ws,net ~ 234J Ws,net ~ 577 J Ws,net ~ 948 J Ws,...

Question

Q16. What is the net work Ws,net Ws,12 + Ws,23 + Ws,34 + Ws,4I done by the gas during the entire cycle?Ws,net 1550 J Ws,net ~ 234J Ws,net ~ 577 J Ws,net ~ 948 J Ws,net ~ 300 J

Q16. What is the net work Ws,net Ws,12 + Ws,23 + Ws,34 + Ws,4I done by the gas during the entire cycle? Ws,net 1550 J Ws,net ~ 234J Ws,net ~ 577 J Ws,net ~ 948 J Ws,net ~ 300 J



Answers

Given the cycle shown, what is the total work done by the gas during the cycle? a.$-10 \mathrm{J}$ b. $0 \mathrm{J}$ c. $7.5 \mathrm{J}$ d. $17.5 \mathrm{J}$

Hi Friends as cylinder figure Bergman on the guest to take from A. To B. It's 400 and from C. To D. It is -4 under the. Hence then can set out from he We see the zero fuck taking due to a Vogel is Plus 414 er morgan on the guests in taking from in taking it from B to the pressure is decreased, the temperature remains him and volume increases. So Burton from B to C. Bergen from due to weigh its culture to learn off. So why would need ar minus 300 art. So for gun from B. To C. Plus data where to learn off to into wondered are decades 400 are into .6 & 93. It's called 2-77 art bogdan along a. V. Um mm City can tell each other because pressure changes but temperature is the same. Network on their guests of whom? All psyllium. Yeah through the whole network His cult of 2 77 hour. Farsighted so it is very close to once a week. That's thanks for watching it.

So her This problem there two parts on in the a part, we have to calculate the net walk down on the gas per cycle. So let us first rodeo figure giving any question so this difficult given Now the world done during each step off the cycle will be equal to the negative area off the section so well done will be equal to the minus off area off this section. If a calculated slimmed it will be three V minus we a that is minus weaken Dicle upstate it's to be a and if we covered this damn engine, it is through B A minus B a It will be to be a from here. Work will be equals two minus four b a v a. So this was the answer for the A pot Next to little is holding the be part on for the be part will be applying the first law of thermodynamics. So they're exchanging internal energy as equals two q plus w So have you to find the value of Cube. But we know that for idle guess since the temperature off initial and final point is seen so again say that the change in internal energy of the process will be zero. So we're left with Q plus W is equal to zero from here. Cuba, because to minus of W you can block the value of the blue here, so we'll get four times B a v a. So this is the answer for the B, but I hope you were the problem.

So this question belongs to the thermodynamics in which in changing the state of gas, A diabetic lee. So processes a diabetic. So changing heat delta ki will be zero from state A. To state B. Okay. The amount of work done is done on the system, so hence DELTA W. This is equal to negative 22.3 jewel because work is done on the system. So from here, first of all, we can calculate from the thermodynamics law. So delta Q. This is equal to delta U. Plus delta W. So changing internal energy delta you. This will be minus of delta W. And delta W. Is this minus value? So this will be 22.3 jewels. So this is a change in internal energy for the eight to be processed now for the it to be process the heat absorbed. If the gas is taken from state A to B via a process in which net heat absorb. That is delta Q. This is equal to 9.35 calorie. Then we have to calculate how much network done is done by the system. So Delta W. We have to determine. So this change in internal energy will remain same because the process is still eight to be so. Hence from the thermodynamics law, This value, so from here Delta W. This will be equals two delta Q minus delta you so delta Q. It is equal to uh this value, so 9.35 calorie and 4.19 Afterward playing this will be into jewel and minus Delta W. Which is 22.3 jewels. So from here Delta W comes out to big 16.9 jewels. So this will be the answer for this problem. Okay and this is the work done by the system. Okay? That's why it is positive. Okay, thank you. Uh Thank you.

Consider an ideal gas that goes through the process as depicted by the figure on the left starting at point A with pressure piece of I volume visa vie and temperature t survive on and it goes from a to B B to C C. D. Indeed ec. And the first thing we're wondering about this ideal gas is what is the network done on it per cycle. And handy trick is that we know for cyclic processes the area enclosed by the path of the process on TV pressure times volume diagram is equal to the network recycle. So this is actually the fastest way to do this is just really the area of this square. So the height free P i minus two p i r three p I must be I just two piece of I on times with of free visa by minus visa buying again to you the survive on due to the direction of the path. This is gonna be negative because you're gon b and then b two c at three p I and then see the d and then d the a p I. So this is gonna be just negative for p i ve I is equal Teoh work per cycle And that's the fastest simplest where? Jenny. But I'm going to show it, just in case you don't believe me. So we can just use our classic equation 20.9 to give us that The work is equal to negative times the inner girl of the initial volume to the final volume of the pressure over Devi. And we're gonna break this into parts. So first is the integral from a B, which is V I Teoh the, uh the ITV I The volume stays constant and e pressure doesn't do anything. There's no work done here. We know that for any ice ivo metric process where the volume is constant, the change internal energy is just a change in heat. So we know there's no work done on this little PDV leg for it being so we can go ahead and say that zero and then we're looking at the leg from B to C and this I'm just distributing this negative to each piece. So this time we're going from Volume V I to three v i. And the pressure is a constant for a piece, Abi or integrate that or TV. And then we're gonna add our leg from CDT, which, just like our like from a to B just like I like from a to B isn't nice suit phone metrical process. Meaning that the only change in internal energy is the heat change. There's no work done in here. So again, going from free V survive to three piece of I of this PDV is we can go ahead and say is here again because even if it waas, something would just be canceling itself just from the integral, even outside of just thinking of the physics. And then finally, from Dita A from three v i v survive to visa by and that's a constant pressure of just piece of by the Wien so we can go ahead on and evaluate our two remaining in the girls to get that evaluates to negative three piece of. I just This got constant carrying through and then are It's just V from, uh, these visa free visa by so free visa by minus visa by minus, and then from this inner girl, just a constant piece of by and this time we're going from free visa by two Visa by So just visa by minus for revista by Andi not just simplifies to negative for peace of i v survive. So we didn't need to do Dina grow, but I just wanted to show you that that trick of the area and closed by the path of the process does work. But this confirms that the work done is negative for peace of by Visa by. So the next question might be what is to change in heat in our ideal gas and recall that are sick. This same concept of the cyclic process being the area enclosed by the path of process on the diagram is mathematically the same as saying that the Delta E internal is equal to zero in that namely, Q equals negative. W as in, if we go back to this diagram, the work done and be the some C and A sub D is equal to the negative dude of the heat acquired in case of A B and C sub B. Because what's going on here like a to be? The volume is constant, while the pressure expanse from the ideal gas law PV equals and R T we know the temperature is going up there. So the accuse being added here pieces be the seeing. We keep the pressure constant, expand the volume. So here we can expect the temperature to you increase and then from see the d. We do the reverse. We keep the volume constant. What would decrease the pressure, lowering the temperature and again from DT A. We decrease the volume while keeping the pressure constant, decreasing temperature so that net you net heat added or lost is equal to negative work just to remind you so we can go ahead and say that our Q is equal to or piece of ivy survive for this just in negative of our work. And then So finally, we're going to take a look at this with some actual numbers. So if we're given that there's exactly 1.0 moles of our ideal gas and that that initial temperature tease of I equals zero degrees Celsius, what is the mathematical value of this four piece of by visa by quantity? And we can calculate this fairly easily, thanks to the ideal gas saw PV equals and R T, where P is the pressure views the volume, and it's a number of moles. Are is the universal gas constant, which, if you call 8.314 and that's in Jules per more Kelvin's and we're gonna have to if we're going to you. This temperature in the ideal gas law is an absolute temperature, so we're gonna have to go ahead and avert this temperature into Kelvin's. And it's the standard Celsius. Suppose to 73.15 is to tell us. So let's go ahead and calculate are our temperature and pressure at point A where everything's our initial contestants. Soapy I visa by equals and our T survive and we now have values for and 1.0 moles are 8.31 for Jules, for mole, Kelvin and T survive. I was just to 17 free 0.15 Um, sorry Kelvin's. And that just calculates out to 22 70 jewels and we're looking. We know that the work is equal to negative, for he survive east of I. So we know that that is just equal to four times our piece of my abusive by of just nine point. Oh, wait. Times 10 to the fourth jewels, and that is the final answer of sorry that was negative of how much work is done, actually, as a number, and therefore we also know the heat is the same number, just positive. And that's all that's going on in this cyclic process where we have the PV diaper.


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