<p>Are we suppose to know Q=nC(Delta)T ?? I'm confused with Q, U, and W. Since they are all expressed in joules, can't they be used interchangably? How come the PR book says that Q=nC(Delta)T and my text says: U=nC(Delta)T.. Confused..</p>
<p>They aren't interchangable.</p>
<p>Q is the heat added or removed, W is the work done and U is the total change in energy. Q = mct. (or n depending on whether you are using mass or number of moles) U = Q+W</p>
<p>And, w = pv or the area under the PV curve.</p>
<p>Personally I think PR has tons of mistakes. I think I've found one in the book and it confuses me. Can someone see if it is a mistake? Thanks</p>
<p>PR says: "The expanding gas did negative work against its surroundings, pushing the piston upward."</p>
<p>--I think it should be: The expanding gas did positive work against its surroundings, pushing the piston upward. I think the book meant that the surroundings did negative work on the system. </p>
<p>This is a typo right?</p>
<p>I think you are correct. Work done by the system is positive and work done on the system is negative.</p>
<p>But, in the formula U = q+w you'll be using work done on the system or you can change it to u = q-w to use work done by the system.</p>
<p>Anyone else (about the piston thing)?</p>
<p>Wait, the piston thing makes sense, you're doing neg work on the piston. The gas does positive work but the work done against the surrounding is negative... It's just too wordy.</p>
<p>lol, I found another mistake on page 180.. The graphs of the question and solution are totally different!! Am I just seeing things or am I really right?</p>
<p>Does anyone know where I can find a list of thermo processes such as isothermal, adiabatic etc.??</p>
<p>Yup.. Koda, you're right about the piston typo. You don't need to know anything with specific heat. You're talking about Cp or Cv which is specific heat at a constant pressure and constant volume right?</p>
<p>Yeah, I've learned it already anyway. I'm starting engines right now. wahoo! =] Do isothermal and adiabatic processes always look like a curve (with the adiabatic curve more steeper than the isothermal one)?? The arrows can be pointed in any direction (clockwise or counterclockwise) right?</p>
<p>Do these objectives follow the equation: PV=nRT??</p>
<p>Relate the pressure and volume of a gas during an isothermal expansion or compression.</p>
<p>Relate the pressure and temperature of a gas during constant-volume heating or cooling, or the volume and temperature during constant-pressure heating or cooling.</p>
<p>Q=nC(delta T) when there is additional work being done in the system, (delta U)= Q+W. U=nC(delta T) is used when there is no heat transferred into a system, such as during an adiabatic process. BTW, work done by a system is -W and work done ON a system is +W.</p>
<p>Koda, those objectives do make use of the ideal gas law. For an isothermal expansion, PV is equal to a constant because temperature is constant. So on a PV diagram, when the volume goes up pressure goes down because, when multiplied together, PV must always equal the constant nRT. You can apply the same reasoning to the isochoric process.</p>
<p>Wow, thanks a lot dahuie1! Your explanations really help me a lot!</p>
<p>I'm also confused with the area within a cyclic curve on a PV graph. Does the area within the curve represent the work done ON the system/gas? </p>
<p>My text says, for heat engines, that: "The work done by the engine for a cyclic process is the area enclosed by the curve representing the process on a PV diagram" --which would equal the work done by the system/gass.</p>
<p>I know that the W(engine)= -W(system) because the system is doing work on the engine. So... I'm kind of confused here, my guess would be that the signs would change?</p>
<p>For the area under a PV diagram, you have to follow the equation W=-P(delta V) . So lets say you have an isobaric process that is a horizontal line from lets say, 4V to 2V under 1P of pressure. Plug in these values into the equation, W=-p(2V-4V), which gives you 2PV. 2 PV of work was done ON the system because work is positive. Remember that the equation is W=-P(final volume - initial volume) and that is why you get a positive W, because the process goes from right to left. So, work done on a gas gives you a positive W and work done by the gas gives you a -W. </p>
<p>You can think of it this way: lets say I have gas inside a piston. If I was to compress the piston, i would be doing work ON the gas, because work is a force over a distance. If there is no heat transferred, the temperature of the gas would go up, because the equation become (delta internal energy) = W. If i lowered the temperature of the room and made the gas expand, the gas would be doing work on the balloon - which means it is losing energy to do that work. (delta U) becomes negative, which means the temperature goes down. Just remember that internal energy is proportional to temperature, so a rise in U raises the temperature and a loss of U lowers the temperature.</p>
<p>Thanks again! =] You're a really good tutor.</p>
<p>What the difference between: 1) Compute the maximum possible efficiency of a heat engine operating between two given temperatures. 2) Comput the actualy efficiency of a heat engine.</p>
<p>I've only learned this equation so far: e=W/Q but I'm not sure if it gives the maximum possible efficiency or the ** actual efficiency**. How is it possible to calculate the actual efficiency anyways?</p>
<p>Also, what are the easiest ways of calculating temperature?? Do you need to use PV=nRT or are there other equations? Thanks.</p>
<p>The equation e = W/Q gives you the actual efficiency. It is pretty intuitive if you think of it as the actual work done by an engine divided by the amount of heat that was put in. As for the maximum efficiency of an engine, you use the equation e = (T(h) - T(c))/T(h). T(h) is the initial temperature of the heat engine and T(h) is the temperature of the exhaust, which should be colder.</p>
<p>The only ways i know of calculating temperature in physics b are using PV=nRT, Q = mc(delta T), U = 3/2 N(Boltzmann constant)T.</p>
<p>Oh okay. Given a heat engine cyclic process on a PV diagram, how do you calculate the Q(hot, heat energy in) and Q(cold, heat enegy out)?? The graph is like a slide shape with a step 1 being isochoric(isovolumetric), step 2 being isothermal, and step three being isobaric. </p>
<p>The book says that Q(hot)= Q(step1)+Q(step2)</p>
<p>I tried using Q=nCv(delta)T to calculate the Q(step 1) but the answer isn't right.</p>
<p>Are we even suppose to know this stuff?</p>
<p>I understood your question as a PV diagram where the process starts as an isochoric process, increases in pressure, becomes an isothermal process, slides down and to the right, and then becomes an isobaric process and travels to the left.</p>
<p>To calculate the heat that is put into the engine, you calculate work on the isobaric process (W=-P(delta V)). It is the work that is done on the gas to compress it. The isothermal process is irrelevent because the net change in internal energy during that process is 0. For the heat energy outputted, you would calculate Q at the top of the isochoric process, because as pressure increases while volume stays constant, temperature increases.</p>
<p>Great anwers! I'm on the Carnot engine right now and it all seems confusing to me. Should I just skip this section? All I know is that it is the most efficient engine and that the processes are: 1.isothermal,2.adiabatic,3.isothermal,4.adiabatic. Is there anything else important that I need to know? </p>
<p>I'm really confused and don't know how to find the temperature in graphs. I mean a PV diagram only shows pressure and volume right? How would you know if the curve shows if the temperature increased or decreased or stayed the same?? Same for heat. I really appreciate your help! =]</p>