Consider a thermal engine filled with N molecules of a monatomic ideal gas. The

Question

Consider a thermal engine filled with N molecules of a monatomic ideal gas. The gas undergoes a cyclic transformation, as shown. Processes AB and CD are isothermal (constant temperature), and processes BC and DA are isochoric (constant volume). The quantities P1,p2, V1, and V2 are defined in the figure. For example, when the gas is in state A, it has pressure P2 and volume. V1; when the gas is in state B. it has volume , but the pressure is not given. What is the change in internal energy UB-UA during the process AB? What is the change in internal UD-UC energy during the process CD? Express the change in internal energies in terms of p1,p2,v1 and/or v2. Separate your answers with a comma. UR-UA UD-UC = What is the change in internal energy Uc - UB during the process BC? What is the change in internal energy UA-UD during the process DA? Express the change in internal energies in terms of P1, P2, V1 and/or V2 . Separate your answers with a comma. UC-UB UA-UD = What is the total change in the gas's internal energy after it has completed a cycle? U end of cycle - U beginning of cycle = Finally, consider the heat absorbed by the gas during its cycle. Define QAB to be heat absorbed by the gas in process AB, etc. If a process is exothermic(i.e., heat is given off by the gas during the process), then the quantity of heat absorbed will be negative. Indicate the processes in this cycle during which heat is flowing into the gas. That is. indicate the segments where Q is positive. DA only CD and DA DA and AB AB and BC

Explanation / Answer

Part E the internal energy of an ideal gas dependds only the temperature of the gas. Since, the processes AB and CD are isothermal. So, change in internal enery in both processes is zero.

UAB = UCD = 0

Part F Also, UBC = 3/2 NkB TBC

Now, by ideal gas law, pV=NkB T, we have

TBC = V2/NkB (p1 - pB)

We need pB in terms of p2 . this can be simply found by considering the process AB which is isothermal. Thus, pV = constant. This gives, p2V1 = pBV2, thus

TBC = 1/NkB (p1V2 - p2V1)

Thus, UBC = 3/2 (p1V2 - p2V1)

Similarly, UDA = 3/2 NkB TDA

TDA = V1/NkB (p2 - pD)

Again by considering the isothermal process CD, we have p1V2 = pDV1

So, TDA = 1/NkB (p2V1 - p1V2)

So, UDA = 3/2 (p2V1 - p1V2)

Part G, We have, final U = UA + UAB + UBC + UCD + UDA = UA + 0 + 3/2 (p1V2 - p2V1) + 0 + 3/2 (p2V1 - p1V2)

So final U = UA

Part H Since, AB and CD are isothermal processes, so, UAB = Q - W = 0

So, Q = W = integral of pdV

So, QAB = NkBTA log (V2/V1).

Since V2 > V1, so QAB will be positive.

Also, in process BC volume is constant, so work is zero, thus QBC = UBC = 3/2 TBC

Since, this process is isochoric expansion (pressure decreases), so temperature decreases, thus QBC < 0

Similarly, QCD = NkBTA log (V1/V2)

This will be negative as V2 > V1

So, QCD < 0

Also, similarly, QDA = UDA = 3/2 TDA

Again, since the process is isochoric compression (pressure increases), this means that temperature increases. So, QDA > 0

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