The working substance of a certain Carnot engine is1.90 mol of an ideal monatomi

Question

The working substance of a certain Carnot engine is1.90 mol of an ideal monatomic gas. During the isothermalexpansion portion of this engine's cycle, the volume of the gasdoubles, while during the adiabatic expansion the volume increasesby a factor of 5.7. The work output of the engine is 990 J ineach cycle. Compute the temperatures of the two reservoirs between whichthis engine operates What is Tlow and Thigh I am stuck I am really confused even and explanation will doand will give you lifesaver THANKS The working substance of a certain Carnot engine is1.90 mol of an ideal monatomic gas. During the isothermalexpansion portion of this engine's cycle, the volume of the gasdoubles, while during the adiabatic expansion the volume increasesby a factor of 5.7. The work output of the engine is 990 J ineach cycle. Compute the temperatures of the two reservoirs between whichthis engine operates What is Tlow and Thigh I am stuck I am really confused even and explanation will doand will give you lifesaver THANKS

Explanation / Answer

Equations we will need:

e = 1 - (TL/TH) for ideal Carnot engine

e = W/QH

U = Q-W (key concept)

Wab = QH = nR(TH)ln(Vb/Va)

~~~

Let e = e. That is, 1-(TL/TH) = W/QH

Isolate W. Multiply QH over. QH - QH(TL/TH). Factor out QH and make the equation a little neater. Same as writing: W = QH[(TH - TL) / TH]. We have W, we just need to rewrite QH and one of the unknown T's.

U = Q - W where U is internal energy.

For isothermal process, U = 0. Therefore 0 = Q-W. So, W = Q.

Work for an isothermal process, for a to b is: Wab = nR(TH)ln2. n is moles (give), R is the gas constant 8.314 J/K moles, ln2 is just some number.

W = Q, thenWab = nR(TH)ln2 = QH

Note we aren't given Wab, but are given W.

Plugin: W = QH [(TH - TL) / TH]

W = nR(TH)ln2 [(TH - TL) / TH]

For adiabatic processes, we know PbVb^gamma = PcVc^gamma. For Ideal monatomic gas, gamma is 5/3.

From PV=nRT, we can rewrite Pb and Pc in terms of the other variables. Namely, Pb = nRTb/Vb and Pc = nRTc/Vc and put it into this formula

In this case, Tb is also TH and Tc is TL.

So

(nRTH/Vb)Vb^gamma =(nRTL/Vc)Vc^gamma

n and R cancel out when you divide over

TH = TL (Vc/Vb)^(gamma minus 1 or 5/3 - 1)

Vc/Vb ratio is 5.7 b/c it "increased by factor 5.7"

TH = TL(5.7)^(2/3)

We are ready to solve using the W equation above

W = nR(TH - TL) ln2

W = nR (TL(5.7)^(2/3) - TL) ln2

W = nR TL(5.7^(2/4) - 1) ln2

Given W = 990 J....solve for TL plug and chug.

Then plug into the TH equation to get TH.

Source of Equations: Giancoli Physics for Scientists and Engineers 5th ed. Look at p. 533 to p. 534 for equation derivations

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