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Consider an Otto cycle (ideal spark-ignition cycle) with a monatomic ideal gas as the working substance. Processes 3->4 and 1->2 are reversible adiabatic. Given: T1 = 300 K, compression ratio V1/V2 = 8, P3/P2 = 4/3. If the ratio W234/W412 = x, find the value of |3x|, where W234 is the net work done during processes 2->3->4 and W412 is the net work done during 4->1->2.
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Correct answer: 4
Solution
Monatomic ideal gas: gamma = 5/3, Cv = 3R/2. State points: 1 (T1, V1), 2 (T2, V2), 3 (T3, V2), 4 (T4, V1). Adiabatic 1->2: T2 = T1*(V1/V2)^(gamma-1) = 300*8^(2/3) = 300*4 = 1200 K. Isochoric 2->3: P3/P2 = T3/T2, so T3 = T2*(P3/P2) = 1200*(4/3) = 1600 K. Adiabatic 3->4: T4 = T3*(V2/V1)^(gamma-1) = 1600*(1/8)^(2/3) = 1600/4 = 400 K. W234 = W23 + W34: W23 = 0 (isochoric), W34 = nCv(T3-T4) = n*(3R/2)*(1600-400) = 1800nR. W412 = W41 + W12: W41 = 0 (isochoric), W12 = nCv(T1-T2) = n*(3R/2)*(300-1200) = -1350nR (negative = work done on gas). W234/W412 = 1800/(-1350) = -4/3. x = -4/3. |3x| = |3*(-4/3)| = |-4| = 4.
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