A 0.20 mol sample of an ideal gas goes through the Carnot cycle of figure below.

Question

A 0.20 mol sample of an ideal gas goes through the Carnot cycle of figure below. (Figure 1)

Figure 1 of 1

Part A

Calculate the heat Qh absorbed.

Express your answer using four significant figures.

Part B

Calculate the heat Qc rejected.

Express your answer using four significant figures.

Part C

Calculate the work done.

Express your answer using four significant figures.

Part D

Use these quantities to determine the efficiency.

Part E

Find the maximum temperature.

Express your answer using four significant figures.

Part F

Find the minimum temperature.

Express your answer using four significant figures.

Explanation / Answer

Part A) . Tmax= PV/nR= 8 /0.0821x0.2 = 487.21 K
Qh = nRT ln (V2/V1) = 0.2x8.314x487.21 ln (2) = 561.54 J

Part B). Tmin= 4.1x1.612/0.2x0.0821 = 402.51 K
Qc = nRT ln (V2/V1) = 0.2x8.314x402.51 ln 2 = 463.92 J

Part C). PVgamma =constant

4x2gamma = 2.05x 3.224gamma
1.951 = 1.612gamma
gamma = 1.4

Work = P1V1-P2V2/(gamma-1) = (4x2-2.05x3.224)/0..4 = 3.477 L atm = 352.22 J

Part D). e= W/Qh = 352.22/561.54 = 0.6272 = 62.72%

Part E). Tmax= PV/nR = 8/0.2x0.0821 = 481.2107 K

Part F) Tmin= PV/nR = 4x1.612/0.2x0.0821= 392.6918 K

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