I already got the answer to be 62% (Wnet/Qin =2.8/4.5 *100) but would like to ch

Question

I already got the answer to be 62% (Wnet/Qin =2.8/4.5 *100) but would like to check my answer.

Consider the P-V diagram of the closed, piston-cylinder system below: Path 1- > 2 is internally reversible and isothermal: Path 2- > 3 is internally reversible and adiabatic: Path 3- > 1 is internally irreversible due to friction that arises in compression The cylinder is filled with 40 g of an ideal, diatomic gas. The gas has a molecular weight of 24 kg/kmol and a C_v of 25 k/kmol-K. The internal energy of an ideal gas is dU = nC_v dT. The following is known about the process: A. At state 1, the temperature is 162 degree C and the volume is 42.0 L. B. The heat added in Path 1- > 2 is 4.50 kJ. C. The change in internal energy between states 2 and 3 is -1.40 kJ. D. The work associated with path 3- > 1 is-3.10 kJ. What is the efficiency of the cycle (W_net/Q_in)?

Explanation / Answer

consider styep 1 to 2
it is isothermal, so, from first law of thermodynamics
work done by system + INternal energy change of the system = heat added to system
W1 = 4.5*10^3 J - 0 = 4500 J ( because Internal energy change is 0 for isothermal process)
Q1 = 4500 kJ

similiarly, for step 2 to 3
U2 = Change in internal energy = -1400 J
Q2 = heat added to system = 0
so, Work done, W2 = 1400 J

similiarly for step 3 to 1
w3 = -3100 j
u3 = ?
now, U1 + U2 + U3 = 0
so. U3 = -U1 -U2 = 0 + 1400 = 1400 J
hence, W3 = Q3 - U3
Q3 = -3100 + 1400 = -1700 J

Hence, Wnet = 4500 + 1400 - 3100 = 2800 J
Qin = 4500 J
efficiency = W/Q = 62 pc

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