Consider a four step cycle which is carried out reversibly. Step 1: State A to S

Question

Consider a four step cycle which is carried out reversibly. Step 1: State A to State B. 2.13 mole of an ideal gas, initially at 1.52 atm, is expanded isothermally at 132.3 degree C to twice its initial volume. Step 2: State B to State C. The temperature of the gas is lowered adiabatically to 106.3 degree C. Step 3: State C to State D. The gas is compressed isothermally to state D. Step 4: State D to State A. The gas is compressed adiabatically. Calculate p, V, and T at each point A, B, C, D. (Note that state D is at the intersection of the isotherm through C and the adiabatic through A.) Calculate q, w, deltaU, deltaH. and AS for each step of the cycle and for the entire cycle. What fraction of the heat transferred in the expansion steps is converted to work over the cycle? Summarize your results in tabular form.

Explanation / Answer

pV = nRT

p=1.52 atm n=2.13 R=8.31J T=405.3 K

1.52 atm x V = 2.13 x 8.31 J x 405.3 K

V=4.7x10-3 m3

State A to State B.(1-->2)

Q=n(2R/1)T

Q = 2.13x2x8.31x405.3

Q= 13427.44 J

State B to State C. (2--->3)

Q=n(3R/2)T

Q = 2.13x(3x8.31/2)x379.3

Q= 10070.58 J

State C to State D. (3--->4)

Q=n(4R/3)T

Q = 2.13x(4x8.31/3)x379.3

Q=8951.63J

State D to State A. (4--->1)

Q=n(1R/4)T

Q = 2.13x(1x8.31/4)x379.3

Q= 1678.43 J

State A to State B.(1-->2) Q= 13427.44 J State B to State C. (2--->3) Q= 10070.58 J State C to State D. (3--->4) Q=8951.63J State D to State A. (4--->1) Q= 1678.43 J

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