03/10/2014

Analyzing the Cycle:

Here comes the detail of whole cycle:

• The work has only be done between point 1 to 2 and 3 to 4 because the change of volume is similar to displacement since W=F*d.
• During the step 2 to 3 and 3 to 4, the heat energy transferred to the gas from a reservoir because delta E(int) = Q - W, 2 -> 3 delta E = -3450 J = Q - W ,but W is 0, so -3450 J = Q also from step 3 to 4 :delta E(int) = Q - W, 3->4 delta E =-2370 J = Q - W, and W = -1580 (P*delta V) -2370 - 1580 = Q = -3950 .
• During the step 1 to 2 and 4 to 1, the heat energy transferred to a reservoir from the gas because delta E(int) = Q -W ,and since the Q is positive after calculation the heat is coming into the reservoir.
• The work done is part A the cycle is positive by W = P*delta V which change V is positive cause the W become positive. However, the part C is negative since its change V is negative.
• The net work is equal to sum of work which 2040 + (-1580) = 460 J.

No. 6

Q=mCpdeltaT

Cp= Q/mdeltaT

Cp = 102 J / (1 mole * 5 K ) 

Cp = 20.4 J / mole

No. 7

Cp = 12465 J / (1 mole * 600 K )

Cp= 20.8 J / mole

No . 8

Here comes the way to derive the constant pressure.

Carnot Cycle:

First we solve E on each points, and get the changing E by subtracting two points.
A to B and C to D are both isothermal expansion and compression.
B to C and D to A are both adiabatic expansion and compression.

For isothermal, we use W = P1V1ln(V2/V1) to solve the work then since Q=W in isothermal , we don't need to solve Q again. Of course the delta U  is zero.

For adiabatic, we use W = (P1V1-P2V2)/(r-1) to solve work, and Q = 0 in adiabatic. So the delta U = Q -W  => delta U = -W