Really stuck on this practice problem, any help would be appreciated! 0.1 mol of
ID: 2121106 • Letter: R
Question
Really stuck on this practice problem, any help would be appreciated!
0.1 mol of steam (c'p = 34.47 J/mol*K c'v = 28.03 J/mol*K) undergoes the following cyclic process. it is first compressed at a constant pressure of 1.00 atm from 6.09 liters to 3.06 liters, then it undergoes a pressure increase to 1.5 atm at constant volume, then undergoes a expansion at constant pressure to 4.5L followed by an adiabatic expansion back to its original state.
a)Find the temperatures at points A, B, C, and D (use PV=nRT) point A is 6.09L and 1 atm, point B is 3.06 L and 1 atm, point C is 3.06L and 1.5 atm, and point D is 4.5L and 1.5 atm
b)Find the change in internal energy for each part of this process and the change in internal energy for the entire cycle
c) Find the Q for each part of the process and the Q for the entire process.
d) Find the work done for each part of this process and the total work done during the cycle.
e) Find the change in entropy for each part of the process and for the entire cycle.
f) is this an engine or a refrigerator?
g) If this is an engine, compute the efficiency. If it's a refrigerator, compute the COP instead.
Explanation / Answer
(a) T(A) = pV/nR = (101325 Pa)(0.000609 m^3) / (0.1mol*8.314 J/molK) = 742 K
T(B) = T(A) times V(B)/V(A) = 742.2 K * (3.06/6.09) = 373 K
T(C) = T(B) times p(C)/p(B) = 1.5 * (372.9K) = 559 K
T(D) = T(C) times V(D)/V(C) = (4.5/3.06) (559.4 K) = 823 K
(b) There is no change in internal energy for the cycle as a whole.
The internal energy change for each part of the process is
(28.03 J/mol K) times (0.1 mol) times (the temperature change on that branch).
The APPROXIMATE values are
-1050 J, +600 J, +700 J, -250 J
but use a calculator (I didn't)
(c,d) From B to C, no work is done, so the heat added is equal to the change in internal energy.
From D to A, there is no heat transfer, so the work done by the steam is equal to the internal energy drop,
possibly about 250 J.
From A to B, the work done ON the steam is (101325 Pa)(0.00609 m^3 - 0.00306 m^3),
about 300 J; the heat rejected is the (absolute value of) the internal energy drop plus
the work done on the steam.
From C to D, the work done BY the steam is (1.5)(101325 Pa)(0.0045 m^3 - 0.00306 m^3),
about 230 J; the heat input is the internal energy increase plus the work done by the steam.
The net work for the cycle can be figured out by adding the four "work" figures above
(positive for CD and DA, negative for AB, zero for BC), possibly about 300 J.
The net heat input for the cycle is the same as the net work done by the steam.
Another way would be to use the Cp to get the heat for CD and AB,
I guess that's why they asked about the heat before the work.
(e) No change for DA.
For AB and CD, use Cp ln(T2/T1)
For BC use Cv ln(Tc/Tb)
Should all add up to zero -- no entropy change for whole cycle.
(f) It's an engine. Heat is added and the steam does net positive work.
(g) efficiency = (net work)/(sum of ADDED heat)
The denominator is the heat addition for BC and CD only
...don't count the exhausted heat one way or the other.
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