A monatomic gas is taken through the sequence of operations connecting the point
ID: 3895467 • Letter: A
Question
A monatomic gas is taken through the sequence of operations connecting the points A-B-C-A (in that order) as shown in PV diagram.
a) given that the temperature at point A is 1200K, find the temperatures of points B and C. Label the temperatures of points A, B, and C on the graph.
b) calculate the number of moles of the gas
c) In the table, enter the following values for each step in the cycle and then sum those values over the entire cycle
- W being work done on the gas
- Q being the change in the thermal energy of the gas
- ?E the heat transfer into the gas
d) suppose that this system represents the operation of a heat engine. What is the efficiency of this engine (defined as the magnitude of the net work for the cycle divided by the heat input for the cycle)?
e) what is the maximum possible efficiency for any heat engine that operates between the same temperatures as this system?
Explanation / Answer
a)
Ta = 1200 K
Pa*Va/Ta = Pb*Vb/Tb = Pc*Vc/Tc
4000*3/1200 = 2000*3/Tb = 2000*1/Tc
Tb = 600 K
Tc = 200 K
b)
n = Pa*Va / (Ru*Ta)
= 4000*3 / (8.314*1200)
n = 1.2028 moles
c)
Wab = 0 (since constant volume process)
Qab = n*Cv*(Tb - Ta) = 1.2028*12.5*(600 - 1200) = -9020.9 J
Eab = Wab + Qab = -9020.9 J
Wbc = Pb*(Vc - Vb) = 2000*(1 - 3) = -4000 J
Qbc = n*Cv*(Tc - Tb) = 1.2028*12.5*(200 - 600) = -6014 J
Ebc = Wbc + Qbc = -10014 J
Wca = (Pc + Pa)/2 *(Va - Vc) = (2000 + 4000)/2 *(3 - 1) = 6000 J
Qca = n*Cv*(Ta - Tc) = 1.2028*12.5*(1200 - 200) = 15035 J
Eca = Wca + Qca = 21035 J
W_cycle = Wab + Wbc + Wca = 2000 J
Q_cycle = Qab + Qbc + Qca = 0
E_cycle = Eab + Ebc + Eca = 2000 J
d)
Efficieny = Net work / Heat input
= W_cycle / Eca
= 2000 / 21035
= 0.095 or 9.5 %
e)
Max. efficiency = 1 - T_low / T_high
= 1 - 200 / 1200
= 0.833 or 83.3 %