Q#9.42
A thermodynamic system is taken from state a to state c in Fig. P19.42 along either path abc or path adc. Along path abc, the work W done by the system is 450 J. Along path adc, W is 120 J. The internal energies of each of the four states shown in the figure are Ua = 150 J, Ub = 240 J, Uc = 680 J and Ud = 330 J. Calculate the heat flow Q for each of the four processes ab, bc, ad, and dc. In each process, does the system absorb or liberate heat?
Answer:
ΔU = Q − W, W = 0 when ΔV = 0.
For each process, Q = ΔU + Q. No work is done in the processes ab and dc, and so
Wab=Wabc = 450 J and Wad=Wadc = 120 J.
The heat flow for each process is: for ab, Q = 90 J.
For bc, Q = 440 J + 450 J = 890 J
For ad, Q = 180 J + 120 J = 300 J
For dc, Q = 350 J.
Heat is absorbed in each process. Note that the arrows representing the processes all point in the direction of increasing temperature (increasing U).
ΔU is path independent so is the same for paths adc and abc. Qadc = 300 J + 350 J = 650 J
Qabc = 90 J + 890 J = 980 J. Q and W are path dependent and are different for these two paths.
Q#19.43
A volume of air (assumed to be an ideal gas) is first cooled without changing its volume and then expanded without changing its pressure, as shown by the path abc in Fig. P19.43.
(a) How does the final temperature of the gas compare with its initial temperature?
(b) How much heat does the air exchange with its surroundings during the process abc? Does the air absorb heat or release heat during this process? Explain.
(c) If the air instead expands from state a to state c by the straight-line path shown, how much heat does it exchange with its surroundings?
Answer:
Use pV = nRT to calculate TcTa. Calculate ΔU and W and use ΔU = Q − W to obtain Q.
For path ac, the work done is the area under the line representing the process in the pV-diagram.
(a) TcTa=pcVcpaVa
TcTa=(1.0×105 Pa)(0.060 m3)(3.0×105 Pa)(0.020 m3)=1.00
So, Tc=Ta
(b) Since , Tc=Ta, ΔU=0 for process abc. For ab, ΔV = 0 and Wab = 0. For bc, p is constant and
Wbc=pΔT
Q#19.44
Three moles of argon gas (assumed to be an ideal gas) originally at a pressure of 1.50 ×105 Pa and a volume of 0.0280 m3 are first heated and expanded at constant pressure to a volume of 0.0435 m3 then heated at constant volume until the pressure reaches 3.50×104 Pa, then cooled and compressed at constant pressure until the volume is again 0.0280 m3 and finally cooled at constant volume until the pressure drops to its original value of 1.50 ×104 Pa.
(a) Draw the pV-diagram for this cycle.
(b) Calculate the total work done by (or on) the gas during the cycle.
(c) Calculate the net heat exchanged with the surroundings. Does the gas gain or lose heat overall?
Answer:
For a cycle, ΔU = 0 and Q = W. Calculate W.
The magnitude of the work done by the gas during the cycle equals the area enclosed by the cycle in the pV-diagram.
(a) The cycle is sketched in Figure 19.44.
|W| = (3.50×104 Pa−1.50×104 Pa)(0.0435 m3−0.0280 m3)=+310 J
More negative work is done for cd than positive work for ab and the net work is negative. W = −310 J
(c) Q = W = −310 J. Since Q < 0, the net heat flow is out of the gas.
During each constant pressure process W = pΔV and during the constant volume process W = 0.
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