WWCY said:
Let me try to rephrase what your explanation to see if I understood it. The process that was described here was irreversible, and consisted of changes ##\Delta U, \Delta V##. Because S is a state variable, it doesn't matter how the system transforms as long as it reaches the same end state. Therefore the
change in entropy can also be described a quasistatic process that consists of the same changes.
Correct
If that is the case, ##\Delta U## suggests a change in T, and ##\Delta V## suggests a change in P. Yet, the final answer shows that ##\Delta S = \frac{1}{T_0} (\Delta U + P_0\Delta V)## where ##T_0## and ## P_0## are the temperature and pressure of the gas before the compression. Why is that?
The answer is not quite correct, but it is a good approximation for this specific problem (since the volume change is very small). So, I am going to do the problem over for you so that we can compare "the final answer" with the exact answer.
Also, if I took a reversible path to the end-state of the system, shouldn't there be no new entropy created, since quasistatic, reversible processes don't create entropy?
Quasistatic reversible processes don't create entropy. But, entropy can be transferred between the system and surroundings if the reversible path is not adiabatic.
Now, let's solve the problem the right way. We are going to assume that the irreversible path is adiabatic. The first step is to determine the final thermodynamic equilibrium state of the system using the first law of thermodynamics. For our irreversible process, we have:
$$nC_v\Delta T=-P_{ext}\Delta V$$where, for this constant-external-force problem (with ##P_{ext}## constant) with $$P_{ext}=2000N/0.01m^2=200000Pa$$ The number of moles of air is given by the ideal gas law: $$n=\frac{P_0V_0}{RT_0}=\frac{10^5(1)}{(8314.)(300)}=0.0401\ moles$$ The molar heat capacity of the air is: $$C_v=2.5R=(2.5)(8.314)=20.785\ \frac{J}{mole.K}$$The work done by the gas on the surroundings is:$$W=P_{ext}\Delta V=-(200000)(0.001)(0.01)=-2 J$$So the temperature rise is:
$$\Delta T=\frac{2}{nC_v}=\frac{2}{(0.0401)(20.785)}=2.40 C$$
The final temperature is then $$T_f=300+2.4=302.4\ K$$and the final volume is $$V_f=1 - 0.01 = 0.99\ liters$$After the gas has equilibrated in the cylinder, its final pressure can be determined from the ideal gas law: $$P_f=\frac{nRT_f}{V_f}=\frac{(0.0401)(0.08314)(302.4)}{0.99}=1.0184\ bars=1.0184\times 10^5 Pa$$
Now let's get the exact answer for the entropy change. If we evaluate the entropy change either by following a reversible path or, equivalently, by integrating the thermodynamic identity between the initial and final states, we obtain:
$$\Delta S=nC_v\ln{\frac{T_f}{T_0}}+nR\ln{\frac{V_f}{V_0}}=nC_v\ln{\left(1+\frac{\Delta T}{T_0}\right)}+nR\ln{\left(1+\frac{\Delta V}{V_0}\right)}$$But we know from what we did above that:
$$nC_v=\frac{\Delta U}{\Delta T}$$ and $$nR=\frac{P_0V_0}{T_0}$$
If we substitute these relationships into our equation for the exact entropy change, we obtain:
$$\Delta S=\frac{\Delta U}{\Delta T}\ln{\left(1+\frac{\Delta T}{T_0}\right)}+\frac{P_0V_0}{T_0}\ln{\left(1+\frac{\Delta V}{V_0}\right)}$$
Note the similarity now to the approximate answer which was portrayed in your course as the exact answer. We know that in this problem, ##\frac{\Delta T}{T_0}<<1## and ##\frac{\Delta V}{V_0}<<1##. If we make use of these conditions to expand the logarithmic terms in our exact answer in a Taylor series, and retain only the first terms in the expansions, we obtain: $$\ln{\left(1+\frac{\Delta T}{T_0}\right)}\approx \frac{\Delta T}{T_0}$$and $$\ln{\left(1+\frac{\Delta V}{V_0}\right)}\approx \frac{\Delta V}{V_0}$$If we substitute these relationships into our exact equation for ##\Delta S##, we obtain the following
approximate equation for ##\Delta S##:
$$\Delta S\approx \frac{\Delta U+P_0\Delta V}{T_0}$$
So the equation they gave in your course is only an approximation to the exact entropy change.
