I. Analysis of the problem. Choosing the air in the cylinder to be the system, recognizing that for an ideal gas at constant temperature U is constant so that ΔU = 0, and neglecting the kinetic and potential energy terms for the gas (since the mass of 1 mol of air is only 29 g), we obtain the
following energy balance equation:
0=Q-\int P d V (a)
The total work done by the gas in lifting and accelerating the piston and the weights against the frictional forces, and in expanding the system volume against atmospheric pressure is contained in the -\int P d V term. To see this we recognize that the laws of classical mechanics apply to the piston and weights, and equate, at each instant, all the forces on the piston and weights to their acceleration,
\text { Forces on piston and weights }=\left(\begin{array}{c}\text { Mass of piston } \\\text { and weights }\end{array}\right) \times \text { Acceleration }
and obtain
\left[P \times A-P_{ atm } \times A-( W +\omega) g+F_{ fr }\right]=( W +\omega) \frac{d v}{d t} (b)
Here we have taken the vertical upward (+z) direction as being positive and used P and Patm to represent the pressure of the gas and atmosphere, respectively; A the piston area; ω its mass; W the mass of the weights on the piston at any time; \upsilon the piston velocity; and F_{ fr } the frictional force, which is proportional to the piston velocity. Recognizing that the piston velocity \upsilon is equal to the rate of change of the piston height h or the gas volume V , we have
\upsilon=\frac{d h}{d t}=\frac{1}{A} \frac{d V}{d t}
Also we can solve Eq. b for the gas pressure:
P=P_{ atm }+\frac{( W +\omega)}{A} g-\frac{F_{ fr }}{A}+\frac{( W +\omega)}{A} \frac{d v}{d t} (c)
At mechanical and thermodynamic equilibrium \text { (i.e., when } d \upsilon / d t=0 \text { and } \upsilon =0 \text { ) }, we have
P=P_{ atm }+\frac{( W +\omega)}{A} g (d)
With these results, the total work done by the gas can be computed. In particular,
\begin{aligned}\int P d V &=\int\left[P_{ atm }+\left(\frac{ W +\omega}{A}\right) g-\frac{F_{ fr }}{A}+\frac{( W +\omega)}{A} \frac{d v}{d t}\right] d V \\&=\left[P_{ atm }+\frac{( W +\omega)}{A} g\right] \Delta V-\frac{1}{A} \int F_{ fr } d V+\frac{( W +\omega)}{A} \int \frac{d v}{d t} d V\end{aligned}
This equation can be simplified by rewriting the last integral as follows:
\frac{1}{A} \int \frac{d \upsilon }{d t} d V=\frac{1}{A} \int \frac{d \upsilon }{d t} \frac{d V}{d t} d t=\int \frac{d \upsilon }{d t} \upsilon d t=\frac{1}{2} \int \frac{d \upsilon ^{2}}{d t} d t=\Delta\left(\frac{1}{2} \upsilon ^{2}\right)
where the symbol Δ indicates the change between the initial and final states. Next, we recall from mechanics that the force due to sliding friction, here F_{ fr }, is in the direction opposite to the relative velocity of the moving surfaces and can be written as
F_{ fr }=-k_{ fr } \upsilon
where k_{ fr } is the coefficient of sliding friction. Thus, the remaining integral can be written as
\frac{1}{A} \int F_{ fr } d V=-\frac{1}{A} \int k_{ fr } \upsilon \frac{d V}{d t} d t=-k_{ fr } \int \upsilon ^{2} d t
The energy balance, Eq. a, then becomes
Q=\int P d V=P_{ atm } \Delta V+( W +\omega) g \Delta h+k_{ fr } \int \upsilon ^{2} d t+( W +\omega) \Delta\left(\frac{1}{2} \upsilon ^{2}\right) (e)
This equation relates the heat flow into the gas (to keep its temperature constant) to the work the gas does against the atmosphere in lifting the piston and weights (hence increasing their potential energy) against friction and in accelerating the piston and weights (thus increasing their kinetic energy).
The work done against frictional forces is dissipated into thermal energy, resulting in a higher temperature at the piston and cylinder wall. This thermal energy is then absorbed by the gas and appears as part of Q. Consequently, the net flow of heat from a temperature bath to the gas is
Q^{ NET }=Q-k_{ fr } \int \upsilon^{2} d t (f)
Since heat will be transferred to the gas, and the integral is always positive, this equation establishes that less heat will be needed to keep the gas at a constant temperature if the expansion occurs with friction than in a frictionless process.
Also, since the expansion occurs isothermally, the total heat flow to the gas is, from Eq. a,
Q=\int_{V_{i}}^{V_{f}} P d V=\int_{V_{i}}^{V_{f}} \frac{N R T}{V} d V=N R T \ln \frac{V_{f}}{V_{i}} (g)
Combining Eqs. e and f, and recognizing that our interest here will be in computing the heat and work flows between states for which the piston has come to rest (v = 0), yields
Q^{ NET }=Q-k_{ fr } \int \upsilon ^{2} d t=P_{ atm } \Delta V+( W +\omega) g \Delta h=P \Delta V=-W^{ NET } (h)
where P is the equilibrium final pressure given by Eq. d. Also from Eqs. f, g, and h, we have
Q^{ NET }=-W^{ NET }=N R T \ln \left(\frac{V_{2}}{V_{1}}\right)-k_{ fr } \int \upsilon ^{2} d t (i)
Here W^{ NET } represents the net work obtained by the expansion of the gas (i.e., the work obtained in raising the piston and weights and in doing work against the atmosphere).
The foregoing equations can now be used in the solution of the problem. In particular, as a weight is removed, the new equilibrium gas pressure is computed from Eq. d, the resulting volume change from the ideal gas law, W^{ NET } \text { and } Q^{ NET } from Eq. h, and the work against friction from Eq. i. There is, however, one point that should be mentioned before we proceed with this calculation. If there were no mechanism for the dissipation of kinetic energy to thermal energy (that is, sliding friction between the piston and cylinder wall, and possibly also viscous dissipation on expansion and compression of the gas due to its bulk viscosity), then when a weight was removed the piston would be put into a perpetual oscillatory motion. The presence of a dissipative mechanism will damp the oscillatory motion. (As will be seen, the value of the coefficient of sliding friction, kfr, does not affect the amount of kinetic energy ultimately dissipated as heat. Its value does, however, affect the dynamics of the system and thus determine how quickly the oscillatory motion is damped.)
II. The numerical solution. First, the mass W of the weights is computed using Eq. d and the fact that the initial pressure is 2.043 bar. Thus,
P=2.043 bar =1.013 bar +\frac{(5+ W ) kg }{0.01 m ^{2}} \times 9.807 \frac{ m }{ s ^{2}} \times \frac{1 Pa }{ kg /\left( m s ^{2}\right)} \times \frac{1 bar }{10^{5} Pa }
or
(5 + W) kg = 105.0 kg
so that W = 100 kg.
The ideal gas equation of state for 1 mol of air at 25°C is
\begin{aligned}P V=N R T &=1 mol \times 8.314 \times 10^{-5} \frac{ bar m ^{3}}{ mol K } \times(25+273.15) K \\&=2.479 \times 10^{-2} \text { bar } m ^{3}=2479 J\end{aligned} (j)
and the initial volume of the gas is
V=\frac{2.479 \times 10^{-2} \text { bar } m ^{3}}{2.043 \text { bar }}=1.213 \times 10^{-2} m ^{3}
Process a
The 100-kg weight is removed. The equilibrium pressure of the gas (after the piston has stopped oscillating) is
P_{1}=1.013 bar +\frac{5 kg \times 9.807 m / s ^{2}}{0.01 m ^{2}} \times \frac{10^{-5} bar }{ kg /\left( m s ^{2}\right)}=1.062 bar
and the gas volume is
V_{1}=\frac{2.479 \times 10^{-2} \text { bar } m ^{3}}{1.062 \text { bar }}=2.334 \times 10^{-2} m ^{3}
Thus
\begin{aligned}\Delta V &=(2.334-1.213) \times 10^{-2} m ^{3}=1.121 \times 10^{-2} m ^{3} \\-W^{ NET } &=1.062 bar \times 1.121 \times 10^{-2} m ^{3} \times 10^{5} \frac{ J }{ bar m ^{3}} \\&=1190.5 J =Q^{ NET }\end{aligned}
and
Q=N R T \ln \frac{V_{1}}{V_{0}}=2479 J \ln \frac{2.334 \times 10^{-2}}{1.213 \times 10^{-2}}=1622.5 J
(since NRT = 2479 J from Eq. j). Consequently, the work done against frictional forces (and converted to thermal energy), which we denote by W_{ fr }, is
-W_{ fr }=Q-Q^{ NET }=(1622.5-1190.5) J =432 J
The total useful work obtained, the net heat supplied, and the work against frictional forces are given in Table 1. Also, the net work, P ΔV , is shown as the shaded area in the accompanying figure, together with the line representing the isothermal equation of state, Eq. j. Note that in this case the net work is that of raising the piston and pushing back the atmosphere.
Table 1
\begin{array}{lccc}\hline & -W^{ NET }=Q^{ NET } & Q & -W_{ fr } \\\text { Process } & ( J ) & ( J ) & ( J ) \\\hline a & 1190.5 & 1622.5 & 432.0 \\b & 1378.7 & 1622.5 & 243.8 \\c & 1493.0 & 1622.5 & 129.5 \\d & 1622.5 & 1622.5 & 0 \\\hline\end{array}
Process b
The situation here is similar to that of process a, except that the weight is removed in two 50-kg increments. The pressure, volume, work, and heat flows for each step of the process are given in Table 2, and -W_{ i }^{ NET }=P_{ i }(\Delta V)_{ i }, the net work for each step, is given in the figure.
Process c
Here the weight is removed in four 25-kg increments. The pressure, volume, work, and heat flows for each step are given in Table 2 and summarized in Table 1. Also, the net work for each step is given in the figure.
Table 2
\begin{array}{lccccc}\hline \text { Process b } & & & & & \\& P & V & -W_{ i }^{ NET }=P_{ i }(\Delta V)_{ i } & Q=N R T \ln \frac{V_{ i }}{V_{ i -1}} & -W_{ fr } \\\text { Stage } & \begin{array}{c}P \\( bar )\end{array} & \begin{array}{c}V \\\left( m ^{3}\right)\end{array} & ( J ) & ( J ) & ( J ) \\\hline 0 & 2.043 & 1.213 \times 10^{-2} & & & \\1 & 1.552 & 1.597 \times 10^{-2} & 596.0 & 681.8 & 85.8 \\2 & 1.062 & 2.334 \times 10^{-2} & 782.7 & 940.7 & 158.0 \\\text { Total } & & & 1378.7 & 1622.5 & 243.8 \\\hline \text { Process c } & & & & & \\& & & -W_{ i }^{ NET }=P_{ i }(\Delta V)_{ i } & Q=N R T \ln \frac{V_{ i }}{V_{ i -1}} & -W_{ fr } \\& P & V & ( J ) & ( J ) & ( J ) \\\text { Stage } & (\text { bar }) & \left( m ^{3}\right) & & & & \\\hline 0 & 2.043 & 1.213 \times 10^{-2} & & & & \\1 & 1.798 & 1.379 \times 10^{-2} & 298.5 & 318.0 & 19.5\\2 & 1.552 & 1.597 \times 10^{-2} & 338.3 & 363.8 & 25.5 \\3 & 1.307 & 1.897 \times 10^{-2} & 392.1 & 426.8 & 34.7 \\4 & 1.062 & 2.334 \times 10^{-2} & 464.1 & 513.9 & 49.8 \\\text { Total } & & & 1493.0 & 1622.5 & 129.5 \\\hline\end{array}
Process d
The computation here is somewhat more difficult since the number of stages to the calculation is almost infinite. However, recognizing that in the limit of the mass of a grain of sand going to zero there is only a differential change in the pressure and volume of the gas and negligible velocity or acceleration of the piston, we have
-W^{ NET }=\sum_{i} P_{ i }\left(\Delta V_{ i }\right) \rightarrow \int P d V=N R T \ln \frac{V_{f}}{V_{i}}=Q^{ NET }
and W_{ fr }=0, since the piston velocity is essentially zero at all times. Thus,
-W^{ NET }=Q^{ NET }=Q=1 mol \times 8.314 \frac{ J }{ mol K } \times 298.15 K \times \ln \frac{2.334 \times 10^{-2}}{1.213 \times 10^{-2}} = 1622.5 J
This result is given in Table 1 and the figure.
Comments
Several points are worth noting about this illustration. First, although the initial and final states of the gas are the same in all three processes, the useful or net work obtained and the net heat required differ. Of course, by the energy conservation principle, it is true that –W NET = QNET for each process. It is important to note that the most useful work is obtained for a given change of state if the change is carried out in differential steps, so that there is no frictional dissipation of mechanical energy to thermal energy (compare process d with processes a, b, and c). Also, in this case, if we were to reverse the process and compress the gas, it would be found that the minimum work required for the compression is obtained when weights are added to the piston in differential (rather than finite) steps. (See Problem 3.29.)
It should also be pointed out that in each of the four processes considered here the gas did 1622.5 J of work on its surroundings (the piston, the weights, and the atmosphere) and absorbed 1622.5 J of heat (from the thermostatic bath maintaining the system temperature constant and from the piston and cylinder as a result of their increased temperature due to frictional heating). We can see this from Table 1, since -\left(W^{ NET }+W_{ fr }\right)=Q=1622.5 J for all four processes. However, the fraction of the total work of the gas obtained as useful work versus work against friction varies among the different processes.
