Polytropic Processes, Free Expansion and Cycles

One law for every process

Real compressors and engines rarely follow an exact isotherm or adiabat. Their paths are well described by \(pV^x = \text{constant}\), where the number \(x\) is the polytropic index. A process of this kind is a polytropic process. The four processes of the last two lessons are all special cases. This lesson uses \(R = 8.31\,\text{J/(mol K)}\), with \(C_v = \tfrac32 R\) for a monatomic gas and \(\tfrac52 R\) for a diatomic gas.

Put in the special values. With \(x = 0\), \(p\) is constant: isobaric. With \(x = 1\), \(pV\) is constant: isothermal. With \(x = \gamma\), the process is adiabatic. As \(x\) grows very large, \(p^{1/x}V = \text{constant}\) tends to \(V = \text{constant}\): isochoric. Other values are allowed too. A gas whose pressure rises in proportion to its volume has \(p/V\) constant, so \(x = -1\).

The work follows as in the adiabatic case. Integrating \(p = KV^{-x}\) gives \(W = (p_1V_1 - p_2V_2)/(x - 1)\), which is \(nR(T_1 - T_2)/(x - 1)\). The same steps as in the last lesson also give \(TV^{x-1} = \text{constant}\) and \(Tp^{(1-x)/x} = \text{constant}\). A compressor with \(x = 1.3\) takes air at \(300\,\text{K}\) from \(1\,\text{atm}\) to \(6\,\text{atm}\). It delivers it at \(T = 300 \times 6^{0.3/1.3} = 454\,\text{K}\).

The heat is found from the first law, one mole at a time. The derivation below gives the molar heat capacity of the process.

The formula passes every test. At \(x = 0\) it gives \(C_v + R = C_p\). At \(x = \gamma\) it gives \(C_v - C_v = 0\), since \(R/(1-\gamma) = -C_v\): no heat, as an adiabat should. As \(x\) grows very large it gives \(C_v\). At \(x = 1\) it blows up, and rightly so. An isothermal gas takes in heat with no change of temperature, so the heat per kelvin is infinite.

Move the slider in the figure above from \(x = 0\) upwards. The curve swings from a flat isobar through the isotherm and the adiabat towards the vertical line. The work from \(25\) to \(50\,\text{L}\) falls from \(2500\,\text{J}\) to \(1733\,\text{J}\) at \(x = 1\), and to \(1513\,\text{J}\) at \(x = 1.4\). Between \(x = 1\) and \(x = 1.4\) the heat capacity is negative: at \(x = 1.2\), \(C = 2.5R - 5R = -2.5R\).

What does a negative heat capacity mean? Between \(x = 1\) and \(x = \gamma\), an expanding gas does more work than the heat it takes in. It makes up the difference from its internal energy, so it cools while heat flows in. Run backwards, the gas warms while it gives out heat. Nothing is wrong with the first law. \(C\) is simply heat per kelvin, and here heat and temperature change in opposite directions.

A gas rushing into nothing

A rigid, insulated box is split in two by a partition. One half holds a gas, and the other half is empty. Now the partition is removed. The gas rushes into the empty half and soon fills the whole box. This is a free expansion.

Apply the first law to Fig 6.1. The gas pushes against nothing as it spreads, because a vacuum offers no pressure. The walls do not move. So the gas does no work: \(W = 0\). The walls are insulated, so \(Q = 0\). Then \(\Delta U = Q - W = 0\). The internal energy of an ideal gas depends on \(T\) alone, so the temperature is unchanged, \(300\,\text{K}\). By \(pV = nRT\), doubling the volume halves the pressure, to \(1\,\text{atm}\).

The start and the end have the same temperature. Is the process isothermal? No. While the gas rushes, it is dense in one place and thin in another, so it has no single pressure or temperature. It is not quasi-static and cannot be drawn on the \(p\)-\(V\) diagram. Only its two end states are points. Its end states happen to lie on one isotherm, but the gas never followed it. It is also an irreversible process. The gas never gathers itself back into one half on its own.

A real gas does cool a little. Its molecules attract one another, and spreading them apart takes energy. With \(Q = 0\) and \(W = 0\), that energy must come from the kinetic energy of the molecules. So the temperature drops slightly. An ideal gas has no attraction and does not cool at all.

A cycle, leg by leg

A cycle problem looks long, but it has one method. Split the cycle into its legs. For each leg find \(W\) from the area, \(\Delta U\) from \(nC_v\,\Delta T\), and \(Q\) from the first law. Put the three numbers in a row of a table. For an ideal gas \(nC_v\,\Delta T = (C_v/R)\,\Delta(pV)\), so the temperatures are not even needed. For a monatomic gas, \(\Delta U = \tfrac32\,\Delta(pV)\).

Run the chips of Fig 6.2 one by one. A to B warms the gas at constant volume: \(W = 0\) and \(Q = \Delta U = \tfrac32(3000 - 1000) = 3000\,\text{J}\). B to C expands at \(300\,\text{kPa}\): \(W = 300 \times 20 = 6000\,\text{J}\) and \(\Delta U = 9000\,\text{J}\), so \(Q = 15\,000\,\text{J}\). C to D cools at constant volume: \(Q = \Delta U = -9000\,\text{J}\). D to A is compressed at \(100\,\text{kPa}\): \(W = -2000\,\text{J}\), \(\Delta U = -3000\,\text{J}\) and \(Q = -5000\,\text{J}\).

Add the columns. The \(\Delta U\) column gives \(3000 + 9000 - 9000 - 3000 = 0\), as it must. The \(W\) column gives \(4000\,\text{J}\), the area of the rectangle, \(200 \times 20\). The \(Q\) column also gives \(4000\,\text{J}\). The gas absorbed \(18\,000\,\text{J}\) on the first two legs and turned \(4000/18\,000 = 22\,\%\) of it into work. The next chapter calls this fraction the efficiency.