Constant volume
The first law, \(Q = \Delta U + W\), holds for every process. It becomes most useful when one quantity is held fixed. This lesson takes three such processes in turn, each for an ideal gas. It uses \(R = 8.31\,\text{J/(mol K)}\) and \(1\,\text{atm} = 1.013 \times 10^5\,\text{Pa}\), with \(C_v = \tfrac32 R\) for a monatomic gas and \(\tfrac52 R\) for a diatomic gas.
In an isochoric process the volume does not change. A gas heated in a rigid steel vessel is the example. The walls do not move, so the gas does no work: \(W = 0\). The first law then gives \(Q = \Delta U = nC_v\,\Delta T\). Every joule of heat raises the internal energy. At a fixed volume, \(pV = nRT\) makes the pressure rise in step with the temperature. On the \(p\)-\(V\) diagram the path is a vertical line.
An empty \(5\,\text{L}\) pressure cooker holds air at \(300\,\text{K}\) and \(1\,\text{atm}\). With the lid locked and the whistle held down, it is heated to \(400\,\text{K}\). Air is diatomic. Find the work done, the heat taken in and the final pressure.
The amount of air is \(n\) \(= pV/RT\) \(= 1.013 \times 10^5 \times 0.005/(8.31 \times 300)\) \(= 0.203\,\text{mol}\).
The volume is fixed, so \(W = 0\).
\(Q\) \(= \Delta U\) \(= nC_v\,\Delta T\) \(= 0.203 \times \tfrac52 \times 8.31 \times 100\) \(= 422\,\text{J}\).
At constant volume \(p \propto T\), so \(p_2 = 1\,\text{atm} \times 400/300 = 1.33\,\text{atm}\).
Two moles of a monatomic ideal gas in a rigid vessel are heated from \(300\,\text{K}\) to \(450\,\text{K}\). How much heat does the gas take in, in joules?
A car tyre is close to a vessel of fixed volume. On a long drive in summer its air warms from \(27\,\text{°C}\) to about \(57\,\text{°C}\), from \(300\,\text{K}\) to \(330\,\text{K}\). The absolute pressure rises by the same factor, \(1.1\). A tyre at \(3.2\,\text{atm}\) absolute (\(2.2\,\text{atm}\) on the gauge) reaches \(3.52\,\text{atm}\), and the gauge reads \(2.52\). So petrol pumps advise checking the pressure before a drive, when the tyres are cold.
Constant pressure
In an isobaric process the pressure stays fixed. A gas under a free piston that carries a steady load is the example. As the gas warms it expands and lifts the load. The work is \(W = p\,\Delta V\), and by \(pV = nRT\) this is \(nR\,\Delta T\). The internal energy still rises by \(nC_v\,\Delta T\). So the heat is \(Q = nC_v\,\Delta T + nR\,\Delta T = nC_p\,\Delta T\). On the \(p\)-\(V\) diagram the path is a horizontal line.
Play the isochoric and isobaric chips in Fig 4.1, one after the other. Both warm the gas by \(50\,\text{K}\), so both raise \(U\) by the same \(\tfrac32 R \times 50 = 623\,\text{J}\). The isochoric gas needs just that much heat. The isobaric gas lifts its loaded piston as well, doing \(R \times 50 = 416\,\text{J}\) of work. So it needs \(Q = \tfrac52 R \times 50 = 1039\,\text{J}\). The \(\Delta U\) bar is the same length in both chips; only the work bar differs.
One mole of a diatomic ideal gas is heated from \(300\,\text{K}\) to \(400\,\text{K}\) at constant pressure. Which set of values is right?
Three moles of an ideal gas are heated by \(40\,\text{K}\) at constant pressure. How much work does the gas do, in joules?
Constant temperature
In an isothermal process the temperature stays fixed. The gas sits in a cylinder with thin conducting walls, on a large body at steady temperature, a heat reservoir. The piston moves slowly, so heat has time to flow in or out and keep \(T\) steady. For an ideal gas \(U\) depends on \(T\) alone, so \(\Delta U = 0\). The first law gives \(Q = W\): all the heat taken in leaves as work.
The path is the curve \(pV = nRT\) with \(T\) fixed, a hyperbola called an isotherm. The pressure is \(p = nRT/V\), so the work is an integral that the derivation below works out.
\[ W = \int_{V_1}^{V_2} p\,dV \]
Why. the work done by a gas in any quasi-static process.
\[ p = \dfrac{nRT}{V} \]
Why. the ideal gas equation, with \(T\) the same at every stage.
\[ W = nRT\int_{V_1}^{V_2} \dfrac{dV}{V} \]
Why. \(n\), \(R\) and \(T\) are constants, so they come out of the integral.
\[ W = nRT\ln\dfrac{V_2}{V_1} = nRT\ln\dfrac{p_1}{p_2} \]
Read it. the integral of \(1/V\) is \(\ln V\); and \(p_1V_1 = p_2V_2\) at fixed \(T\), so \(V_2/V_1 = p_1/p_2\).
One mole at \(300\,\text{K}\) that doubles its volume does \(W\) \(= 8.31 \times 300 \times \ln 2\) \(= 2493 \times 0.693\) \(= 1728\,\text{J}\). It takes in the same \(1728\,\text{J}\) of heat from the reservoir. In a compression the logarithm is negative. The gas does negative work, and it gives the same amount of heat back to the reservoir.
Fig 4.2 shows why two isotherms never cross. At any one volume, the pressure is proportional to \(T\), so the hotter curve always lies above the colder one. If two isotherms met, one state would have two temperatures. In the isothermal chip of Fig 4.1 the state point slides down the \(300\,\text{K}\) curve. The \(Q\) and \(W\) bars grow together, and the \(\Delta U\) bar stays empty.
Two moles of an ideal gas at \(300\,\text{K}\) are compressed isothermally from \(20\,\text{L}\) to \(5\,\text{L}\). How much heat does the gas give out, in joules?
One mole of an ideal gas at \(400\,\text{K}\) expands isothermally until its pressure falls from \(2\,\text{atm}\) to \(1\,\text{atm}\). How much work does it do, in joules?
Find \(\Delta U\) in three cases. In an isochoric process the gas takes in \(500\,\text{J}\) of heat. In an isobaric process it takes in \(700\,\text{J}\) and does \(200\,\text{J}\) of work. In an isothermal process it takes in \(400\,\text{J}\). Which list is right, in that order?
Think of the internal energy as a bank balance, heat as deposits and work as withdrawals. At constant volume there are no withdrawals, so every deposit adds to the balance. At constant pressure a fixed share is withdrawn at once. At constant temperature the balance must not change, so every deposit is withdrawn in full.
- Writing \(\Delta U = nC_p\,\Delta T\) at constant pressure.\(\Delta U\) is always \(nC_v\,\Delta T\) for an ideal gas. \(C_p\) belongs to the heat, \(Q = nC_p\,\Delta T\).
- Thinking an isothermal gas takes in no heat because its temperature does not rise.It takes in heat all the time, exactly as much as the work it does. Only \(\Delta U\) is zero.
- Using \(\ln(V_1/V_2)\) for the work of expansion.\(W = nRT\ln(V_2/V_1)\), final over initial. Expansion gives a positive logarithm and positive work.