Every Indian kitchen knows the sound. The cooker sits quietly on the flame for a few minutes. Then the whistle lifts, steam shrieks out, and the whistle drops back. A few minutes later it happens again. The flame has been giving heat the whole time. Where does that heat go?
Energy that belongs to the state
A gas stores energy in the random motion of its molecules: their kinetic energy, and any energy of rotation, vibration and attraction. This total, measured in the frame of the gas as a whole, is its internal energy \(U\). For an ideal gas the last chapter found \(U = \tfrac f2 nRT\), where \(f\) is the number of degrees of freedom of a molecule. This lesson uses \(R = 8.31\,\text{J/(mol K)}\), \(1\,\text{atm} = 1.013 \times 10^5\,\text{Pa}\) and \(L_v = 2.26 \times 10^6\,\text{J/kg}\).
The internal energy depends only on the state of the gas. Two samples of the same gas in the same state have the same \(U\), whatever their histories. So \(U\) is a state function, like pressure and temperature. Heat and work are different: the last lesson showed that work depends on the path. So between two states, \(\Delta U\) is fixed, but \(Q\) and \(W\) are not.
For an ideal gas this has a sharp consequence. \(U = \tfrac f2 nRT\) depends on the temperature alone. So \(\Delta U = \tfrac f2 nR\,\Delta T = nC_v\,\Delta T\) for every process, not only at constant volume. The symbol \(C_v\) appears because a constant-volume process is where it was first measured. The formula itself holds whether the volume changes or not.
A gas is taken from state A to state B by two different paths. Which quantities must be the same for both paths?
Energy is conserved
Give a gas some heat \(Q\). Energy cannot vanish, so it must go somewhere. Part of it stays in the gas and raises its internal energy by \(\Delta U\). The rest leaves as the work \(W\) the gas does on its surroundings. So \(Q = \Delta U + W\). This is the first law of thermodynamics. It is the conservation of energy, written for a system that exchanges heat as well as work.
The heat \(Q\) given to a system equals the increase \(\Delta U\) in its internal energy plus the work \(W\) done by the system: \(Q = \Delta U + W\).
The signs are the ones fixed in the last lesson: \(Q\) is heat given to the gas, and \(W\) is work done by the gas. Some books write \(\Delta U = Q + W\), with \(W\) the work done on the gas. The physics is the same; only the sign of \(W\) is flipped. Keep one convention and never mix the two in a problem.
Try some settings in the figure above. With \(Q = +500\,\text{J}\) and \(W = +200\,\text{J}\), the gas keeps \(300\,\text{J}\) and warms. With \(Q = 0\) and \(W = +300\,\text{J}\), the work comes out of the gas's own store, so \(\Delta U = -300\,\text{J}\) and it cools. With \(Q = -300\,\text{J}\) and \(W = -600\,\text{J}\), the gas loses heat but gains more as work done on it, so \(U\) rises by \(300\,\text{J}\).
A gas gives out \(200\,\text{J}\) of heat while \(300\,\text{J}\) of work is done on it. Find the change in its internal energy, in joules.
Heat depends on the path too
Return to the two paths of the last lesson, now for a monatomic ideal gas. From state 1 (\(10\,\text{L}\), \(300\,\text{kPa}\)) to state 2 (\(30\,\text{L}\), \(150\,\text{kPa}\)), \(U = \tfrac32 pV\) rises from \(4500\,\text{J}\) to \(6750\,\text{J}\). So \(\Delta U = 2250\,\text{J}\) by every path. Path A does \(6000\,\text{J}\) of work and so takes in \(Q = 2250 + 6000 = 8250\,\text{J}\). Path B does \(3000\,\text{J}\) and takes in only \(5250\,\text{J}\).
A gas goes from state A to state B by path 1, taking in \(800\,\text{J}\) of heat and doing \(500\,\text{J}\) of work. By path 2 it does only \(200\,\text{J}\) of work. How much heat does it take in along path 2?
Along path 1, \(\Delta U = Q - W = 800 - 500 = 300\,\text{J}\).
\(U\) is a state function, so \(\Delta U = 300\,\text{J}\) along path 2 as well.
Along path 2, \(Q = \Delta U + W = 300 + 200 = 500\,\text{J}\).
After a cycle the gas is back in its starting state, so \(\Delta U = 0\). The first law then says \(Q_\text{net} = W_\text{net}\): over a cycle, the net heat taken in equals the net work done. An engine that absorbs \(1000\,\text{J}\) of heat and rejects \(600\,\text{J}\) in each cycle does \(400\,\text{J}\) of work. It turns \(40\,\%\) of the heat absorbed into work.
In one cycle an engine absorbs \(1000\,\text{J}\) of heat and rejects \(600\,\text{J}\). What is the change in the internal energy of its gas over the cycle?
Where the heat of boiling goes
Boil one gram of water at \(100\,\text{°C}\) and \(1\,\text{atm}\). The chapter on calorimetry says it takes \(Q\) \(= mL_v\) \(= 0.001 \times 2.26 \times 10^6\) \(= 2260\,\text{J}\). But the steam takes up far more room than the water. One gram of water fills \(1.00\,\text{cm}^3\); one gram of steam at \(100\,\text{°C}\) fills \(1671\,\text{cm}^3\). To make that room, the steam pushes back the atmosphere, and that is work.
The work is \(W\) \(= p\,\Delta V\) \(= 1.013 \times 10^5 \times (1671 - 1) \times 10^{-6}\) \(= 169\,\text{J}\). The first law gives the rest: \(\Delta U = Q - W = 2260 - 169 = 2091\,\text{J}\). So only \(7.5\,\%\) of the heat of boiling does work on the air. The other \(92.5\,\%\) stays in the steam as internal energy. It pulls the molecules apart against their attraction.
Now the cooker. With its lid locked and the whistle down, the cooker is a closed system of nearly fixed volume. The water and steam can push back nothing, so the flame's heat all goes into internal energy. The temperature and the pressure inside rise together. The whistle lifts when the steam's push on it beats its weight, about one atmosphere above the air outside. Escaping steam then does work, lifting the whistle and pushing back the air: that is the shriek. Between whistles, every joule of heat stays inside and cooks the dal.
One kilogram of water boils at \(1\,\text{atm}\) and becomes \(1.671\,\text{m}^3\) of steam. Its volume as water was \(0.001\,\text{m}^3\). Find the increase in its internal energy, in megajoules. Take \(L_v = 2.26 \times 10^6\,\text{J/kg}\).
Mayer's relation from the first law
The last chapter found \(C_p - C_v = R\) by an energy argument. The first law now makes that argument formal. Warm \(n\) moles of an ideal gas by \(\Delta T\), once at constant volume and once at constant pressure. Because \(U\) depends on \(T\) alone, \(\Delta U\) is the same both times. The two heats differ only by the work.
\[ Q_v = \Delta U = nC_v\,\Delta T \]
Why. at constant volume \(W = 0\), so the first law gives \(Q = \Delta U\); \(C_v\) is defined by this heat.
\[ Q_p = \Delta U + p\,\Delta V \]
Why. the first law at constant pressure, where the work is \(p\,\Delta V\).
\[ p\,\Delta V = nR\,\Delta T \]
Why. \(pV = nRT\) with \(p\) fixed, applied at the start and the end and subtracted.
\[ nC_p\,\Delta T = nC_v\,\Delta T + nR\,\Delta T \]
Why. \(Q_p = nC_p\,\Delta T\) by definition, and \(\Delta U = nC_v\,\Delta T\) in every process of an ideal gas.
\[ C_p - C_v = R \]
Read it. divide by \(n\,\Delta T\).
For an ideal gas the molar heat capacities at constant pressure and at constant volume differ by the gas constant: \(C_p - C_v = R\).
At constant pressure the heat splits in a fixed ratio. The internal energy takes \(C_v/C_p\) of it and the work takes \(R/C_p\). For a monatomic gas that is \(3/5\) and \(2/5\); for a diatomic gas, \(5/7\) and \(2/7\). One mole of oxygen warmed by \(20\,\text{K}\) at constant pressure takes \(Q = \tfrac72 R \times 20 = 582\,\text{J}\). Of this, \(\Delta U = \tfrac52 R \times 20 = 416\,\text{J}\) and \(W = R \times 20 = 166\,\text{J}\).
One mole of a monatomic ideal gas is warmed at constant pressure and takes in \(1000\,\text{J}\) of heat. How much work does it do, in joules?
In 1840 Julius Robert Mayer was a German ship's doctor on a voyage to Java. Bleeding sick sailors in the tropics, he saw that their venous blood was bright red, almost like arterial blood. The body in a hot climate needed less internal heat, he reasoned, so it burned less oxygen. Heat and work, he decided, were forms of one conserved thing. In 1842 he used the gap between \(C_p\) and \(C_v\) of air to estimate how much work equals a unit of heat.
- Using \(\Delta U = nC_v\,\Delta T\) only for constant-volume processes.\(U\) of an ideal gas depends on \(T\) alone, so \(\Delta U = nC_v\,\Delta T\) in every process.
- Mixing sign conventions in one problem.In this book \(Q\) is heat given to the gas and \(W\) is work done by the gas, so \(Q = \Delta U + W\). Write each sign before substituting.
- Thinking a cycle leaves some energy stored in the gas.After a cycle the gas is in its starting state, so \(\Delta U = 0\) and the net heat equals the net work.