Warming without heat#

Pump up a bicycle tyre quickly and the pump's barrel gets hot. Friction is the tempting culprit, yet a perfectly lubricated pump still gets hot.

Everything in this lesson is the first law of thermodynamics: conservation of energy, with the two ways energy crosses a boundary written out separately.

$$\Delta U = Q - W$$

$Q$ is heat into the gas; $W$ is work done by the gas.

One law, four processes#

For an ideal gas the internal energy depends on temperature alone, $U = \tfrac{f}{2}\,n R T$, with $f = 3$ for a monatomic gas and $f = 5$ for a diatomic one. The work is an area:

$$\kin{W} = \int_{V_1}^{V_2} \frc{P}\, d\pos{V}$$

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The area is the work#

Going right along any path, the strip under the curve down to $P = 0$ is positive work, done by the gas. Coming back left along a lower path, the strip under that one is work done on the gas, drawn fainter with a dashed edge. What survives is the area enclosed by the loop:

$$W_{\text{net}} = \oint P\,dV$$

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So an engine has to come back along a lower-pressure path, which means a lower temperature, which means dumping heat somewhere colder. The cold reservoir is the return path.

The bound, and what falls short of it#

A Carnot cycle reaches the bound because all its heat is exchanged at the two extreme temperatures. Any cycle that takes in heat at some intermediate temperature buys that heat at a worse rate, and falls short.

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What comes next#

You heated a gas with no heat, doubled the work of a loop, and found the ceiling no engine can pass. The first law says an engine cannot make energy. It does not say why heat has to go down the temperature scale, or why no engine can beat $1 - T_c/T_h$. That takes a second quantity, one which, unlike energy, is not conserved and only ever increases.