first law of thermodynamics for gas processes

The first law of thermodynamics is the law of energy conservation for processes with a gas: the heat $Q$ transferred to the gas changes its internal energy $\Delta U$ and does work $W$ by the gas: $Q=\Delta U+W,\qquad \Delta U=Q-W.$ The internal energy of an ideal gas depends only on temperature; for a monatomic ideal gas $U=\frac32\nu RT$, so $\Delta U=\frac32\nu R\Delta T$, while for an arbitrary ideal gas $\Delta U=\nu C_V\Delta T$. The work done by a gas when its volume changes is $W=\int p\,dV$; on a $p$-$V$ diagram it is numerically equal to the area under the process curve, with its sign taken into account. For an isochoric process $dV=0$, so $W=0$ and $Q=\Delta U$; for an isothermal process $\Delta U=0$, so $Q=W$; for an adiabatic process $Q=0$, so $\Delta U=-W$. If the gas expands, then $W>0$; during compression $W<0$. The heat is $Q>0$ if the gas receives energy and $Q<0$ if it releases energy.

A gas receives $Q=500$ J of heat. It expands and does $W=200$ J of work. Find the change in the gas internal energy.

An ideal gas expands isobarically at pressure $p=2.0\cdot10^5$ Pa from $V_1=1.0\cdot10^{-3}$ m$^3$ to $V_2=2.5\cdot10^{-3}$ m$^3$. The gas receives $Q=900$ J of heat. Find the work done by the gas and the change in its internal energy.

In a cyclic process $A\to B$ the gas receives $400$ J of heat isochorically; $B\to C$ it expands adiabatically and does $400$ J of work; $C\to A$ it is compressed isothermally and releases $150$ J of heat. Find the net work done by the gas per cycle and verify the first law for the whole cycle.

Perform a safe computer experiment in an ideal-gas simulation (for example, PhET “Gas Properties”). For the same amount of gas, set initial $p$, $V$, and $T$. Carry out four processes in turn: isochoric heating, isothermal expansion, adiabatic compression, and isobaric expansion. For each process record $Q$, $W$, and $\Delta T$, then verify $\Delta U=Q-W$ and compare it with $\Delta U=\nu C_V\Delta T$. Do not heat closed vessels and do not use household pressurized cylinders; make all observations only in the simulation.

Show that the first law of thermodynamics for a gas is an energy balance: heat, work, and internal-energy change are linked by signs and depend on the type of process.

A $p$-$V$ sketch: four process curves with arrows — a vertical isochore ($W=0$), a horizontal isobar ($W=p\Delta V$), an isotherm hyperbola ($\Delta U=0$), and an adiabat ($Q=0$); the areas under the curves are shaded as the work done by the gas. Beside it is a balance block: $Q=\Delta U+W$ with signed labels $Q>0$, $W>0$, $\Delta U>0$.

In a diesel cylinder the air is compressed from $1.0\ \mathrm{L}$ to $0.05\ \mathrm{L}$ (compression ratio 20) while the pressure rises from $1\ \mathrm{atm}$ to $\approx 50\ \mathrm{atm}$. Compression lasts about $20\ \mathrm{ms}$, so heat exchange with the walls is negligible: $Q \approx 0$. The temperature rises from $300\ \mathrm{K}$ to $\approx 900\ \mathrm{K}$, and the injected fuel ignites without a spark plug.

The first law in the form $Q = \Delta U + W$, with $W$ the work done by the gas, is at work here. During compression the piston does work on the gas, so $W < 0$, while $Q \approx 0$, hence $\Delta U = -W > 0$: nearly all the work supplied turns into internal energy. This is the limiting case $Q = 0$, and it produces the heating that makes self-ignition possible.

A $0.5\ \mathrm{L}$ aerosol can lies in the sun. The air inside at $20\,^{\circ}\mathrm{C}$ ($293\ \mathrm{K}$) is at $1\ \mathrm{atm}$ and contains $n \approx 0.021\ \mathrm{mol}$ of gas. The can is rigid, so its volume does not change and $W = 0$. By noon the gas has warmed to $60\,^{\circ}\mathrm{C}$ ($333\ \mathrm{K}$), the pressure inside has risen to $\approx 1.14\ \mathrm{atm}$, and the heat supplied, $Q = nC_V\Delta T \approx 0.021 \cdot 20.8 \cdot 40 \approx 17\ \mathrm{J}$, goes entirely into internal energy.

This is an isochoric process: $V = const$, so the work $W = \int p\,\mathrm{d}V = 0$ and the first law reduces to $Q = \Delta U = nC_V\Delta T$. All the heat goes into raising the internal energy, and through the equation of state $pV = nRT$ at constant $V$ and $n$ the temperature rise shows up as a pressure rise, exactly in the ratio $333/293 \approx 1.14$.

On a summer day air near the ground at $20\,^{\circ}\mathrm{C}$ and $1000\ \mathrm{hPa}$ is carried up $1.5\ \mathrm{km}$ in $10\ \mathrm{min}$ by a rising current from a sun-warmed clearing. At that height the pressure is $840\ \mathrm{hPa}$, the parcel's volume has grown by about $13\ \%$, and the temperature has dropped by about $10\,^{\circ}\mathrm{C}$ per kilometre (the dry adiabatic lapse rate), that is to $\approx 5\,^{\circ}\mathrm{C}$.

Expanding against the surrounding pressure, the air does positive work $W = \int p\,\mathrm{d}V > 0$. Heat exchange with the surroundings does not have time to occur, so $Q \approx 0$, and the first law $\Delta U = Q - W$ gives $\Delta U = -W < 0$: internal energy and temperature fall at the expense of expansion work. Cooling to the dew point is what makes the water vapour condense and the cloud appear.