First Law of Thermodynamics for Non-Flow Processes

A closed system exchanges energy with its surroundings, but no matter enters or leaves it. Energy can be transferred to or from the system by heating or by work. The first law of thermodynamics accounts for these transfers: energy is conserved.

For a non-flow process, the energy transferred to the system by heating is balanced by the increase in its internal energy and the work it does on its surroundings:

Here, $$Q$$ is the energy transferred to the system by heating, $$\Delta U$$ is the change in its internal energy, and $$W$$ is the work done by the system on its surroundings. Internal energy is the energy associated with the microscopic motion and interactions of the system’s particles.

Rearranging the balance gives $$\Delta U = Q - W$$. Heating the system increases its internal energy, while work done by the system transfers energy to the surroundings and reduces the increase in internal energy.

Sign convention for non-flow processes

• $$Q$$ is positive when energy is transferred to the system by heating and negative when energy is transferred from it by heating.
• $$W$$ is positive when the system does work on its surroundings. When work is done on the system, $$W$$ is negative.
• $$\Delta U$$ is positive for an increase in internal energy and negative for a decrease.

For example, if $$20,\text{J}$$ of work is done on a system, then $$W=-20,\text{J}$$. If the system also receives $$80,\text{J}$$ by heating, its internal energy increases by $$\Delta U=80-(-20)=100,\text{J}$$.

Process dependence of heating and work

The amounts transferred by heating and work depend on how the process occurs. A different process could produce the same increase of $$100,\text{J}$$ in internal energy: the system could receive $$140,\text{J}$$ by heating while doing $$40,\text{J}$$ of work on its surroundings. Then $$\Delta U=140-40=100,\text{J}$$.

The two processes have different values of $$Q$$ and $$W$$, but both obey the same energy balance.

Isothermal Change of an Ideal Gas

An isothermal change takes place at constant temperature. A fixed amount of ideal gas can expand or be compressed isothermally: its pressure and volume change, but its temperature stays the same.

The pressure, volume and temperature of an ideal gas are related by:

Here, $$p$$ is pressure, $$V$$ is volume, $$n$$ is the amount of gas, $$R$$ is the molar gas constant and $$T$$ is absolute temperature. For a fixed amount of gas undergoing an isothermal change, $$n$$ and $$T$$ remain constant, so:

Pressure is therefore inversely proportional to volume: if the volume doubles, the pressure halves. If the gas is compressed to half its original volume, its pressure doubles.

Energy in an isothermal change

The internal energy of an ideal gas depends only on its temperature. Since temperature does not change during an isothermal change, the change in internal energy is $$\Delta U=0$$.

Let $$Q$$ be energy transferred to the gas by heating and $$W$$ be work done by the gas. The first law of thermodynamics gives $$\Delta U=Q-W$$. Because $$\Delta U=0$$:

During isothermal expansion, energy supplied by heating equals the work done by the gas. During isothermal compression, work is done on the gas and an equal amount of energy leaves it by heating. In both cases, its internal energy remains unchanged.

Adiabatic Change of an Ideal Gas

An adiabatic change takes place without energy transfer by heating between a gas and its surroundings. The gas can still transfer energy by doing work or having work done on it.

For an adiabatic change, $$Q=0$$, where $$Q$$ is the energy transferred to the gas by heating. If $$W$$ is the work done by the gas, the first law of thermodynamics gives:

Here, $$\Delta U$$ is the change in the gas’s internal energy. For an ideal gas, internal energy depends on temperature. The direction of work therefore determines the temperature change:

• In an adiabatic expansion, if the gas does work on its surroundings, $$W$$ is positive. Its internal energy decreases, so the gas cools.
• In an adiabatic compression, work is done on the gas, so $$W$$ is negative. Its internal energy increases, so the gas warms.

Reversible adiabatic changes

For a reversible adiabatic change of an ideal gas with constant heat capacities, the pressure and volume are related by:

Here, $$p$$ is the gas pressure, $$V$$ is its volume, and $$\gamma=C_p/C_V$$. The quantities $$C_p$$ and $$C_V$$ are the heat capacities at constant pressure and constant volume, respectively.

The product $$pV^{\gamma}$$ remains the same throughout a change that meets these conditions. As the volume increases, the pressure decreases; as the volume decreases, the pressure increases.

Constant-Pressure Change of an Ideal Gas

When a fixed amount of ideal gas changes temperature while its pressure remains constant, its volume changes in proportion to its absolute temperature. Heating the gas increases its volume; cooling it decreases its volume.

The ideal gas equation is $$pV=nRT$$. If the pressure $$p$$ and amount of gas $$n$$ remain constant, then:

Here, $$V_1$$ and $$T_1$$ are the initial volume and temperature, and $$V_2$$ and $$T_2$$ are the final volume and temperature. Temperatures must be measured on the absolute scale, in kelvin. For example, doubling the kelvin temperature doubles the volume at constant pressure.

Work done during a constant-pressure change

As the gas changes volume against a constant pressure, work is transferred. The work done by the gas is:

During expansion, $$\Delta V$$ is positive, so the work done by the gas is positive. During compression, $$\Delta V$$ is negative, so the work done by the gas is negative: work is done on the gas instead.

Energy transfer at constant pressure

Energy transferred to the gas by heating can increase its internal energy and be transferred as work. With work done by the gas taken as positive, the first law gives:

Here, $$Q$$ is the energy transferred to the gas by heating, $$\Delta U$$ is its change in internal energy, and $$p\Delta V$$ is the work done by the gas. The signs show the direction of each change: for example, work done by the gas is negative during compression.

Constant-Volume Change of an Ideal Gas

When a fixed amount of ideal gas is kept at a constant volume, its pressure changes in direct proportion to its absolute temperature. Heating the gas raises its pressure; cooling it lowers its pressure.

The ideal gas equation is $$pV=nRT$$. Since the volume $$V$$ and amount of gas $$n$$ are fixed, and $$R$$ is constant, the ratio of pressure to temperature stays constant:

The temperature $$T$$ must be measured on the absolute temperature scale, in kelvin. For example, doubling the temperature in kelvin doubles the pressure, provided the volume and amount of gas remain unchanged.

Energy change at constant volume

Pressure–volume work requires a change in volume. Because the volume does not change, the work $$W$$ done by the gas is zero. Using the first law in the form $$\Delta U=Q-W$$, where $$Q$$ is energy transferred to the gas by heating, gives:

Thus, energy transferred to the gas by heating increases its internal energy. If energy is transferred out by heating, its internal energy decreases.