# The First Law of Thermodynamics

Chemistry I · Energy, Equilibrium and Electrochemistry · https://tryals.app/learn/chemistry-i/the-first-law-of-thermodynamics

## Energy Cannot Be Created, Only Moved

Thermodynamics starts by drawing a boundary. The **system** is what we study; the **surroundings** are everything else. An **open** system exchanges matter and energy, a **closed** system only energy, an **isolated** system neither.

Energy crosses that boundary in exactly two forms. **Heat** $q$ flows because of a temperature difference; **work** $w$ is everything else, and for a gas it is usually expansion against a pressure:

$$w = -P\Delta V$$

The sign convention matters. Both $q$ and $w$ are positive when energy enters the system. A gas that expands does work on its surroundings, so $\Delta V > 0$ makes $w$ negative and the system loses energy.

The **first law** states that the internal energy $U$ changes only by what crosses the boundary:

$$\Delta U = q + w$$

$U$ is a **state function**: it depends only on the current state, never on how that state was reached. Pressure, volume, temperature, enthalpy, entropy and Gibbs energy are all state functions. Heat and work are emphatically **not**: you can move between the same two states by many routes, each with a different split of $q$ and $w$, yet always the same $\Delta U$. This is the whole reason Hess's law works.

A **reversible** process proceeds through a continuous sequence of equilibrium states, infinitesimally slowly. It is an idealisation, but it is the limit that delivers the maximum possible work, and it is the route entropy must be computed along.

> **Common pitfall:** calling heat a property of a system. A system does not "contain" heat, it contains internal energy. Heat is energy *in transit* across a boundary, which is why $q$ has no value until you specify a process.

## Practice questions

8 of this lesson's 12 practice questions, with answers. The full set is in the app.

### 1. A system absorbs 150 J of heat and has 60 J of work done on it. Compute the change in internal energy, in joules.

**Answer:** 210

**Why:** $\Delta U = q + w = 150 + 60 = 210$ J. Both terms are positive because both represent energy entering the system across the boundary.

Page: https://tryals.app/practice/chemistry-i/the-first-law-of-thermodynamics/a-system-absorbs-150-j-of-heat-and-has-60-j-of-work-done-on-it

### 2. Heat and work are both state functions of a system.

**Answer:** False

**Why:** False, neither is a state function. A system holds internal energy, not heat or work; both are modes of *transfer* whose values depend on the path, which is why the same $\Delta U$ can be reached with completely different splits of $q$ and $w$.

Page: https://tryals.app/practice/chemistry-i/the-first-law-of-thermodynamics/heat-and-work-are-both-state-functions-of-a-system

### 3. Which quantities are state functions, that is, which depend only on the current state rather than on the route taken to reach it?

A. Work
B. Internal energy
C. Enthalpy
D. Temperature

**Answer:** B. Internal energy; C. Enthalpy; D. Temperature

**Why:** Internal energy, enthalpy and temperature are fully determined by the current state. Work is a transfer quantity and depends on the path, so it is not a state function.

Page: https://tryals.app/practice/chemistry-i/the-first-law-of-thermodynamics/which-quantities-are-state-functions-that-is-which-depend-only-on

### 4. Arrange these processes in order of increasing work extracted from the same expansion, from least to most.

**Answer:**

1. Expansion into a vacuum, against zero external pressure
2. Expansion against a constant low external pressure
3. Expansion against a constant high external pressure
4. Reversible expansion, matching the external pressure at every step

**Why:** Expanding into a vacuum does no work at all. Work grows with the opposing pressure, and the reversible route, where the external pressure matches the internal one at every instant, extracts the theoretical maximum.

Page: https://tryals.app/practice/chemistry-i/the-first-law-of-thermodynamics/arrange-these-processes-in-order-of-increasing-work-extracted-from

### 5. Match each term to its precise meaning.

**Answer:**

- System → The part of the universe under study
- Isolated system → Exchanges neither matter nor energy
- Reversible process → Passes through a continuous sequence of equilibrium states
- State function → Depends only on the current state, not the route

**Why:** The system is what you draw a boundary around; an isolated one lets nothing cross; a reversible process is the infinitely slow equilibrium limit; and a state function forgets the path entirely.

Page: https://tryals.app/practice/chemistry-i/the-first-law-of-thermodynamics/match-each-term-to-its-precise-meaning

### 6. Sort each sign convention by what it represents for the system.

**Answer:**

- Energy entering the system: q is positive, w is positive
- Energy leaving the system: The gas expands against an external pressure, The system releases heat to the surroundings

**Why:** The convention is uniform: positive $q$ and positive $w$ both mean energy entering. Expansion and heat release are the system paying energy out, so both carry negative signs.

Page: https://tryals.app/practice/chemistry-i/the-first-law-of-thermodynamics/sort-each-sign-convention-by-what-it-represents-for-the-system

### 7. Internal energy is a state function whereas heat is a path-dependent process. What follows from this distinction when evaluating a gas brought between two fixed endpoints?

A. The heat exchanged is strictly independent of the chosen external route
B. The sum of heat and work remains identical across all possible pathways
C. The reversible pathway will require zero net heat to complete the change
D. The final internal energy depends entirely on whether work was performed

**Answer:** B. The sum of heat and work remains identical across all possible pathways

**Why:** Confusing path-dependent transfers with state variables leads to the error of treating heat as a stored property. While individual values of q and w vary along different trajectories, their sum is constrained to Delta U, which is fixed entirely by the boundary states.

Page: https://tryals.app/practice/chemistry-i/the-first-law-of-thermodynamics/internal-energy-is-a-state-function-whereas-heat-is-a-path-dependent

### 8. A gas expands from 1.0 L to 4.0 L against a constant external pressure of 3.0 atm. Compute the work done on the gas, in L atm.

**Answer:** -9 (within ±0.1)

**Why:** $w = -P\Delta V = -3.0 \times 3.0 = -9.0$ L atm. The gas does work on its surroundings, so from the system’s point of view the value is negative.

Page: https://tryals.app/practice/chemistry-i/the-first-law-of-thermodynamics/a-gas-expands-from-1-0-l-to-4-0-l-against-a-constant-external