# The First Law and Heat Capacities

Physics II · Thermodynamics · https://tryals.app/learn/physics-ii/the-first-law-and-heat-capacities

## Two Ways to Change the Energy

The **first law** states that internal energy changes only through heat and work:

$$dU = \delta Q + \delta W = \delta Q - P\,dV$$

$U$ is a state function while $Q$ and $W$ are not, the whole content of the law is that this particular *combination* is path-independent. It rules out perpetual motion of the first kind: a machine producing work from nothing.

**Heat capacity** is the energy needed per degree, and it depends on what is held fixed:

$$C_V = \left(\frac{\partial U}{\partial T}\right)_V, \qquad C_P = \left(\frac{\partial H}{\partial T}\right)_P$$

where $H = U + PV$ is the enthalpy. At constant volume no work is done, so all the heat raises $U$. At constant pressure the gas also expands and does work, so more heat is needed for the same temperature rise, hence $C_P > C_V$ always. For an ideal gas the excess is exactly the work of expansion:

$$C_P - C_V = nR$$

The ratio $\gamma = C_P/C_V$ characterises the molecule: $5/3$ for a monatomic gas, $7/5$ for a diatomic one at room temperature. **Equipartition** explains why, each quadratic degree of freedom contributes $\tfrac{1}{2}k_BT$ per molecule, and a diatomic molecule has two rotational modes a monatomic one lacks.

Applied to the standard processes, the first law gives:

| Process | Constraint | Consequence |
|---|---|---|
| Isochoric | $dV = 0$ | $\Delta U = Q$ |
| Isobaric | $P$ fixed | $Q = \Delta H$ |
| Isothermal, ideal | $\Delta U = 0$ | $Q = -W$ |
| Adiabatic | $Q = 0$ | $\Delta U = W$, and $PV^\gamma$ constant |

> **Common pitfall:** assuming an adiabatic process is also isothermal. In an adiabatic expansion no heat enters, yet the gas does work, so its internal energy, and therefore its temperature, must **fall**. Adiabatic and isothermal are different curves, and the adiabat is the steeper of the two.

## Practice questions

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

### 1. One mole of a monatomic ideal gas has $C_V = 12.47$ J/mol/K. Compute its $C_P$ in J/mol/K, using $R = 8.314$ J/mol/K, to two decimal places.

**Answer:** 20.79 (within ±0.05)

**Why:** $C_P = C_V + R = 12.47 + 8.314 = 20.79$ J/mol/K. The extra $R$ is exactly the work the gas does expanding against constant pressure while being warmed.

Page: https://tryals.app/practice/physics-ii/the-first-law-and-heat-capacities/one-mole-of-a-monatomic-ideal-gas-has-cv-12-47-j-mol-k-compute-its

### 2. Why is $C_P$ always larger than $C_V$ for a gas?

A. At constant volume some supplied heat is converted directly into internal work
B. Enthalpy depends on temperature whilst internal energy remains strictly constant
C. At constant volume extra thermal energy is needed to maintain fixed pressure
D. At constant pressure the gas also does expansion work, so more heat is needed

**Answer:** D. At constant pressure the gas also does expansion work, so more heat is needed

**Why:** At constant volume every joule raises $U$. At constant pressure the gas also expands and does work on its surroundings, so it needs extra heat for the same temperature rise, precisely $nR$ per degree for an ideal gas.

Page: https://tryals.app/practice/physics-ii/the-first-law-and-heat-capacities/why-is-cp-always-larger-than-cv-for-a-gas

### 3. An adiabatic expansion leaves the temperature of an ideal gas unchanged.

**Answer:** False

**Why:** False, with $Q = 0$ the work must come from the internal energy, so $U$ and therefore $T$ **fall**. This is why compressed air warms and expanding gas cools, and it is what distinguishes an adiabat from an isotherm.

Page: https://tryals.app/practice/physics-ii/the-first-law-and-heat-capacities/an-adiabatic-expansion-leaves-the-temperature-of-an-ideal-gas

### 4. A gas absorbs 800 J of heat while doing 300 J of work on its surroundings. Compute the change in its internal energy, in J.

**Answer:** 500 (within ±5)

**Why:** $\Delta U = Q + W = 800 - 300 = 500$ J. The gas keeps only what it does not spend pushing on its surroundings.

Page: https://tryals.app/practice/physics-ii/the-first-law-and-heat-capacities/a-gas-absorbs-800-j-of-heat-while-doing-300-j-of-work-on-its

### 5. Internal energy $U$ is a state function whereas heat $Q$ and work $W$ are path-dependent. What is the fundamental physical implication of this distinction for a closed cyclic process?

A. The gas cannot perform any mechanical work during the cycle
B. Heat and work are individually conserved throughout the cycle
C. The net change in enthalpy must exceed the internal energy change
D. The total work output must exactly equal the net heat supplied

**Answer:** D. The total work output must exactly equal the net heat supplied

**Why:** Because $U$ is a state function, its net change around any closed cycle is zero ($\Delta U = 0$). By the first law, $Q = -W$, meaning heat converted into work must balance precisely. Treating $Q$ or $W$ as conserved individually mistakes path quantities for state properties.

Page: https://tryals.app/practice/physics-ii/the-first-law-and-heat-capacities/internal-energy-u-is-a-state-function-whereas-heat-q-and-work-w-are

### 6. Which statements about the first law are correct?

A. The law forbids a machine that produces work from nothing
B. Heat and work are each separately conserved
C. Internal energy is a state function
D. Heat and work are individually path-dependent

**Answer:** A. The law forbids a machine that produces work from nothing; C. Internal energy is a state function; D. Heat and work are individually path-dependent

**Why:** The law’s content is that the *combination* $Q + W$ is path-independent even though neither term is. Nothing conserves heat or work separately, a fact the caloric theory got wrong.

Page: https://tryals.app/practice/physics-ii/the-first-law-and-heat-capacities/which-statements-about-the-first-law-are-correct

### 7. An ideal gas absorbs 600 J of heat during an isothermal expansion. Set the work it does on its surroundings, in J.

**Answer:** 600 (within ±50)

**Why:** For an ideal gas $U$ depends only on $T$, so $\Delta U = 0$ and the first law gives $Q = -W$: all **600 J** emerges as work. None of the heat is retained, because the temperature has not moved.

Page: https://tryals.app/practice/physics-ii/the-first-law-and-heat-capacities/an-ideal-gas-absorbs-600-j-of-heat-during-an-isothermal-expansion