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First Law of Thermodynamics

CSCA First Law of Thermodynamics study guide organized around the publicly available CSCA syllabus. Practice Physics questions on aicsca.com.

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First Law of Thermodynamics

1. Core Concepts & Sign Convention

The First Law of Thermodynamics is essentially the **Law of Conservation of Energy** applied to thermal processes. For a system (like gas in a cylinder), the change in internal energy $\Delta U$ equals the sum of heat $Q$ supplied to the system and work $W$ done *on* the system.

**Formula**:

$$ \Delta U = Q + W $$

**The sign convention is the most common pitfall. Memorize this with the diagram below:**

* **$\{Q\}$ (Heat)**:

* **Absorbed** (Heat In): System gains energy $\rightarrow Q > 0$

* **Released** (Heat Out): System loses energy $\rightarrow Q < 0$

* **$\{W\}$ (Work)**:

* **Work done ON gas** (Compression): System gains energy $\rightarrow W > 0$

* **Work done BY gas** (Expansion): System spends energy $\rightarrow W < 0$

* **$\{\Delta U\}$ (Internal Energy)**:

* Increase $\rightarrow \Delta U > 0$

* Decrease $\rightarrow \Delta U < 0$

2. Judgment Criteria for Variables

For Ideal Gases (common in CSCA exams), follow this sequence:

1. **Internal Energy $\Delta U$** $\rightarrow$ **Check Temperature $T$**

* Depends ONLY on Temperature. $T \uparrow \Rightarrow \Delta U > 0$; $T \downarrow \Rightarrow \Delta U < 0$; Isothermal $\Rightarrow \Delta U = 0$.

2. **Work $W$** $\rightarrow$ **Check Volume $V$**

* $V$ decreases (Compression) $\Rightarrow$ Work done ON gas ($W > 0$).

* $V$ increases (Expansion) $\Rightarrow$ Work done BY gas ($W < 0$).

* $V$ constant $\Rightarrow W = 0$.

* **Graphical Method**: The **Area** under the curve in a $p-V$ diagram represents the magnitude of Work.

3. **Heat $Q$** $\rightarrow$ **Calculate Last**

* Usually calculated using the law: $Q = \Delta U - W$.

* Exception: Adiabatic process $Q=0$.

3. Summary of Special Processes

| Process | Condition | $\Delta U$ | $W$ (Work on gas) | $Q$ (Heat In) | Energy Conversion |

| :--- | :--- | :--- | :--- | :--- | :--- |

| **Isochoric** | Const $V$ ($W=0$) | $\Delta U > 0$ (if heating) | $0$ | $Q = \Delta U$ | Heat entirely increases Internal Energy |

| **Isothermal**| Const $T$ ($\Delta U=0$) | $0$ | $W < 0$ (Expansion) | $Q = -W$ | Heat entirely converted to Work Out |

| **Adiabatic** | No Heat ($Q=0$) | $\Delta U = W$ | $W > 0$ (Compression) | $0$ | Work done on gas increases Internal Energy (T rises) |

| **Isobaric** | Const $p$ | $\Delta U \propto \Delta T$ | $W = -p\Delta V$ | $Q = \Delta U - W$ | Mixed conversion |

4. Typical Examples

**Example 1: Qualitative Analysis**

An ideal gas undergoes an **Adiabatic Expansion**. How does its temperature change?

**Solution**:

1. **Adiabatic** $\Rightarrow Q = 0$.

2. **Expansion** $\Rightarrow$ Gas does work, so Work done *on* gas is negative ($W < 0$).

3. From $\Delta U = Q + W = 0 + W < 0$, Internal Energy decreases.

4. For ideal gases, decreased Internal Energy means Temperature **drops**.

**Example 2: Quantitative Calculation**

2 moles of monatomic ideal gas at initial temperature $300 \text{K}$ expand isobarically to double their volume. Find the heat $Q$ absorbed. (Given $C_{V,m} = \frac{3}{2}R$)

**Solution**:

1. **Process**: Isobaric, $V_2 = 2V_1$. From $V_1/T_1 = V_2/T_2$, we get $T_2 = 2T_1 = 600 \text{K}$, so $\Delta T = 300 \text{K}$.

2. **Find $\Delta U$**: $\Delta U = \nu C_{V,m} \Delta T = 2 \times \frac{3}{2}R \times 300 = 900R$.

3. **Find $W$**: Work done BY gas $W' = p\Delta V = \nu R \Delta T = 2R \times 300 = 600R$. Thus work ON gas $W = -600R$.

4. **Find $Q$**: Using $Q = \Delta U - W$.

$Q = 900R - (-600R) = 1500R$.

With $R \approx 8.31$, $Q \approx 12465 \text{J}$.