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Application of the Ideal Gas Law

CSCA Application of the Ideal Gas Law study guide organized around the publicly available CSCA syllabus. Practice Chemistry questions on aicsca.com.

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Core Concepts: Application of the Ideal Gas Law

This section is the "Master Key" of chemical calculations. It is used not only to calculate the physical state of a gas but also as the core tool for deriving molar mass and density. In the CSCA exam, **matching units** is the biggest trap.

#### 1. Core Formula and Physical Meaning

The Ideal Gas Law describes the relationship between $P, V, T, n$ of an ideal gas in equilibrium:

$$PV = nRT$$

* **$P$ (Pressure)**: Common units: $atm, kPa, Pa, mmHg$.

* **$V$ (Volume)**: Common units: $L, mL, m^3$.

* **$n$ (Moles)**: Amount of substance, unit $mol$.

* **$T$ (Temperature)**: **Absolute Temperature**. Unit $K$ (Kelvin).

* **Conversion**: $T(K) = t(^{\circ}C) + 273.15$ (Usually approximate to +273).

#### 2. Choosing "R": The Key to Success

The numerical value of the universal gas constant $R$ depends **entirely** on the units selected for $P$ and $V$. You must memorize the two most common combinations:

| Value of R | Unit of P | Unit of V | Usage Context |

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

| **0.082** (0.0821) | **atm** | **L** | Common in Chemistry |

| **8.314** | **Pa** ($N/m^2$) | **m³** | Common in Physics (SI) |

| **8.314** | **kPa** | **L** | Useful alternative (since $Pa \cdot m^3 = kPa \cdot L$ ) |

#### 3. Two Important Derivatives (High Frequency)

1. **Relation between Molar Mass and Density** ($PM = \rho RT$)

* Substitute $n = \frac{m}{M}$ into the equation $\rightarrow PV = \frac{m}{M}RT \rightarrow PM = \frac{m}{V}RT$.

* Since density $\rho = \frac{m}{V}$, we get:

$$PM = \rho RT$$

* *Mnemonic*: **P**er**M**eate = **D**ir**T** $\rightarrow$ $PM = DRT$ (where $D$ is Density).

2. **Combined Gas Law**

* When the amount of gas $n$ is constant (closed system):

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

#### Typical Example Analysis

**Example 1: Calculating Molar Mass**

The density of a gas is $2.62 \text{ g/L}$ at $27^{\circ}C$ and $1.00 \text{ atm}$. Calculate its molar mass. Could it be $SO_2$?

**Solution**:

1. **Knowns**: $T = 300 \text{ K}$; $P = 1 \text{ atm}$; $\rho = 2.62 \text{ g/L}$.

2. **Formula**: $PM = \rho RT \rightarrow M = \frac{\rho RT}{P}$.

3. **Calculation** (Choose $R=0.082$):

$$M = \frac{2.62 \times 0.082 \times 300}{1.00} \approx 64.45 \text{ g/mol}$$

4. **Conclusion**: Molar mass of $SO_2$ is $64 \text{ g/mol}$. The result is very close, so it could be $SO_2$.

**Example 2: Gas Compression**

A cylinder contains $2.0 \text{ L}$ of gas at $25^{\circ}C$ and $100 \text{ kPa}$. If compressed to $1.0 \text{ L}$ and heated to $50^{\circ}C$, what is the new pressure?

**Solution**:

1. **Identify**: $n$ is constant. Use Combined Gas Law.

2. **Convert**: $T_1 = 298 \text{ K}$, $T_2 = 323 \text{ K}$.

3. **Solve**:

$$\frac{100 \times 2.0}{298} = \frac{P_2 \times 1.0}{323}$$

$$P_2 = \frac{100 \times 2.0 \times 323}{298 \times 1.0} \approx 216.8 \text{ kPa}$$

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#### Pitfall Guide

1. **Temperature Unit**: Always, ALWAYS use **Kelvin**! Even if given Celsius, add 273 immediately.

2. **R Units**: If $P$ is $Pa$ and $V$ is $L$, you cannot use $8.314$ directly. Convert $L$ to $m^3$ or $Pa$ to $kPa$ first.

3. **STP**: $0^{\circ}C (273 K), 1 \text{ atm}$.

**SATP**: $25^{\circ}C (298 K), 1 \text{ bar}$.

*Be careful not to blindly apply the STP volume ($22.4 \text{ L/mol}$) unless conditions match.*