Question

The decomposition of baking soda produces carbon dioxide gas. If \(42.5 \mathrm{~mL}\) of wet gas is collected over water at \(22{ }^{\circ} \mathrm{C}\) and \(764 \mathrm{~mm} \mathrm{Hg}\), what is the volume of dry gas at STP conditions? (The vapor pressure of water at \(22^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~mm}\) Hg. .

Short answer

The volume of dry gas at STP is approximately 38.56 mL.

Step-by-step solution

Understand STP Conditions

Standard Temperature and Pressure (STP) are defined as a temperature of \(0^{\circ}C\) (273.15 K) and a pressure of \(1 \text{ atm}\) (760 mmHg). This means we need to find the volume of dry gas under these specified conditions.

Use Dalton's Law of Partial Pressures

The total pressure of the collected gas is given as 764 mmHg. However, this includes the pressure exerted by the water vapor. Using Dalton's Law,\[\text{Pressure of dry gas} = \text{Total Pressure} - \text{Vapor Pressure of Water}\]\[\text{Pressure of dry gas} = 764 \text{ mmHg} - 19.8 \text{ mmHg} = 744.2 \text{ mmHg}\].

Apply the Combined Gas Law

The Combined Gas Law is\[\frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \]where \(P_1\) is the initial pressure of the dry gas (744.2 mmHg), \(V_1\) the initial volume (42.5 mL), \(T_1\) the initial temperature in Kelvin (295.15 K),\(P_2\) the standard pressure (760 mmHg), and \(T_2\) the standard temperature in Kelvin (273.15 K).

Convert Temperatures to Kelvin

Convert the temperature from Celsius to Kelvin using \[T (K) = T (^{\circ}C) + 273.15\]Thus, \(T_1 = 22^{\circ}C + 273.15 = 295.15\ \text{K}\) and \(T_2 = 0^{\circ}C + 273.15 = 273.15\ \text{K}\).

Rearrange and Solve the Combined Gas Law

Rearrange the equation for \(V_2\):\[V_2 = \frac{P_1 \cdot V_1 \cdot T_2}{P_2 \cdot T_1} \]Substitute in the known values:\[V_2 = \frac{744.2 \text{ mmHg} \times 42.5 \text{ mL} \times 273.15 \text{ K}}{760 \text{ mmHg} \times 295.15 \text{ K}}\]\[V_2 \approx 38.56 \text{ mL}\].

Interpret the Final Volume

The volume of the dry gas at STP conditions is approximately \(38.56\ \text{mL}\). This calculation accounts for the temperature and pressure alterations affecting the gas's behavior in a contained environment.

Key concepts

Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures is a fundamental principle when dealing with gases, especially in scenarios where gases are collected over water, like in the exercise you encountered. This law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases.

In simpler terms, if you have a container filled with more than one gas, each gas exerts its pressure independent of the others. The total pressure you measure is just adding up the pressures from each gas.

• **Total Pressure**: The sum of all partial pressures of gases in a mixture.
• **Partial Pressure**: The pressure that one gas would exert if it alone occupied the entire volume.

In your problem, the total pressure (764 mmHg) includes both the carbon dioxide and the water vapor's pressures. By subtracting the vapor pressure of water (19.8 mmHg) from the total pressure, you isolate the pressure the dry gas (carbon dioxide) is exerting alone (744.2 mmHg). Dealing with actual experiments, this approach allows accurate measurements even when gases are combined or collected over a liquid such as water.

Combined Gas Law

The Combined Gas Law merges three individual gas laws into one formula. These are Boyle's law, Charles's law, and Gay-Lussac's law. It is incredibly useful when you have a gas involved in changing conditions of temperature, volume, and pressure.

Mathematically, it is expressed as: \[ \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \] Where:

• \( P_1 \), \( V_1 \), and \( T_1 \) are the initial pressure, volume, and temperature,
• \( P_2 \), \( V_2 \), and \( T_2 \) are the final conditions after some change.

This equation is indispensable when no gas leaves or enters the system, but its conditions change. By rearranging the equation to solve for \( V_2 \), you can predict how the volume of gas changes with shifts in temperature and pressure.

In the given exercise, you used the Combined Gas Law to find the volume of the dry carbon dioxide gas at Standard Temperature and Pressure (STP). You started with initial conditions and recalculated based on the new STP values. This allows you to see how the gas behaves under different environmental conditions, which is crucial for laboratory and industrial applications.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure (STP) are commonly used benchmarks in chemistry to allow for consistent and comparable data. STP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atmosphere (atm), which is equivalent to 760 mmHg.

The purpose of these standard conditions is to provide a baseline so that scientists can compile and compare experimental data without the variability introduced by differing experimental conditions. This can be especially helpful in gas calculations, as gas behavior is significantly influenced by changes in pressure and temperature.

• **Standard Temperature**: 0°C or 273.15 K
• **Standard Pressure**: 1 atm or 760 mmHg

In solving the exercise, moving from the initial measured conditions to STP allowed you to find out what the volume of the carbon dioxide gas would be if measured at this standardized set of conditions. The conversion, aided by the Combined Gas Law, ensured that you were comparing equivalent conditions across different scenarios.