Problem 3 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

A sample of gas has a volume of \(346 \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C}\) when the pressure is 1.00 atm. What volume will it occupy if the conditions are changed to \(35^{\circ} \mathrm{C}\) and 1.25 atm? (Section 8.2)

Step-by-Step Solution

Answer

The new volume of the gas is approximately 277 cm³.

Step 1: Understand the Ideal Gas Law

The Ideal Gas Law is given by the equation \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles of gas, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. For problems with changing conditions, we often use the Combined Gas Law, \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).

Step 2: Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin. Use the formula \( T_K = T_C + 273.15 \).- Initial temperature: \( T_1 = 25^{\circ}C + 273.15 = 298.15 \, K \)- Final temperature: \( T_2 = 35^{\circ}C + 273.15 = 308.15 \, K \)

Step 3: Apply the Combined Gas Law

Substitute the known values into the Combined Gas Law equation:\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]\((P_1 = 1.00 \, atm, \, V_1 = 346 \, cm^3, \, T_1 = 298.15 \, K, \, P_2 = 1.25 \, atm, \, T_2 = 308.15 \, K)\)\[ \frac{1.00 \, atm \times 346 \, cm^3}{298.15 \, K} = \frac{1.25 \, atm \times V_2}{308.15 \, K} \]

Step 4: Solve for the Unknown Volume \( V_2 \)

Rearrange the equation to solve for \( V_2 \).\[ V_2 = \frac{1.00 \, atm \times 346 \, cm^3 \times 308.15 \, K}{1.25 \, atm \times 298.15 \, K} \]Calculate \( V_2 \) to find the final volume.

Step 5: Calculate the Result

Plug the numbers into the equation.\[ V_2 = \frac{1.00 \times 346 \times 308.15}{1.25 \times 298.15} = 276.97 \, cm^3 \]The sample of gas will occupy approximately \( 276.97 \, cm^3 \) under the new conditions.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the relationship between pressure, volume, temperature, and the number of moles of a gas. It is expressed as:\[ PV = nRT \]- **P** is the pressure.- **V** is the volume.- **n** is the amount of gas in moles.- **R** is the ideal gas constant (usually 0.0821 L atm/mol K).- **T** is the absolute temperature in Kelvin.
This equation is extremely useful in predicting the behavior of gases under different conditions. However, when dealing with changes in conditions such as pressure, volume, and temperature, we often employ a derived form of the Ideal Gas Law called the Combined Gas Law:\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]This version helps us solve for one variable when the others change, as it accommodates shifts in pressure, volume, and temperature without calculating moles.

temperature conversion

When working with gas laws, ensuring all temperatures are in Kelvin is crucial, as the Ideal Gas Law and its variations require absolute temperature. Conversion between Celsius and Kelvin is straightforward using the formula:\[ T_K = T_C + 273.15 \]- **Example:** If the temperature is \(25^{\circ}C\), convert it to Kelvin by adding 273.15, giving us 298.15 K.
Kelvin is an absolute scale with absolute zero, the lowest possible temperature, as its zero point. It avoids negative values, making calculations simpler and more universally applicable.
In our exercise, converting temperatures from Celsius to Kelvin was necessary to utilize the Combined Gas Law correctly.

pressure-volume relationship

The pressure-volume relationship is a crucial part of understanding gas behavior. According to Boyle's Law, for a given amount of gas at constant temperature, the pressure of a gas is inversely proportional to its volume:\[ P \times V = \, constant \]- **Higher pressure** decreases volume (if temperature remains constant).- **Lower pressure** increases volume.
In our exercise, both pressure and volume change, so we use the Combined Gas Law. This relationship reflects how gases contract and expand with pressure changes, foundational to our Combined Gas Law calculations.
This understanding helps predict what will happen to a gas if pressure changes while keeping the number of moles and temperature constant, or how these variables are interlinked when additional factors vary.

gas volume calculation

Calculating gas volume when given different pressures and temperatures involves applying the Combined Gas Law:\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]The formula relates the initial and final states of the gas, making it particularly useful for dynamics problems in chemistry.
**To solve for the new volume \(V_2\):**- Isolate \(V_2\) in the equation.- Substitute the known values.- Solve for \(V_2\).
For the given exercise:1. Initial volume \(V_1 = 346 \, \mathrm{cm}^3\), initial pressure \(P_1 = 1.00 \, \mathrm{atm}\), initial temperature \(T_1 = 298.15 \, \mathrm{K}\).2. Final pressure \(P_2 = 1.25 \, \mathrm{atm}\), final temperature \(T_2 = 308.15 \, \mathrm{K}\).3. Plug these values into the equation and solve for \(V_2\):\[ V_2 = \frac{1.00 \, \mathrm{atm} \times 346 \, \mathrm{cm}^3 \times 308.15 \, \mathrm{K}}{1.25 \, \mathrm{atm} \times 298.15 \, \mathrm{K}} \]Calculating yields \(V_2 \approx 276.97 \mathrm{cm}^3\), demonstrating how conditions impact gas volume.