Problem 11 - General Chemistry: Principles and Modern Applications

Question

If 3.0 L of oxygen gas at \(177^{\circ} \mathrm{C}\) is cooled at constant pressure until the volume becomes \(1.50 \mathrm{L}\), then what is the final temperature?

Step-by-Step Solution

Answer

The final temperature of the gas is \(225.075K\).

Step 1: Conversion to Kelvin

First, convert the original temperature from Celsius to Kelvin using the relation: \( T(K) = T(°C) + 273.15 \). The original temperature of \(177^{\circ}C\) becomes \(177 + 273.15 = 450.15K \).

Step 2: Apply Charles's Law

Apply Charles's Law, which can be written: \( \frac{V_1}{T_1} = \frac{V_2}{T_2}\). Here, \(V_1 = 3.0L\) (initial volume), \(T_1 = 450.15K\) (initial temperature), and \(V_2 = 1.5L\) (final volume). The only missing variable is \(T_2\) which we are solving for (the final temperature in Kelvin)

Step 3: Solve for the Final Temperature

Rearrange the equation from Step 2 to solve for \(T_2\): \(T_2 = \frac{V_2 × T_1}{V_1}\). Substituting the known values, we find \(T_2 = \frac{1.5L × 450.15K}{3.0L}= 225.075K\). The final temperature is \(225.075K\)

Key Concepts

Gas Laws

Gas laws are a set of fundamental principles that describe the behavior of gases under various conditions. These laws help us understand how gases respond to changes in pressure, volume, and temperature.

One of the key gas laws is Charles's Law, which is crucial when studying gases. It focuses on the relationship between volume and temperature. Charles's Law states that, at constant pressure, the volume of a gas is directly proportional to its absolute temperature in Kelvin. This means as temperature increases, the volume of the gas increases as well, and vice-versa. This relationship is mathematically expressed as:

• \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\]

Here, \( V_1 \) and \( V_2 \) represent the initial and final volumes, while \( T_1 \) and \( T_2 \) are the initial and final temperatures respectively. This concise formula allows you to predict how a gas will behave when subjected to temperature changes, provided the pressure remains unchanged.

Temperature Conversion

Temperature conversion is an essential step when working with gas laws as these laws require temperature to be expressed in Kelvin. Kelvin is the absolute temperature scale used in scientific calculations because it starts at absolute zero, where molecular motion ceases.

To convert Celsius to Kelvin, which is often required like in our original exercise, you can use the simple formula:

• \[T(K) = T(°C) + 273.15\]

This formula adds 273.15 to the Celsius temperature, transitioning it into the Kelvin scale. For example, converting \(177^{\circ}C\) becomes \(450.15K\), as shown in the solution. Each step of this process is important for achieving accurate results in calculations involving temperature.

Volume and Temperature Relationship

The volume and temperature relationship is at the heart of Charles's Law. This relationship tells us that the volume of a gas tends to expand when warmed and contract when cooled, provided the pressure remains constant.

When solving problems based on this principle, it's important to consider this relationship carefully. Before solving for unknowns, both the initial and final temperatures should be measured in Kelvin.

• The initial state is represented by \( V_1 \) and \( T_1 \).
• The final state is represented by \( V_2 \) and \( T_2 \).

By rearranging the formula derived from Charles's Law, you can solve for the unknowns, making predictions on how gases behave when experiencing various temperature changes. Through this relationship, you can clearly see that when the volume of the gas decreases, the temperature also decreases, as demonstrated in our initial problem statement when cooled from 450.15K to 225.075K, resulting in a reduced volume from 3.0L to 1.5L.