Problem 8 - Chemistry: Principles and Reactions

Question

I A sample of air is originally at \(32^{\circ} \mathrm{C}\). If \(P\) and \(n\) are kept constant, to what temperature must the air be cooled to (a) decrease its volume by \(25 \%\) ? (b) decrease its volume to \(25 \%\) of its original volume?

Step-by-Step Solution

Answer

Answer: (a) To decrease the volume by 25%, the air must be cooled to approximately -44.29°C. (b) To decrease the volume to 25% of its original volume, the air must be cooled to approximately -196.86°C.

Step 1: Convert temperatures to Kelvin

Before applying Charles's law, we need to convert the temperatures to Kelvin. \(T_1\) = 32°C + 273.15 = 305.15 K

Step 2: Find the final temperature using Charles's Law (scenario a)

For scenario a, we need to decrease the volume by 25%, so \(V_2 = 0.75V_1\). We can substitute this into Charles's law: \(V_1/T_1 = V_2/T_2\) We want to find \(T_2\). Plug in the known values: \(V_1/305.15 = 0.75V_1/T_2\) Now, we can remove \(V_1\) from the equation since it exists on both sides: \(1/305.15 = 0.75/T_2\) Solving for \(T_2\): \(T_2 = 305.15 * 0.75\) \(T_2 \approx 228.86\) K Finally, convert this back to Celsius: \(T_2 ≈ 228.86 - 273.15\) \(T_2 ≈ -44.29\) °C (a) To decrease the volume by 25%, the air must be cooled to approximately -44.29°C.

Step 3: Find the final temperature using Charles's Law (scenario b)

For scenario b, we need to decrease the volume to 25% of its original volume, so \(V_2 = 0.25V_1\). Substitute this into Charles's law: \(V_1/T_1 = V_2/T_2\) We want to find \(T_2\). Plug in the known values: \(V_1/305.15 = 0.25V_1/T_2\) Now, remove \(V_1\) from the equation since it exists on both sides: \(1/305.15 = 0.25/T_2\) Solving for \(T_2\): \(T_2 = 305.15 * 0.25\) \(T_2 \approx 76.29\) K Finally, convert this back to Celsius: \(T_2 ≈ 76.29 - 273.15\) \(T_2 ≈ -196.86\) °C (b) To decrease the volume to 25% of its original volume, the air must be cooled to approximately -196.86°C.

Key Concepts

Temperature Conversion

Temperature conversion is a crucial aspect when working with gas laws, particularly in Charles's Law calculations involving temperature and volume changes. To comprehend this concept, one must understand that temperature can be measured in Celsius, Fahrenheit, and Kelvin. Gas laws, however, require the use of the Kelvin scale to avoid negative temperature values that could disrupt calculations.

The conversion from Celsius to Kelvin is straightforward with the addition of 273.15 to the Celsius temperature. For example, a temperature of 32°C is equivalent to 305.15 K, which becomes the starting point for applying Charles's Law in gas behavior analysis. This simple yet essential step ensures precision in subsequent mathematical manipulations involved in the volume-temperature relationship.

Gas Laws

Gas laws are fundamental principles that describe the behaviors of gases under various conditions. These laws, including Boyle's Law, Charles's Law, and Avogadro's Law, encapsulate how changes in volume, temperature, and pressure affect a gas. Charles's Law specifically establishes the direct proportionality between the volume of a gas and its temperature, provided that pressure and number of particles (n) are kept constant.

Charles's Law is depicted mathematically as \( V_1/T_1 = V_2/T_2 \), where \(V_1\) and \(T_1\) represent the initial volume and temperature, and \(V_2\) and \(T_2\) represent the final volume and temperature. Understanding this relationship allows one to predict how gas will expand when heated or contract when cooled, hence the relevance in various scientific and real-world applications such as predicting weather balloons' behavior or understanding a car engine's thermodynamics.

Volume-Temperature Relationship

The volume-temperature relationship, derived from Charles's Law, explains how a temperature change reflects in the gas volume when pressure is constant. Imagine a balloon filled with air; heating it causes the air to expand, thus increasing the balloon's volume. Conversely, cooling the gas leads to contraction and a decrease in volume.

In a practical scenario, if we have a gas at a certain volume and we want to decrease this volume by a percentage, we can use the formula from Charles's Law to find the new required temperature. Taking a volume decrease by 25%, one would calculate the new temperature (\(T_2\)) by multiplying the initial temperature (\(T_1\)) by 0.75. Analogously, to decrease the volume to 25% of its initial volume, \(T_1\) would be multiplied by 0.25. This mathematical approach is not just theoretical; it has real-life implications, such as refrigeration processes, where controlling temperature is essential for volume management of gases.