Problem 46 - General Chemistry: Principles and Modern Applications

Question

You decide to establish a new temperature scale on which the melting point of ammonia \(\left(-77.75^{\circ} \mathrm{C}\right)\) is \(0^{\circ}\) A and the boiling point of ammonia \(\left(-33.35^{\circ} \mathrm{C}\right)\) is \(100^{\circ}\) A. What would be (a) the boiling point of water in \(^{\circ}$$\text{A}\); and (b) the temperature of absolute zero in \(^{\circ}$$\text{A}\)?

Step-by-Step Solution

Answer

The boiling point of water and the temperature of absolute zero on the new temperature scale are approximately \(491.4^{\circ}\) A and \(-96708.05^{\circ}\) A, respectively.

Step 1: Convert Reference Points to Kelvin

The reference points given are in Celsius. Convert these to Kelvin using the formula K \(= ^{\circ}\mathrm{C} + 273.15\). This gives \(0^{\circ}\) A \(= -77.75^{\circ}\mathrm{C} + 273.15 = 195.4 K\) and \(100^{\circ}\) A \(= -33.35^{\circ}\mathrm{C} + 273.15 = 239.8 K\).

Step 2: Formulate the Conversion Formula

The relationship between Kelvin and degree A is linear, which can be described by a linear equation. This equation can be formulated as \(A = aK + b\). To solve this equation two sets of points are substituted \( (195.4, 0) \) and \( (239.8, 100)\) respectively and a system of linear equations is solved to find a and b coefficients.

Step 3: Find the Coefficients a and b

The linear conversion formula uses the equation \(A = aK + b\). Solving the system of linear equations using the given points we get the result \( a = 494.8164908616189 \) and \( b = -96708.04827807982 \). Thus the formula becomes \( A = 494.8164908616189 * K - 96708.04827807982\).

Step 4: Find the Boiling Point of Water in Degree A

First we convert the boiling point of water from Celsius to Kelvin, which gives us \(373.15 K\). We then apply this result into the conversion formula, \( A = 494.8164908616189 * K - 96708.04827807982\), and reach the result \( A = 491.4^{\circ}\) A.

Step 5: Find the Temperature of Absolute Zero in Degree A

Knowing that the absolute zero in Kelvin is \(0 K\) we apply this value into the conversion formula, \( A = 494.8164908616189 * K - 96708.04827807982\), and get the result \( A = -96708.05^{\circ}\) A.

Key Concepts

Kelvin Scale

The Kelvin scale is a thermodynamic temperature scale where absolute zero—the theoretical point at which particles stop moving—is defined as 0 Kelvin (K). It is an absolute scale used widely in scientific calculations. This scale is directly linked to Celsius since they share a degree increment, but they start at different places. For instance, while 0°C is the freezing point of water, on the Kelvin scale it is 273.15 K. One key conversion to remember is that to convert Celsius to Kelvin, you add 273.15 to the Celsius temperature.

The Kelvin scale essentially removes the inconvenience of negative numbers, common in the Celsius scale, especially useful in scientific experiments and calculations involving temperatures below freezing. In our problem, we utilize this scale to convert reference points from Celsius as a foundational step for understanding a new, imaginary temperature scale.

Celsius Scale

The Celsius scale, also known as centigrade, is a temperature scale that is based on the freezing point of water at 0°C and the boiling point at 100°C under standard atmospheric conditions. It is commonly used around the world for everyday temperature measurements.

The scale is convenient for experiments and daily life since its increments are direct, allowing swift and straightforward measurement and conversion. It acts as a middle ground for temperature interpretation between Fahrenheit and Kelvin scales.

• To convert Celsius to another scale like Kelvin, one can use straightforward formulas.
• The direct nature of Celsius creates an ease of transition into scientific and practical applications.

In the context of our exercise, knowing how to transform Celsius to Kelvin facilitates further conversion into our novel temperature scale, Degree A.

Linear Equation

Linear equations are mathematical representations of a straight line and are integral in various scientific computations. They take the form of \(y = mx + b\), where \(m\) represents the slope and \(b\) represents the y-intercept.

In the exercise, the relationship between Kelvin and the newly defined temperature scale of Degree A is described using a linear equation. We set points based on temperature conversions to find the specific equation for converting Kelvin to Degree A. By resolving points such as \((195.4, 0)\) K and \((239.8, 100)\) K, the process employs linear equations to deduce the coefficients that define this unique conversion formula.

This method directly links temperature scales with mathematical techniques, showcasing the power of linear relationships to provide practical outcomes in seemingly complex temperature conversions and translations between differing scales.