Problem 3 - Master Resource Book in JEE Main Physics

Question

The triple point of neon and carbon dioxide are \(24.57 \mathrm{~K}\) and \(216.55 \mathrm{~K}\) respectively. These temperatures on the celsius and fahrenheit scales are respectively [NCERT] (a) \(-415.44^{\circ} \mathrm{F}\) and \(-69.88^{\circ} \mathrm{F}\) (b) \(415.44^{\circ} \mathrm{F}\) and \(69.88^{\circ} \mathrm{F}\) (c) \(-315.44^{\circ} \mathrm{F}\) and \(69.8 \mathrm{~F}^{\circ} \mathrm{F}\) (d) \(-69.88^{\circ} \mathrm{F}\) and \(415.44^{\circ} \mathrm{F}\)

Step-by-Step Solution

Answer

The correct answer is (a) \(-415.44^{\circ} \mathrm{F}\) and \(-69.88^{\circ} \mathrm{F}\).

Step 1: Convert Kelvin to Celsius

To convert from Kelvin to Celsius, we use the formula: \[ T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15 \]. First, let's convert the triple point of neon: \[ T_{\text{Celsius, Neon}} = 24.57 - 273.15 = -248.58^{\circ} \text{C} \]. Next, convert the triple point of carbon dioxide: \[ T_{\text{Celsius, CO_2}} = 216.55 - 273.15 = -56.6^{\circ} \text{C} \].

Step 2: Convert Celsius to Fahrenheit

To convert from Celsius to Fahrenheit, we use the formula: \[ T_{\text{Fahrenheit}} = \frac{9}{5}T_{\text{Celsius}} + 32 \]. For neon, \[ T_{\text{Fahrenheit, Neon}} = \frac{9}{5}(-248.58) + 32 = -415.44^{\circ} \text{F} \]. For carbon dioxide, \[ T_{\text{Fahrenheit, CO_2}} = \frac{9}{5}(-56.6) + 32 = -69.88^{\circ} \text{F} \].

Step 3: Match with given options

We now compare our calculated Fahrenheit values with the given options. For neon, we found \(-415.44^{\circ} \text{F}\) and for carbon dioxide, \(-69.88^{\circ} \text{F}\). This matches option (a): \(-415.44^{\circ} \text{F}\) and \(-69.88^{\circ} \text{F}\).

Key Concepts

Kelvin to Celsius conversion

When we talk about temperature conversions, the Kelvin to Celsius conversion is one of the simplest and most common. It is essential in science because Kelvin is the standard unit for temperature in the scientific community. The formula to convert Kelvin to Celsius is straightforward: subtract 273.15 from the Kelvin temperature.

For example, if we were to convert the triple point of neon, which is given as 24.57 K, the conversion process is as follows:
- Use the formula: \( T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15 \)
- Plug in the Kelvin temperature: \( T_{\text{Celsius}} = 24.57 - 273.15 = -248.58^{\circ} \text{C} \).

This illustrates the simplicity and efficiency of using this formula. Remembering this conversion is valuable for both academic exercises and real-world applications where temperature measurements from scientific literature must be interpreted or used in different contexts.

Celsius to Fahrenheit conversion

Converting temperatures from the Celsius scale to the Fahrenheit scale involves using a specific formula that captures the differences between these two units. This is especially useful in fields like meteorology or any context that uses the Fahrenheit scale. The formula is \( T_{\text{Fahrenheit}} = \frac{9}{5}T_{\text{Celsius}} + 32 \). Using this formula is crucial when readings need to be consistent with regions or scientific requirements that use Fahrenheit.

To illustrate, let's convert the Celsius temperature of neon, \(-248.58^{\circ} \text{C}\), to Fahrenheit.
- Apply the formula: \( T_{\text{Fahrenheit}} = \frac{9}{5}(-248.58) + 32 \).
- Perform the multiplication: \(-448.44 + 32 = -415.44^{\circ} \text{F} \).

By converting temperatures correctly, we ensure precision in measurements, especially crucial in various scientific and practical fields where specific temperature thresholds are pivotal.

Triple point temperatures

Triple point temperatures are a fascinating concept in physics and chemistry. A substance's triple point is where its three phases—solid, liquid, and gas—coexist in equilibrium. Each substance has a unique triple point, exemplified by the triple points of neon and carbon dioxide given as 24.57 K and 216.55 K, respectively. Understanding these points aids in calibrating temperature measurement systems and contributes to the study of thermodynamics.

Here’s why they are important:

• They provide a standard reference point for defining thermodynamic temperature scales.
• They highlight the concept of phase equilibrium—a core principle in studying matter.
• Industries such as cryogenics and gas engineering might rely on precise knowledge of these temperatures for operations such as gas liquefaction and storage.

Overall, the study of triple point temperatures enriches our understanding of physical principles and supports technological and scientific advancements across various fields.