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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 \(415.44^{\circ} \mathrm{F}\)

Short Answer

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

Step by step solution

01

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: \[ T_{\text{Celsius, CO_2}} = 216.55 - 273.15 = -56.6^{\circ} \text{C} \].
02

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, \[ T_{\text{Fahrenheit, CO_2}} = \frac{9}{5}(-56.6) + 32 = -69.88^{\circ} \text{F} \].
03

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 \(-69.88^{\circ} \text{F}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Water of volume \(2 \mathrm{~L}\) in a container is heated with a coil of \(1 \mathrm{~kW}\) at \(27^{\circ} \mathrm{C}\). The lid of the container is open and energy dissipates at rate of \(160 \mathrm{~J} / \mathrm{s}\). In how much time temperature will rise from \(27^{\circ} \mathrm{C}\) to \(77^{\circ} \mathrm{C}\) ? [Given specific heat of water is \(4.2 \mathrm{~kJ} / \mathrm{kg}\) ] (a) \(8 \min 20 \mathrm{~s}\) (b) \(6 \min 2 \mathrm{~s}\) (c) 7 min (d) \(14 \mathrm{~min}\)

A substance of mass \(m \mathrm{~kg}\) requires a power input of \(P\) watts to remain in the molten state at its melting point. When the power is turned off, the sample completely solidifies in time \(t\) sec. What is the latent heat of fusion of the substance? (a) \(\frac{P m}{t}\) (b) \(\frac{P t}{m}\) (c) \(\frac{m}{P t}\) (d) \(\frac{t}{P m}\)

A surface at temperature \(T_{0} \mathrm{~K}\) receives power \(P\) by radiation from a small sphere at temperature \(T>T_{0}\) and at a distance \(d\). If both \(T\) and \(d\) are doubled, the power received by the surface will become (a) \(\underline{P}\) (b) \(2 P\) (c) \(4 P\) (d) \(16 P\)

The volume of a metal sphere increases by \(0.24 \%\) when its temperature is raised by \(40^{\circ} \mathrm{C}\). The coefficient of linear expansion of the metal is ... \({ }^{\circ} \mathrm{C}\). (a) \(2 \times 10^{-5} \mathrm{per}^{\circ} \mathrm{C}\) (b) \(6 \times 10^{-5}\) per \(^{\circ} \mathrm{C}\) (c) \(2.1 \times 10^{-5} \mathrm{per}^{\circ} \mathrm{C}\) (d) \(1.2 \times 10^{-5} \mathrm{per}^{\circ} \mathrm{C}\)

A body cools from \(80^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in \(5 \mathrm{~min} .\) Calculate the time it takes to cool from \(60^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\). The temperature of the surroundings is \(20^{\circ} \mathrm{C}\). [NCERT] (a) \(9 \mathrm{~min}\) (b) \(7 \mathrm{~min}\) (c) \(8 \mathrm{~min}\) (d) \(10 \mathrm{~min}\)

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