Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Devise a temperature scale, abbreviated G, for which the magnitude of the ideal gas constant is \(5.52 \mathrm{JG}^{-1} \mathrm{mol}^{-1}\).

Short Answer

Expert verified
The new temperature scale G has an ideal gas constant of \(5.52 JG^{-1}mol^{-1}\). The relationship between G and Kelvin (K) scale is given by \(K = 1.506 G\). To convert a temperature in Kelvin to the G scale, use the equation: \(G = \frac{K}{1.506}\).

Step by step solution

01

Understanding the Ideal Gas Constant and Temperature Scale

The ideal gas constant is given by \(R = 8.314 J K^{-1} mol^{-1}\). The value of the ideal gas constant depends on the temperature scale used. In our case, we want to create a new temperature scale G such that the ideal gas constant value is 5.52 JG^{-1}mol^{-1}.
02

Converting Temperatures between Different Scales

To create a new temperature scale G, we need to understand the relationship between different temperature scales. We know that the Celsius and Kelvin temperature scales have the following relationship: \[K = °C + 273.15\] We can use this relationship to devise a new temperature scale G, specifically, we can transform the Kelvin scale to the G scale.
03

Creating the New Temperature Scale G

Let G be the temperature in the new temperature scale. Given that the ideal gas constant value in G scale is \(5.52 J K^{-1} mol^{-1}\), we can express this relationship between temperature in the Kelvin scale (K) and that of the G scale as follows: \[K = AG + B\] Here, A and B are constants that we need to find. We relate Kelvin and G scale by the ratio of the ideal gas constants in each scale, that is: \[A = \frac{R_{Kelvin}}{R_{G Scale}}\] Substitute the values of the ideal gas constants: \[A = \frac{8.314}{5.52} \approx 1.506\] Now, let's find the value of the constant B by considering absolute zero, which is the lowest possible temperature in any scale. In the Kelvin scale, absolute zero is 0 K; while in Celsius, it is -273.15 °C. In the G scale, we can denote the absolute zero as 0 G. Substitute the values of K and G in the equation: \[0 = 1.506 * 0 + B\] Therefore, we find that B is 0. Now, we have our complete relationship between Kelvin scale and our new G scale: \[K = 1.506 G\]
04

Inverting the Relationship to Express G in Terms of K

To find the temperature in G scale, we simply invert the relationship found in Step 3: \[G = \frac{K}{1.506}\] Now we have established the relationship between the Kelvin scale and our new temperature scale G. The new temperature scale G is defined with the magnitude of the ideal gas constant being 5.52 JG^{-1}mol^{-1}, and the conversion from Kelvin to G scale given by the equation above.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Temperature Scale Conversion
Temperature scale conversion is a method to transform one temperature measurement into another. The most commonly used scales are Kelvin, Celsius, and Fahrenheit. Each scale uses different starting points and units of measurement, which is why conversion formulas are important.

For example, the conversion between Kelvin and Celsius is straightforward, given by the equation: \[ K = °C + 273.15 \] This formula signifies that the Kelvin scale starts 273.15 units above the Celsius scale's zero point, which is why Kelvin values are typically larger than Celsius values.

Understanding the conversion between scales allows scientists and students to compare temperatures without confusion, ensuring that they can work consistently in various scientific computations and experiments.
  • Easy to Use: Once the basic formula is memorized or understood, converting between scales becomes a simple calculation.
  • Universal Understanding: Temperature conversions help create a universal language of temperature in scientific communities.
Kelvin to Celsius Conversion
The Kelvin to Celsius conversion is vital in scientific research, allowing temperatures in absolute measurements (Kelvin) to be related to familiar daily values (Celsius). Kelvin is the SI unit for temperature and starts at absolute zero, the point at which molecular motion ceases.

The conversion formula, \[ K = °C + 273.15 \], reflects the offset between the zero points of the two scales. Here are some key points that make this conversion important and simple:
  • Absolute Measurements: Kelvin allows for measurements without negative numbers, which is crucial for physical sciences.
  • Simple Arithmetic: It involves simply adding 273.15 to a Celsius measurement to convert it into Kelvin, which is manageable for most calculations.
The Kelvin scale's position on absolute zero makes it one of the most used temperature scales in physics and chemistry, as it provides a basis for thermodynamic temperature measurements.
New Temperature Scale Development
Creating a new temperature scale involves establishing a correspondence to known scales like Kelvin or Celsius. This task often arises when one needs a specific scale, as shown in our exercise by forming a scale 'G' from the ideal gas constant adjustment.

In the provided scenario, the goal is to set the ideal gas constant to a value of 5.52 JG\(^{-1}\)mol\(^{-1}\). To achieve this, we figured the scale conversion constant between Kelvin and the new scale G. This conversion required knowing the original ideal gas constants, enabling us to calculate:

\[ K = 1.506 G \]This formula means one unit in the G scale equals approximately 1.506 units in Kelvin. Finding constants A and B depends on ensuring the scale satisfies absolute zero, similar to understanding Kelvin-Celsius shifts.

Steps for Developing New Scales:
  • Identify the need for the scale, like adjusting constant values.
  • Determine the conversion from known scales.
  • Ensure absolute zero correspondence for consistency.
This tailored scale method reflects how scientific endeavors require specific conditions, offering a blend of theoretical development and practical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Many processes such as the fabrication of integrated circuits are carried out in a vacuum chamber to avoid reaction of the material with oxygen in the atmosphere. It is difficult to routinely lower the pressure in a vacuum chamber below \(1.0 \times 10^{-10}\) Torr. Calculate the molar density at this pressure at \(300 .\) K. What fraction of the gas phase molecules initially present for \(1.0 \mathrm{atm}\) in the chamber are present at \(1.0 \times 10^{-10}\) Torr?

The total pressure of a mixture of oxygen and hydrogen is 1.65 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.190 atm when measured at the same values of \(T\) and \(V\) as the original mixture. What was the composition of the original mixture in mole percent?

A typical diver inhales 0.450 liters of air per breath and carries a 25 L breathing tank containing air at a pressure of \(300 .\) bar. As she dives deeper, the pressure increases by 1 bar for every \(10.08 \mathrm{m}\). How many breaths can the diver take from this tank at a depth of \(35 \mathrm{m} ?\) Assume that the temperature remains constant.

A gas sample is known to be a mixture of ethane and butane, A bulb having a \(230.0 \mathrm{cm}^{3}\) capacity is filled with the gas to a pressure of \(97.5 \times 10^{3}\) Pa at \(23.1^{\circ} \mathrm{C}\). If the mass of the gas in the bulb is \(0.3554 \mathrm{g}\). what is the mole percent of butane in the mixture?

Consider a \(31.0 \mathrm{L}\) sample of moist air at \(60 .^{\circ} \mathrm{C}\) and one atm in which the partial pressure of water vapor is 0.131 atm. Assume that dry air has the composition 78.0 mole percent \(\mathrm{N}_{2}, 21.0\) mole percent \(\mathrm{O}_{2},\) and 1.00 mole percent Ar. a. What are the mole percentages of each of the gases in the sample? b. The percent relative humidity is defined as \(\% \mathrm{RH}=\) \(P_{H_{2}} o / P_{H_{2} O}^{*}\) where \(P_{H_{2}, O}\) is the partial pressure of water in the sample and \(P_{H, O}^{*}=0.197\) atm is the equilibrium vapor pressure of water at \(60 .^{\circ} \mathrm{C}\). The gas is compressed at \(60 .^{\circ} \mathrm{C}\) until the relative humidity is \(100 . \% .\) What volume does the mixture contain now? c. What fraction of the water will be condensed if the total pressure of the mixture is isothermally increased to 81.0 atm?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free