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Chapter 1: Problem 5

Which temperature is higher? (a) \(0 \mathrm{~K}\) or \(0^{\circ} \mathrm{C}\) (b) \(300 \mathrm{~K}\) or \(0^{\circ} \mathrm{C}\) (c) \(250 \mathrm{~K}\) or \(-20^{\circ} \mathrm{C}\)

Short Answer

Expert verified
(a) 0掳C, (b) 300 K, (c) -20掳C.

Step by step solution

01

Understanding Absolute Zero

Kelvin (K) is an absolute temperature scale where 0 K is the lowest possible temperature, known as absolute zero. Celsius (掳C) is relative to the freezing point of water at 0掳C, which is 273.15 K.
02

Comparing 0 K to 0掳C

Convert 0掳C to Kelvin: \\[ 0^{ ext{o}} ext{C} = 0 + 273.15 = 273.15 ext{ K} \.\] Therefore, 0 K is lower than 273.15 K.
03

Conclusion for Part (a)

0掳C is higher than 0 K.
04

Convert 0掳C to Kelvin for Part (b)

Again, convert 0掳C to Kelvin: \\[ 0^{ ext{o}} ext{C} = 273.15 ext{ K} \.\]
05

Comparing 300 K to 0掳C (273.15 K)

Since 300 K is higher than 273.15 K (which is 0掳C), 300 K is the higher temperature.
06

Conclusion for Part (b)

300 K is higher than 0掳C.
07

Convert -20掳C to Kelvin for Part (c)

Convert -20掳C to Kelvin: \\[ -20^{ ext{o}} ext{C} = -20 + 273.15 = 253.15 ext{K} \.\]
08

Comparing 250 K to -20掳C (253.15 K)

250 K is less than 253.15 K, thus -20掳C is higher than 250 K.
09

Conclusion for Part (c)

-20掳C is higher than 250 K.

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Key Concepts

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

Kelvin to Celsius conversion
Temperature conversion between Kelvin and Celsius is straightforward. You simply add or subtract 273.15, which is the difference in their starting points.
For instance, to convert from Celsius to Kelvin, add 273.15. So, \( 0^{\circ}C \) becomes \( 273.15 \text{ K} \).
Conversely, to change Kelvin to Celsius, subtract 273.15. Hence, \( 300 \text{ K} \) equals \( 26.85^{\circ}C \).
These conversions allow us to switch between scales easily, ensuring accurate temperature readings in different contexts.
Absolute zero
Absolute zero is a fundamental concept in thermodynamics. It represents the lowest limit of the thermodynamic temperature scale, set at 0 Kelvin (K).
At absolute zero, molecular motion ceases, achieving a state where no heat energy remains in a system. It marks the point where a material has the minimum possible internal energy.
This temperature, therefore, cannot be achieved in practice, but it is crucial for various scientific calculations and understanding the nature of temperature itself.
Temperature comparison
Understanding temperature comparison involves knowing how to convert and interpret readings from different scales. For example:
  • 0 K versus 0掳C: After conversion, 0 K (absolute zero) is significantly lower than 0掳C (273.15 K).
  • 300 K versus 0掳C: 300 K (which equals 26.85掳C) is warmer than 0掳C.
  • 250 K versus -20掳C: Converting gives -20掳C as 253.15 K, making it warmer than 250 K.
These comparisons help in various practical applications, from scientific experiments to everyday weather predictions.
Thermodynamic temperature scale
The thermodynamic temperature scale, including Kelvin, is absolute, meaning it begins at the theoretical limit of absolute zero. Unlike Celsius or Fahrenheit scales, it doesn't rely on the physical properties of any material, like the freezing point of water.
This scale is beneficial because it simplifies the laws of thermodynamics, especially when dealing with gas laws and other calculations.
Kelvin is the primary unit for scientists working with temperatures, since it ensures consistency and avoids negative numbers, which could complicate formulas.

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Most popular questions from this chapter

Pressures of gases in mixtures are referred to as partial pressures and are additive. \(1.00 \mathrm{~L}\) of He gas at \(0.75 \mathrm{~atm}\) is mixed with \(2.00 \mathrm{~L}\) of Ne gas at \(1.5 \mathrm{~atm}\) at a temperature of \(25.0^{\circ} \mathrm{C}\) to make a total volume of \(3.00 \mathrm{~L}\) of a mixture. Assuming no temperature change and that He and Ne can be approximated as ideal gases, what are (a) the total resulting pressure, (b) the partial pressures of each component, and (c) the mole fractions of each gas in the mix?

A pot of cold water is heated on a stove, and when the water boils, a fresh egg is placed in the water to cook. Describe the events that are occurring in terms of the zeroth law of thermodynamics.

The Mount Pinatubo volcano eruption in 1991 released an estimated \(1.82 \times 10^{13} \mathrm{~g}\) of \(\mathrm{SO}_{2}\) into the atmosphere. If the gas had an average temperature of \(-17.0^{\circ} \mathrm{C}\) and filled the troposphere, whose approximate volume is \(8 \times 10^{21} \mathrm{~L}\), what is the approximate partial pressure of \(\mathrm{SO}_{2}\) caused by the eruption?

(a) Rewrite the ideal gas law as volume being a function of pressure and temperature. (b) What is the expression for the total derivative \(\mathrm{d} V\) as a function of pressure and temperature? (c) At a pressure of \(1.08 \mathrm{~atm}\) and \(350 \mathrm{~K}\) for one mole of ideal gas, what is the predicted change in volume if the pressure changes by \(0.10\) atm (that is, \(d p=0.10 \mathrm{~atm}\) ) and the temperature change is \(10.0 \mathrm{~K}\) ?

What are the slopes of the following lines at the point \(x=5\) ? at \(x=10\) ? (a) \(y=5 x+7\) (b) \(y=3 x^{2}-5 x+2\) (c) \(y=7 / x\).
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