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Chapter 2: Problem 139

Solve each equation for the underlined quantity: (a) \({ }^{\circ} \mathrm{F}=9 / 5^{\circ} \mathrm{C}+32\) (b) \(P V=n R T\) (c) \(E=h c / \underline{\lambda}\)

Short Answer

Expert verified
\( ^\circ C= \frac{5}{9}({ ^\circ F}-32) \), \( P= \frac{nRT}{V} \), and \( \lambda= \frac{hc}{E} \).

Step by step solution

01

Identify the parts of the equations

This step is essential to understand each equation and what each symbol represents. (a) The given equation is the temperature conversion formula between degrees Fahrenheit (掳F) and degrees Celsius (掳C). (b) The given equation is the ideal gas equation, where P represents pressure, V represents volume, n represents the amount of gas in moles, R is the ideal gas constant, and T is the temperature. (c) The given equation represents the energy of a photon. The energy is E, h is Planck's constant, c is the speed of light, and 位 is the wavelength of the photon. Now that we understand each equation, we can begin solving them.
02

Solve the equation (a) for 掳C

To solve the equation for 掳C, follow these steps: 1. Subtract 32 from both sides of the equation. 2. Multiply both sides of the equation by 5/9. \( ^\circ C= \frac{5}{9}({ ^\circ F}-32) \)
03

Solve the equation (b) for P

To solve the equation for P, follow these steps: 1. Divide both sides of the equation by V. \( P= \frac{nRT}{V} \)
04

Solve the equation (c) for 位

To solve the equation for 位, follow these steps: 1. Multiply both sides of the equation by 位. 2. Divide both sides of the equation by E. \( \lambda= \frac{hc}{E} \) The solutions for each equation are: (a) \( ^\circ C= \frac{5}{9}({ ^\circ F}-32) \) (b) \( P= \frac{nRT}{V} \) (c) \( \lambda= \frac{hc}{E} \)

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Key Concepts

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

Temperature Conversion
Temperature conversion between degrees Fahrenheit (掳F) and degrees Celsius (掳C) can seem tricky at first.
But once you know the formula, it's quite simple! The formula used is:
  • 掳F = \( \frac{9}{5} \cdot 掳C + 32 \)
To convert from 掳F to 掳C, you'll rearrange the formula to solve for 掳C using the steps:
  • Subtract 32 from the Fahrenheit temperature.
  • Multiply the result by \( \frac{5}{9} \).
This gives us: \[掳C = \frac{5}{9}(掳F - 32)\]This conversion is essential for scientific calculations, especially in regions that use different temperature scales. Understanding this process can make your science or math classes a lot smoother!
Ideal Gas Law
The Ideal Gas Law is one of the cornerstones of physical chemistry. It combines several physical laws into a single equation:
  • \( PV = nRT \)
Here:
  • \( P \) is pressure
  • \( V \) is volume
  • \( n \) is the amount of gas in moles
  • \( R \) is the ideal gas constant
  • \( T \) is the temperature in Kelvin
To solve for pressure \( P \), you can rearrange the equation:\[P = \frac{nRT}{V}\]The Ideal Gas Law is very useful for estimating the behavior of gases under different conditions like temperature or pressure changes.
Just keep in mind that real gases can deviate from this behavior when subjected to extreme conditions, like super high pressures or very low temperatures, making accurate measurements necessary.
Photon Energy
Photon energy is a crucial concept in quantum mechanics, where it describes how light and electromagnetic waves behave. The energy of a photon \( E \) is given by the equation:
  • \( E = \frac{hc}{\lambda} \)
In this equation:
  • \( E \) is the energy of the photon
  • \( h \) is Planck's constant \( 鈮� 6.626 \times 10^{-34} \text{Js} \)
  • \( c \) is the speed of light \( 鈮� 3 \times 10^8 \text{m/s} \)
  • \( \lambda \) is the wavelength of the photon
Solving for the wavelength \( \lambda \) gives:\[\lambda = \frac{hc}{E}\]This helps in understanding how different wavelengths correspond to different energy levels, which is one of the key ideas in studying the electromagnetic spectrum.
Different forms of radiation (visible light, X-rays, etc.) vary because of their wavelengths and energies.

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Most popular questions from this chapter

In the United States, car fuel efficiency is expressed in miles per gallon of gasoline. However, fuel efficiency can also be expressed in kilometers per liter of gasoline. If the fuel efficiency of a car is \(11.0 \mathrm{~km} / \mathrm{L}\), what is its fuel efficiency in miles per gallon? \([1\) mile \(=1.61 \mathrm{~km}, 1\) gallon \(=3.79 \mathrm{~L}]\)

Without doing any numerical calculations, determine which would have the smallest volume: (a) \(50 \mathrm{~g}\) of water (density \(=1.0 \mathrm{~g} / \mathrm{mL}\) ) (b) \(50 \mathrm{~g}\) of salt water (density \(=2.3 \mathrm{~g} / \mathrm{mL}\) ) (c) \(50 \mathrm{~g}\) of mercury (density \(=13.6 \mathrm{~g} / \mathrm{mL}\) ) (d) \(50 \mathrm{~g}\) of alcohol (density \(=0.89 \mathrm{~g} / \mathrm{mL}\) ) Explain your reasoning.

When \(10 \mathrm{~kJ}\) of heat energy is added to a beaker containing \(250 \mathrm{~g}\) of water initially at \(23.0{ }^{\circ} \mathrm{C}\), what is the final temperature of the water?

You measure one edge of a cube using a meterstick marked in centimeters. Unfortunately, the edge is longer than \(1 \mathrm{~m}\). You mark the \(1-\mathrm{m}\) point on the cube edge with a pen and then, using a \(15-\mathrm{cm}\) ruler marked in millimeters, measure the remaining distance to be \(1.40 \mathrm{~cm}\). (a) What is the length of the edge in centimeters? (b) What is the volume of the cube in cubic centimeters? (Remember, the lengths of all edges of a cube are equal.) Watch your significant figures. Use scientific notation if you have to. (c) The cube has a mass of \(111 \mathrm{~kg} .\) What is its density in grams per milliliter? Watch your significant figures.

Indicate whether the trailing zero in each value is significant, not significant, or possibly significant: (a) \(540 \pm 0.5\) (b) \(540 \pm 5\) (c) \(0.540\) (d) \(0.000540\) (e) 540
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