13.3 The ideal gas law  (Page 6/11)

 Page 6 / 11

Section summary

• The ideal gas law relates the pressure and volume of a gas to the number of gas molecules and the temperature of the gas.
• The ideal gas law can be written in terms of the number of molecules of gas:
$\text{PV}=\text{NkT},$
where $P$ is pressure, $V$ is volume, $T$ is temperature, $N$ is number of molecules, and $k$ is the Boltzmann constant
$k=1\text{.}\text{38}×{\text{10}}^{–\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}.$
• A mole is the number of atoms in a 12-g sample of carbon-12.
• The number of molecules in a mole is called Avogadro’s number ${N}_{\text{A}}$ ,
${N}_{\text{A}}=6\text{.}\text{02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}{\text{mol}}^{-1}.$
• A mole of any substance has a mass in grams equal to its molecular weight, which can be determined from the periodic table of elements.
• The ideal gas law can also be written and solved in terms of the number of moles of gas:
$\text{PV}=\text{nRT},$
where $n$ is number of moles and $R$ is the universal gas constant,
$R=8\text{.}\text{31}\phantom{\rule{0.25em}{0ex}}\text{J/mol}\cdot \text{K}.$
• The ideal gas law is generally valid at temperatures well above the boiling temperature.

Conceptual questions

Find out the human population of Earth. Is there a mole of people inhabiting Earth? If the average mass of a person is 60 kg, calculate the mass of a mole of people. How does the mass of a mole of people compare with the mass of Earth?

Under what circumstances would you expect a gas to behave significantly differently than predicted by the ideal gas law?

A constant-volume gas thermometer contains a fixed amount of gas. What property of the gas is measured to indicate its temperature?

Problems&Exercises

The gauge pressure in your car tires is $2\text{.}\text{50}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ at a temperature of $\text{35}\text{.}0\text{º}\text{C}$ when you drive it onto a ferry boat to Alaska. What is their gauge pressure later, when their temperature has dropped to $–\text{40}\text{.}0\text{º}\text{C}$ ?

1.62 atm

Convert an absolute pressure of $7\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ to gauge pressure in ${\text{lb/in}}^{2}\text{.}$ (This value was stated to be just less than $\text{90}\text{.}{\text{0 lb/in}}^{2}$ in [link] . Is it?)

Suppose a gas-filled incandescent light bulb is manufactured so that the gas inside the bulb is at atmospheric pressure when the bulb has a temperature of $\text{20}\text{.}0\text{º}\text{C}$ . (a) Find the gauge pressure inside such a bulb when it is hot, assuming its average temperature is $\text{60}\text{.}0\text{º}\text{C}$ (an approximation) and neglecting any change in volume due to thermal expansion or gas leaks. (b) The actual final pressure for the light bulb will be less than calculated in part (a) because the glass bulb will expand. What will the actual final pressure be, taking this into account? Is this a negligible difference?

(a) 0.136 atm

(b) 0.135 atm. The difference between this value and the value from part (a) is negligible.

Large helium-filled balloons are used to lift scientific equipment to high altitudes. (a) What is the pressure inside such a balloon if it starts out at sea level with a temperature of $\text{10}\text{.}0\text{º}\text{C}$ and rises to an altitude where its volume is twenty times the original volume and its temperature is $–\text{50}\text{.}0\text{º}\text{C}$ ? (b) What is the gauge pressure? (Assume atmospheric pressure is constant.)

Confirm that the units of $\text{nRT}$ are those of energy for each value of $R$ : (a) $8\text{.}\text{31}\phantom{\rule{0.25em}{0ex}}\text{J/mol}\cdot \text{K}$ , (b) $1\text{.}\text{99 cal/mol}\cdot \text{K}$ , and (c) $0\text{.}\text{0821 L}\cdot \text{atm/mol}\cdot \text{K}$ .

(a) $\text{nRT}=\left(\text{mol}\right)\left(\text{J/mol}\cdot \text{K}\right)\left(\text{K}\right)=\text{J}$

(b) $\text{nRT}=\left(\text{mol}\right)\left(\text{cal/mol}\cdot \text{K}\right)\left(\text{K}\right)=\text{cal}$

(c) $\begin{array}{lll}\text{nRT}& =& \left(\text{mol}\right)\left(\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(\text{K}\right)\\ & =& \text{L}\cdot \text{atm}=\left({\text{m}}^{3}\right)\left({\text{N/m}}^{2}\right)\\ & =& \text{N}\cdot \text{m}=\text{J}\end{array}$

Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation? can someone please explain this i need it for my final exam
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