# 13.3 The ideal gas law  (Page 2/11)

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At room temperatures, collisions between atoms and molecules can be ignored. In this case, the gas is called an ideal gas, in which case the relationship between the pressure, volume, and temperature is given by the equation of state called the ideal gas law.

## Ideal gas law

The ideal gas law    states that

$\text{PV}=\text{NkT},$

where $P$ is the absolute pressure of a gas, $V$ is the volume it occupies, $N$ is the number of atoms and molecules in the gas, and $T$ is its absolute temperature. The constant $k$ is called the Boltzmann constant    in honor of Austrian physicist Ludwig Boltzmann (1844–1906) and has the value

$k=1\text{.}\text{38}×{\text{10}}^{-\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J}/\text{K}.$

The ideal gas law can be derived from basic principles, but was originally deduced from experimental measurements of Charles’ law (that volume occupied by a gas is proportional to temperature at a fixed pressure) and from Boyle’s law (that for a fixed temperature, the product $\text{PV}$ is a constant). In the ideal gas model, the volume occupied by its atoms and molecules is a negligible fraction of $V$ . The ideal gas law describes the behavior of real gases under most conditions. (Note, for example, that $N$ is the total number of atoms and molecules, independent of the type of gas.)

Let us see how the ideal gas law is consistent with the behavior of filling the tire when it is pumped slowly and the temperature is constant. At first, the pressure $P$ is essentially equal to atmospheric pressure, and the volume $V$ increases in direct proportion to the number of atoms and molecules $N$ put into the tire. Once the volume of the tire is constant, the equation $\text{PV}=\text{NkT}$ predicts that the pressure should increase in proportion to the number N of atoms and molecules .

## Calculating pressure changes due to temperature changes: tire pressure

Suppose your bicycle tire is fully inflated, with an absolute pressure of $7\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}$ (a gauge pressure of just under $\text{90}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{lb/in}}^{2}$ ) at a temperature of $\text{18}\text{.}0\text{º}\text{C}$ . What is the pressure after its temperature has risen to $\text{35}\text{.}0\text{º}\text{C}$ ? Assume that there are no appreciable leaks or changes in volume.

Strategy

The pressure in the tire is changing only because of changes in temperature. First we need to identify what we know and what we want to know, and then identify an equation to solve for the unknown.

We know the initial pressure ${P}_{0}=7\text{.00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}$ , the initial temperature ${T}_{0}=\text{18}\text{.}0ºC$ , and the final temperature ${T}_{\text{f}}=35\text{.}0ºC$ . We must find the final pressure ${P}_{\text{f}}$ . How can we use the equation $\text{PV}=\text{NkT}$ ? At first, it may seem that not enough information is given, because the volume $V$ and number of atoms $N$ are not specified. What we can do is use the equation twice: ${P}_{0}{V}_{0}={\text{NkT}}_{0}$ and ${P}_{\text{f}}{V}_{\text{f}}={\text{NkT}}_{\text{f}}$ . If we divide ${P}_{\text{f}}{V}_{\text{f}}$ by ${P}_{0}{V}_{0}$ we can come up with an equation that allows us to solve for ${P}_{\text{f}}$ .

$\frac{{P}_{\text{f}}{V}_{\text{f}}}{{P}_{0}{V}_{0}}=\frac{{N}_{\text{f}}{\text{kT}}_{\text{f}}}{{N}_{0}{\text{kT}}_{0}}$

Since the volume is constant, ${V}_{\text{f}}$ and ${V}_{0}$ are the same and they cancel out. The same is true for ${N}_{\text{f}}$ and ${N}_{0}$ , and $k$ , which is a constant. Therefore,

$\frac{{P}_{\text{f}}}{{P}_{0}}=\frac{{T}_{\text{f}}}{{T}_{0}}\text{.}$

We can then rearrange this to solve for ${P}_{\text{f}}$ :

${P}_{\text{f}}={P}_{0}\frac{{T}_{\text{f}}}{{T}_{0}},$

where the temperature must be in units of kelvins, because ${T}_{0}$ and ${T}_{\text{f}}$ are absolute temperatures.

Solution

1. Convert temperatures from Celsius to Kelvin.

$\begin{array}{}{T}_{0}=\left(\text{18}\text{.}0+\text{273}\right)\text{K}=\text{291 K}\\ {T}_{\text{f}}=\left(\text{35}\text{.}0+\text{273}\right)\text{K}=\text{308 K}\end{array}$

2. Substitute the known values into the equation.

${P}_{\text{f}}={P}_{0}\frac{{T}_{\text{f}}}{{T}_{0}}=7\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}\left(\frac{\text{308 K}}{\text{291 K}}\right)=7\text{.}\text{41}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}$

Discussion

The final temperature is about 6% greater than the original temperature, so the final pressure is about 6% greater as well. Note that absolute pressure and absolute temperature must be used in the ideal gas law.

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