# 9.2 Relating pressure, volume, amount, and temperature: the ideal  (Page 2/19)

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Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the P - T relationship for gases is known as either Amontons’s law    or Gay-Lussac’s law . Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant . Mathematically, this can be written:

$P\propto T\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}P=\text{constant}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}T\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}P=k\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}T$

where ∝ means “is proportional to,” and k is a proportionality constant that depends on the identity, amount, and volume of the gas.

For a confined, constant volume of gas, the ratio $\frac{P}{T}$ is therefore constant (i.e., $\frac{P}{T}\phantom{\rule{0.2em}{0ex}}=k$ ). If the gas is initially in “Condition 1” (with P = P 1 and T = T 1 ), and then changes to “Condition 2” (with P = P 2 and T = T 2 ), we have that $\frac{{P}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=k$ and $\frac{{P}_{2}}{{T}_{2}}\phantom{\rule{0.2em}{0ex}}=k,$ which reduces to $\frac{{P}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{{P}_{2}}{{T}_{2}}.$ This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called absolute zero    ). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)

## Predicting change in pressure with temperature

A can of hair spray is used until it is empty except for the propellant, isobutane gas.

(a) On the can is the warning “Store only at temperatures below 120 °F (48.8 °C). Do not incinerate.” Why?

(b) The gas in the can is initially at 24 °C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 °C on a hot day, what is the new pressure in the can?

## Solution

(a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)

(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons’s/Gay-Lussac’s law. Taking P 1 and T 1 as the initial values, T 2 as the temperature where the pressure is unknown and P 2 as the unknown pressure, and converting °C to K, we have:

$\frac{{P}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{{P}_{2}}{{T}_{2}}\phantom{\rule{0.4em}{0ex}}\text{which means that}\phantom{\rule{0.4em}{0ex}}\frac{360\phantom{\rule{0.2em}{0ex}}\text{kPa}}{297\phantom{\rule{0.2em}{0ex}}\text{K}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{{P}_{2}}{323\phantom{\rule{0.2em}{0ex}}\text{K}}$

Rearranging and solving gives: ${P}_{2}=\phantom{\rule{0.2em}{0ex}}\frac{360\phantom{\rule{0.2em}{0ex}}\text{kPa}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}323\phantom{\rule{0.2em}{0ex}}\overline{)\text{K}}}{297\phantom{\rule{0.2em}{0ex}}\overline{)\text{K}}}\phantom{\rule{0.2em}{0ex}}=390\phantom{\rule{0.2em}{0ex}}\text{kPa}$

A sample of nitrogen, N 2 , occupies 45.0 mL at 27 °C and 600 torr. What pressure will it have if cooled to –73 °C while the volume remains constant?

400 torr

## Volume and temperature: charles’s law

If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let’s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up.

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