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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

PV = NkT , size 12{ ital "PV"= ital "NkT"} {}

where P size 12{P} {} is the absolute pressure of a gas, V size 12{V} {} is the volume it occupies, N size 12{N} {} is the number of atoms and molecules in the gas, and T size 12{T} {} is its absolute temperature. The constant k size 12{k} {} is called the Boltzmann constant    in honor of Austrian physicist Ludwig Boltzmann (1844–1906) and has the value

k = 1 . 38 × 10 23 J / K . size 12{k=1 "." "38" times "10" rSup { size 8{ - "23"} } " J"/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 PV size 12{ ital "PV"} {} is a constant). In the ideal gas model, the volume occupied by its atoms and molecules is a negligible fraction of V size 12{V} {} . The ideal gas law describes the behavior of real gases under most conditions. (Note, for example, that N size 12{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 size 12{P} {} is essentially equal to atmospheric pressure, and the volume V size 12{V} {} increases in direct proportion to the number of atoms and molecules N size 12{N} {} put into the tire. Once the volume of the tire is constant, the equation PV = NkT size 12{ ital "PV"= ital "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 . 00 × 10 5 Pa size 12{7 "." "00" times "10" rSup { size 8{5} } " Pa"} {} (a gauge pressure of just under 90 . 0 lb/in 2 size 12{"90" "." 0`"lb/in" rSup { size 8{2} } } {} ) at a temperature of 18 . 0 º C size 12{"18" "." 0°C} {} . What is the pressure after its temperature has risen to 35 . 0 º C size 12{"35" "." 0°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 .00 × 10 5 Pa , the initial temperature T 0 = 18 . 0ºC , and the final temperature T f = 35 . 0ºC . We must find the final pressure P f . How can we use the equation PV = 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 = NkT 0 and P f V f = NkT f . If we divide P f V f by P 0 V 0 we can come up with an equation that allows us to solve for P f .

P f V f P 0 V 0 = N f kT f N 0 kT 0

Since the volume is constant, V f size 12{V rSub { size 8{f} } } {} and V 0 size 12{V rSub { size 8{0} } } {} are the same and they cancel out. The same is true for N f size 12{N rSub { size 8{f} } } {} and N 0 size 12{N rSub { size 8{0} } } {} , and k size 12{k} {} , which is a constant. Therefore,

P f P 0 = T f T 0 . size 12{ { {P rSub { size 8{f} } } over {P rSub { size 8{0} } } } = { {T rSub { size 8{f} } } over {T rSub { size 8{0} } } } "." } {}

We can then rearrange this to solve for P f size 12{P rSub { size 8{f} } } {} :

P f = P 0 T f T 0 , size 12{P rSub { size 8{f} } =P rSub { size 8{0} } { {T rSub { size 8{f} } } over {T rSub { size 8{0} } } } ,} {}

where the temperature must be in units of kelvins, because T 0 size 12{T rSub { size 8{0} } } {} and T f size 12{T rSub { size 8{f} } } {} are absolute temperatures.

Solution

1. Convert temperatures from Celsius to Kelvin.

T 0 = 18 . 0 + 273 K = 291 K T f = 35 . 0 + 273 K = 308 K alignl { stack { size 12{T rSub { size 8{0} } = left ("18" "." 0+"273" right )" K"="291 K"} {} #T rSub { size 8{f} } = left ("35" "." 0+"273" right )" K"="308 K" {} } } {}

2. Substitute the known values into the equation.

P f = P 0 T f T 0 = 7 . 00 × 10 5 Pa 308 K 291 K = 7 . 41 × 10 5 Pa size 12{P rSub { size 8{f} } =P rSub { size 8{0} } { {T rSub { size 8{f} } } over {T rSub { size 8{0} } } } =7 "." "00" times "10" rSup { size 8{5} } " Pa" left ( { {"308 K"} over {"291 K"} } right )=7 "." "41" times "10" rSup { size 8{5} } `"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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Questions & Answers

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hamidat
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Practice Key Terms 4

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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