4.2 The separate & Ideal gas laws  (Page 6/19)

Gases whose properties of P , V , and T are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas    . An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures.

The ideal gas equation contains five terms, the gas constant R and the variable properties P , V , n , and T . Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.

If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$ using units of atm, L, and K. Both sets of conditions are equal to the product of n $\times$ R (where n = the number of moles of the gas and R is the ideal gas law constant).

Using the combined gas law

When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm ( [link] ). If the water temperature is 27 °C, how many liters of air will such a tank provide to a diver’s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm?

Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 °C, we have:

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}⟶\frac{\left(153\text{atm}\right)\left(13.2\text{L}\right)}{\left(300\text{K}\right)}=\frac{\left(3.13\text{atm}\right)\left(V_2\right)}{\left(310\text{K}\right)}$

Solving for V ₂ :

$V_2=\frac{\left(153\sout{\text{atm}}\right)\left(13.2\text{L}\right)\left(310\sout{\text{K}}\right)}{\left(300\sout{\text{K}}\right)\left(3.13\sout{\text{atm}}\right)}=667\text{L}$

(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good “ballpark” estimate.)

Check your learning

A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sample at 0 °C and 1.00 atm.

Answer:

0.193 L

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