The general gas law relates the pressure |(P)|, the volume |(V)|, the temperature |(T)| and the amount of gas |(n)| by comparing an initial situation with a final situation. ||\displaystyle \frac{P_{1}V_{1}}{n_{1}T_{1}}=\frac{P_{2}V_{2}}{n_{2}T_{2}}||
By combining the simple gas laws, we can establish a relationship that allows us to compare two series of variables after a gas has undergone changes. The simple laws used are as follows:
| Law | Formula | Units of Measurement |
| |P_{1}V_{1} = P_{2}V_{2}| | |P_{1}| and |P_{2}| in |\text{kPa}| or |\text{mm Hg}| |V_{1}| and |V_{2}| in |\text{mL}| or |\text{L}| |
|
| |\displaystyle \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}| | |V_{1}| and |V_{2}| in |\text{mL}| or |\text{L}| |T_{1}| and |T_{2}| in |\text{K}| |
|
| |\displaystyle \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}| | |P_{1}| and |P_{2}| in |\text{kPa}| or |\text{mm Hg}| |T_{1}| and |T_{2}| in |\text{K}| |
|
| |\displaystyle \frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}| | |V_{1}| and |V_{2}| in |\text{mL}| or |\text{L}| |n_{1}| and |n_{2}| in |\text{mol}| |
The following formula transforms degrees Celcius (|\text{°C}|) into kelvins (|\text{K}|):
||T_{°C}+ 273{.}15 = T_{K}||
From these laws, a general law can be deduced. It can be expressed as follows:
||\displaystyle \frac{P_{1}V_{1}}{n_{1}T_{1}}=\frac{P_{2}V_{2}}{n_{2}T_{2}}||where
|P_{1}| represents the initial pressure (in |\text{kPa}| or |\text{mm Hg}|)
|V_{1}| represents the initial volume (in |\text{mL}| or |\text{L}|)
|n_{1}| represents the initial amount of gas (in |\text{mol}|)
|T_{1}| represents the initial temperature (in |\text{K}|)
|P_{2}| represents the final pressure (in |\text{kPa}| or |\text{mm Hg}|)
|V_{2}| represents the final volume (in |\text{mL}| or |\text{L}|)
|n_{2}| represents the final amount of gas (in |\text{mol}|)
|T_{2}| represents the final temperature (in |\text{K}|)
This law is very useful when the conditions of a gas vary. It makes it possible to compare the same gas at two different times under two different subassemblies of conditions, the initial (1) and final (2) conditions. It can also be used to deduce all the simple gas laws, since the formula can be simplified by eliminating the variables that remain constant.
A sounding balloon is filled with helium at a temperature of |25\ °\text{C}| and a pressure of |120\:\text{kPa}|. The balloon rises to an altitude of |1,850\:\text{m}|, where the pressure is |80\:\text{kPa}| and the temperature |14\ °\text{C}|. What is the volume of the sounding balloon compared with its initial volume?
- Identifying the problem data or values
||\begin{align}P_{1} &= 120\: \text{kPa} & &\quad & P_{2} &= 80\:\text{kPa}\\
V_{1} &= x & & & V_{2} &= \: ?\\
T_{1} &= 25\ °\text{C} + 273{.}15 = 298{.}15\: \text{K} & & & T_{2} &=14\ °\text{C} + 273{.}15 = 287{.}15\:\text{K} \end{align}|||n_{1}| and |n_{2}| are equal; they can therefore be eliminated from the formula - Final volume calculation||\begin{align} \displaystyle \frac{P_{1}V_{1}}{T_{1}} =\frac{P_{2}V_{2}}{T_{2}} \quad \Rightarrow \quad V_{2} &=
\displaystyle \frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}\\ \\
&= \displaystyle \frac{120\: \text{kPa}\times x\times 287{.}15 \ \text{K}}{298.15 \ \text{K}\times 80\ \text{kPa}}\\ \\
&= 1{.}44\ x \end{align}||
Answer: The sounding balloon's final volume is |1.44| times greater than its initial volume.
A balloon containing |18{.}2\:\text{g}| of gaseous nitrogen at |20\ °\text{C}|, occupies a volume of |16\:\text{L}| at a pressure of |99{.}3\:\text{kPa}|. What will the pressure be if we increase the temperature to |50\ °\text{C}|, reduce the volume to |5\:\text{L}| and add |12{.}8\:\text{g}| of dioxygen?
- Identification of problem data
||\begin{align}P_{1} &= 99{.}3\:\text{kPa} & P_{2} &=\: ?\\V_{1} &= 16\:\text{L} & V_{2} &= 5\:\text{L}\\T_{1} &= 20°\ \text{C} + 273{.}15 = 293{.}15\:\text{K} & T_{2} &= 50\ °\text{C} + 273{.}15 = 323{.}15\:\text{K}\\n_{1} &= 18{.}2\:\text{ g of } N_{2} = 0{.}65 \text { mol} & n_{2} &= 18{.}2 \text { g of } N_{2} \text { and } 12{.}8 \text { g of } O_{2} \\ & & &= 1.05 \text { mol in total} \end{align}|| - Final pressure calculation||\begin{align}\displaystyle \frac{P_{1}V_{1}}{n_{1}T_{1}}=\frac{P_{2}V_{2}}{n_{2}T_{2}} \quad \Rightarrow \quad P_{2} &= \displaystyle \frac{P_{1}V_{1}n_{2}T_{2}}{n_{1}T_{1}V_{2}}\\
&= \displaystyle \frac{99{.}3\:\text{kPa}\times 16\:\text{L}\times 1{.}05\:\text{mol}\times 323{.}15\:\text{K}}{0{,}65\:\text{mol}\times 293{.}15\:\text{K}\times 5\:\text{L}}\\
&= 565{.}83\:\text{kPa}\end{align}||
Answer: The final pressure will be |565{.}86\:\text{kPa}|. This is the total pressure of the system, i.e. it is the sum of the partial pressures of the two gases.