*10.2 Gas Laws
- Page ID
- 158471
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Gas Phase Chemistry
Gasses are compressible fluids which form homogenous solutions. An ideal gas is a gas where there are no attractive (or repulsive) forces between the particles and their kinetic energy (the energy of motion) is linearly proportional to the absolute temperature (K). This relationship is expressed by the Ideal Gas Law as presented in the last section.
PV = nRT
Where, P = Pressure V = Volume
n = moles T = Temperature
R = Ideal Gas Constant (value depends on units)
| R=0.082057 \(\left (\frac{L\cdot atm}{mol\cdot K} \right) \) | R=62.36367 \(\left ( \frac{L\cdot torr}{mol\cdot K} \right )\) |
| R=8.3144598 \(\left ( \frac{J}{mol\cdot K} \right )\) | R=8.3144598 \(\left ( \frac{kg(m^{2})}{(s^{2})mol\cdot K} \right )\) |
Table 10.2.1: Different Values of R. Note, R has units of energy/(mol-K). Note how this can be expressed in terms of the SI base units (bottom right value).
A closer look at the ideal gas law shows and a review of section 5.1, when we introduced the energy of expansions (-PV), shows that the ideal gas constant relates the PV work on a per mol basis to the absolute temperature.
\[R=0.082057\left ( \frac{L\cdot atm}{mol\cdot K} \right )=8.314472\left ( \frac{J}{mol\cdot K} \right )=62.36367\left ( \frac{L\cdot torr}{mol\cdot K} \right )\]
Empirical Gas Laws
Empirical data is experimental data, and an experiment typically involves two variables, the independent variable, which we change, and the dependent variable, whose value changes as the independent variable is changed. We generically define these variables as (X,Y) where X is the independent and Y is the dependent variable, and then set up an equation where Y is a function of X, or Y =fct(X). A common function is the linear function as expressed in fig. 10.2.1
Fig. 10.2.1: Linear function where the value of Y (dependent variable) depends on the value of X (independent variable)
If we take a closer look at the ideal gas law, we see there are four variables, (P, T, V and n) and 1 constant (R). To make a plot of n variables you need n-dimensions. So two variables gives a line, three gives a surface (x,y,x axes), and there is no way we can draw 4 variables. What the early scientists could do is hold 2 of the 4 variables constant, and make 2D plots, and the empirical gas laws are the plots that result in linear equations. These were historically determined, and then later the Ideal gas law was developed. But we can use each of the empirical gas laws as a special case of the ideal gas law, defined by which variables are constant, and which are measured. So the idea is simple, you hold two variables constant, change the third (independent variable) and observe how the fourth adjusts (dependent variable).
Boyle's Law (Constant n,T)
This describes the relationship between the pressure and volume of a gas with a constant number of particles at constant temperature. From the Ideal Gas Law we would predict:
\[PV=nRT\Rightarrow V=\left ( nRT \right )\frac{1}{P}\]
since n, R and T are all constants,
\[V=\(k\frac{1}{P}~~where,k=\left ( nRT \right )\]
or switching dependent and independnt variables
\[P=k'\frac{1}{V}~~where, k=\frac{1}{k'}\]
Part (a) of figure 10.2.2 shows that P and V are not linear, but as predicted above and shown in part (b), V is linear to the reciprocal of P.
Fig. 10.2.2: Plot (a) shows a non linear relationship of V as a function ofP, while (b) shows V and P are inversely related.
Here the number of particles and the temperature are constant. The pressure (force per unit area) is a function of the kinetic energy of the gas and the frequency of collisions. We will shortly learn that the average kinetic energy of a gaseous system is proportional to the absolute temperature, and so at constant T the energy profile is constant (realize a gas has many particles moving at a variety of speeds). What the data says is that as P goes down, 1/P goes up, and V goes up.
It is easiest to visualize the Pressure dependence on the the volume \(P=k'\frac{1}{V}~~where, k=\frac{1}{k'}\). If you increase the volume, you increase the free flight time as the particles move across a larger container before impacting the wall, which in turn reduces the frequency of collisions with the wall, and thus reduces the pressure. The inverse is also easy to see. If you make the container smaller, you increase the collision frequency, so the pressure goes up.
Gay-Lussac's Law (Constant n,V)
This describes the relationship between the pressure and temperature of a gas with a constant number of particles at constant temperature. From the Ideal Gas Law we would predict:
PV=nRT \(\Rightarrow P=\left ( \frac{nR}{V} \right )T\)
since n,V & R are constants
P=kT where, k=\(\left ( \frac{nR}{V} \right )\)
or switching dependent and independent variables
T=k'P where, k=\(\frac{1}{k'}\)
This describes the relationship between the pressure and the temperature of a gas with a constant number of particles at constant volume. This describes a rigid closed container containing a fixed number of gas particles. As the temperature goes up the particles move faster, hit the wall more frequently at higher velocities and so the pressure goes up. Likewise, as the temperature goes down the molecules slow down, hit the wall less frequently at lower velocities and so the pressure drops. Note, there is a lower limit to this equations, which is defined as the boiling point, and at temperatures below that Temperature the particle condenses out as a liquid (it is no longer a gas), and the Ideal gas law does not describe the system.
Charles' Law (Constant n,P)
This describes the relationship between the volume and the temperature of a gas with a constant number of particles at constant pressure.
\[PV=nRT\Rightarrow V=\left ( \frac{nR}{P} \right )T \Rightarrow V=kT~~where,k=\left ( \frac{nR}{P} \right )~~or~~T=k'V~~where, k=\frac{1}{k'}\]
Avogadro's Law (Constant P,T)
This describes the relationship between the volume and the number of particles in a gas at constant Pressure and Temperature.
\[PV=nRT\Rightarrow V=\left ( \frac{RT}{P} \right )n \Rightarrow V=kn~~where,k=\left ( \frac{RT}{P} \right )~~or~~n=k'V~~where, k=\frac{1}{k'}\]
Ideal Gas Law Calculations
There are three basic types of calculations involving the ideal gas law
- PV=nRT calculations.
- Calculations involving Density and formula weights.
- Two State Problems
1. PV=nRT Calculations:
These are simple algebraic problems where you are given 3 of the 4 variables. The trick is to keep your units in your equations and be sure the units of the answer make sense. If you think about it, the numerical value of R depends on the units of R (see table 10.2.1), and your variables must have the units of R or they will not cancel.
2. Calculations involving density and formula weight.
By now you should know the equations for formula weight (molar mass) and density
\[\ fw(\frac{g}{mol})=\frac{m(g)}{n(mol)} \; \;\;\;\;\;\;\;\; d=\frac{m(g)}{v(l)}\]
This does two things.
a. Express ideal gas constant in terms of mass and formula weight by substituting \( n(mol)=\frac{m(g)}{fw(\frac{g}{mol})} \) into the ideal gas law, giving:
\[ PV=\left (\frac{m}{fw} \right )RT\]
b. noting that \( d=\frac{m}{V}\), this can be rearranged to:
\[ P=\left (\frac{m}{V} \right )\frac{RT}{fw}\; \; \; \; or\; \; \; P=\frac{dRT}{fw}\]
Example \(\PageIndex{1}\): Calculating the formula weight from density, pressure and temperature.
What is the formula weight of an unknown gas if it has a density of 4.08g/l at 88 °C and 1.4 atm?
Solution
NOTE: There are many ways you can solve this problem, and in video we note that we are looking for the formula weight, and so we start with an equation that has the formula weight, and solve it for the formula weight by identifying what variables we do not have, and substituting for them.
Example \(\PageIndex{2}\): Calculating the density of a known gas at a given temperature and pressure
What is the density of sulfur dioxide gas at 88 deg C and 1.4 atm?
Solution
Note, since we know what the gas is, we know its formula weight. Since we are solving for density, we start with the density equation.
3. Two State Equations
If you have a gaseous system and then perturb it by changing one of more variables the other variables adjust to that change. The two state approach is a quick way to solve these problems. The idea is simple, you have a relationship PV=nRT, so at one state, P1V1 =n1RT1 and P2V2=n2RT2
so we can set up the equivalence that"
\[ \frac{P_{1}V_{1}}{P_{2}V_{2}}=\frac{n_{1}RT_{1}}{n_{2}RT_{2}}=\frac{n_{1}T_{1}}{n_{2}T_{2}}\]
Two State Approach
What is the volume of a perfectly elastic balloon if it rises to an altitude where the pressure drops to 0.100 atm and the temperature drops to 173 K if it had a volume of 5.00 liter at sea level where the pressure was 1.00 atm and the temperature was 273 K?
Note: Always check the units to make sure they match the units for the formula. Example: for this equation temperature must be in units of Kelvin or the answer obtained will be incorrect.
Practice
Gas Laws
Use the ideal gas law to derive an equation that relates the remaining variables for a sample of an ideal gas if the following are held constant.
- amount and volume
- pressure and amount
- temperature and volume
- temperature and amount
- pressure and temperature
Ideal Gas Law
1. A sample of gaseous Cl2 has a volume of 12.4 L at 500.0 K and 0.521 atm. How many moles are present?
2. What is the mass of chlorine gas in a 12.4L container at 500.0K and 0.521 atm?
3. What volume is occupied by 1.06 mol of CO2 gas at 299 K and a pressure of 0.89 atm?
4. What is the temperature of 1.41 mol of methane gas in a 5.0 L container at 1.00 atm?
5. What is the pressure exerted by a 1.75 mol sample of water at 7.0L and 20oC?
Two State
1. What pressure would 6.1 mole of a gas in a rigid container have at 20.0 oC have if it had a pressure of 0.45 atm at -45.0 oC?
2. A pressure tank containing chlorine gas at a pressure of 2.00 atm and a temperature of 40oC is set with a pressure relief valve set to open at a pressure of 10.0 atm. At what temperature will the relief valve open?
3. A gas is in a sealed cylinder at 25oC with a piston which can expand or contract to change the volume. The initial volume and pressure are 75.0 L and 980.0 torr. A force is applied to the piston and the volume adjusts to a final pressure and temperature of 5.00 atm at 188oC. What is the final volume?
Gas Phase Reactions
1. 5.00 g of Mg is added to 50.0 mL of 0.800M HCl. A double displacement reaction occurs in a sealed container at 25.0oC. What is the pressure of the hydrogen gas generated if the volume above the solution is 100.0 ml, and we ignore any pressure due to the evaporated water?
2. Nitrogen dioxide is formed in a closed container at 90oC and 1.00 atm when 1.50 g NO and 2.00 mole of O2 are mixed. After the reaction is finished the pressure changes to 3.90 atm. What is the final temperature?
Answers
Gas Laws
- P/T = constant
- V/T = constant (Charles’ law)
- P/n = constant
- PV = constant (Boyle’s law)
- V/n = constant (Avogadro’s law)
Ideal Gas Law
1. \[n=\frac{PV}{RT}=\frac{0.521atm\left ( 12.4L \right )}{0.08206\frac{L\cdot atm}{mol\cdot K}\left ( 500K \right )}=0.157mol\]
2. \[n=\frac{PV}{RT}=\frac{0.521atm\left ( 12.4L \right )}{0.08206\frac{L\cdot atm}{mol\cdot K}\left ( 500K \right )}=0.157mol\]
\[0.157mol~Cl_{2}\left ( \frac{70.90g~Cl_{2}}{mol} \right )=11.1g\]
3. \[V=\frac{nRT}{P}=\frac{1.06mol\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )299K}{0.89atm}=29L\]
4. \[T=\frac{PV}{nR}=\frac{1.00atm\left ( 5.0L \right )}{1.41mol\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )}=43K\]
5. \[P=\frac{nRT}{V}=\frac{1.75mol\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )293.15K}{7.0L}=5.8atm\]
Two State
1. \[P_{1}=P_{2}\frac{T_{1}}{T_{2}}=\left ( 0.45atm \right )\left ( \frac{293.15K}{228.15K} \right )=0.58atm\]
2. \[T_{2}=T_{1}\frac{P_{2}}{P_{1}}= \left ( 313.15K \right )\left ( \frac{10.0atm}{2.00atm} \right )=1570K\]
3. \[V_{2}=V_{1}\left ( \frac{P_{1}}{P_{2}} \right )\left ( \frac{T_{2}}{T_{1}} \right )=75.0L\left ( \frac{980torr\left ( \frac{1atm}{760torr} \right )}{5.00atm} \right )\left ( \frac{461.15K}{298K} \right )=29.9L\]
Gas Phase Reactions
1. \[P_{H_{2}}=\frac{n_{H_{2}}RT}{V}=\frac{0.0200mol~H_{2}\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )298K}{0.100L}=4.89atm\]
2. \[T_{2}=T_{1}\left ( \frac{n_{1}}{n_{2}} \right )\left ( \frac{P_{2}}{P_{1}} \right )=363.15K\left ( \frac{2.05mol}{2.025mol} \right )\left ( \frac{3.90atm}{1.00atm} \right )=1433.77K =1400K\]
Contributors
- Bob Belford (UALR) and November Palmer (UALR)