10.1: Gas Properties
- Page ID
- 170545
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Conversion between Gas Pressure Units
Exercise \(\PageIndex{1}\)
A pressure of 1.00 atm has a metric equivalent of 1.01 × 105 ________.
- Answer
-
Pascals
Exercise \(\PageIndex{2}\)
Which of the following is the largest value of pressure?
a. 0.716 atm b. 18.3 in Hg c. 972 mm Hg d. 14.9 psi e. 86572 Pa
- Answer
-
c. 972 mm Hg
Convert all to same units.
a. 0.716 atm
b. \(\large \frac{18.3 in Hg}{29.9213 (\frac{atm}{in Hg})} = 0.612\;atm\)
c. \(\large \frac{972 mm Hg}{760 (\frac{atm}{mm Hg})}=1.28\;atm\)
d. \(\large \frac{14.9 psi}{14.7\frac{atm}{psi}}=1.01\;atm\)
e. \(\large \frac{86572 Pa}{1.01*10^{5}\frac{atm}{Pa}}=0.857\;atm\)
Exercise \(\PageIndex{3}\)
A particular gas exerts a pressure of 356 mm Hg. What is this pressure in units of bar? (1 atm = 760 mm Hg = 101.3 kPa = 1.013 bar)
- Answer
-
0.473 bar
Exercise \(\PageIndex{4}\)
The local weather forecaster reports that the current barometric pressure is 15.9 inches of mercury. What is the current pressure in atmospheres?
- Answer
-
0.531 atm
Exercise \(\PageIndex{5}\)
It is possible to make a barometer using a liquid other than mercury. What would be the height (in meters) of a column of dichloromethane at a pressure of 0.790 atm, given that 0.790 atm is equal to a 0.758 m column of mercury and the densities of mercury and dichloromethane are 13.5 g/cm3 and 1.33 g/cm3, respectively.
- Answer
-
7.69 m
\(\large d=\frac{P_{atm}}{hg}\) can be used to find the height in a column.
Use a ratio with the pressure in atm is equal to one.
\(\large \frac{P_{CH_{2}Cl_{2}}}{P_{Hg}}=\frac{d_{CH_{2}Cl_{2}}*g*h_{CH_{2}Cl_{2}}}{d_{Hg}*g*h_{Hg}}\)
\(\large \frac{0.790}{0.790}=\frac{1.33*981*h_{CH_{2}Cl_{2}}}{13.5*981*60}\) Note: Converted to cm from m.
\(\large h_{CH_{2}Cl_{2}}=609.02 cm\)