Chapter 1 Notes/Questions/Answers
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Equation of State (EoS)- any gas can have its state specified by the following properties: Volume, Pressure, Temperature, N (number of moles);
Very few EoS are known, only:
- Low pressure gases
- Water and dilute aqueous solutions
Perfect- individual particles don’t interact at all, no chemical or physical interactions; most likely to be approached at low pressures.
Ideal Gas Law- ideal mixtures are mixtures in which all the interactions (particle-to-particle) are equal, but not necessarily zero. Perfect gases have 0 interactions, ideal gases have equal interactions.
- Boyle’s Law- Pressure and volume are inversely proportional (at constant T, P ∝1/V).
- Charle’s Law- Volume and temperature are directly proportional (V=A+BT, where A, B are constants).
- Avogadro’s Principle- Volume and moles are directly related. All gases contain an equal number of molecules at a given volume.
-Vm(molar volume)=Vgas/number of moles; Vm=24.8 L at 1 bar.
Uses of Perfect Gas Law- PV=nRT-used to calculate P,V,n, or T when other 3 are given
- P1V1/T1=P2V2/T2- combined gas equation; used when you have a known set of conditions that change
- Vm=RT/P- used to calculate molar volume of a perfect gas at any P,T
Partial Pressures of Gas Mixtures
PT=PA+PB...Pn; Total Pressure= Sum of Partial Presures;
PA=XA*PTotal, XA is the mole fraction (moles of A/total moles);
-what influences pressure of a certain gas is not its molecular weight (SF6 and H2 contribute equally), but moles of the gas.
Kinetic Model of Gases- Gas consists of molecules in ceaseless random motion
- Size of molecules is negligible compared to distance traveled
- Molecules don’t interact except during collision
-infinite collisions;
-F=ma,
thus heavier mass = higher concentration or higher speed of molecules
(related to temperature), all causing more gas pressure
Maxwell Distribution of Speeds of Molecules
-at lower temperatures, the average speed is low and the range is small
-at higher temperatures, the average speed is higher and the range is broad
-there’s a threshold of energy level for chemical reactions, thus, at higher temperatures, molecules are more likely to reach that threshold and more likely to participate in that chemical reaction
-heavy molecules move slower and have smaller range, lighter molecules move faster and have a broad range
Diffusion-process by which molecules of different substances mingle with each other
-rates in solids are very low, liquids intermediate, and gases quick
Effusion-escape of gas through small opening
-both processes are temperature dependant (as temperature rises, rate rises)
Separation-distance between molecules; Forces almost exclusively electrostatic; Too close, they push. Too far, pull.
-Equilibrium exists between 2 molecules when minimum amount of energy consumed in keeping molecule stabilized via push and pull forces.
Molecular interactions affect bulk properties of gases, modify EoS
-isotherms (temp stays the same along line)are similar to ideal gases at high temps (ex.CO2 @ 50°C), very diff at low temps.
From A→B→C: CO2 is gas, as piston is pushed down, P increases and the gas compresses (Boyle’s Law);
From C→D→E: CO2 condenses. @ E it has condensed to a compact liquid, and piston rests on surface of liquid. No P change since it’s condensing, and the molecules are close enough that they attract each other and cohere into a liquid.
From E→F: CO2 is a already a very compact liquid, need a lot of P to further compress it (trying to force the molecules closer together when they’re already in contact/trying to overcome strong repulsive interactions
Gas can’t be condensed to liquid by P unless under the critical temperature.
-when you heat a liquid in a container, the pressure of the container increases, and the liquid becomes less dense and the gas becomes more dense. After a certain temperature, Tc, the densities of the liquid and gas are the same, and a single state of matter (super-critical fluid) exists (for all real gases)
Virial EoS: P=(nRT/V) (1+nB/V+nC2/V2...nXn/Vn), where B, C are empirical coefficients that fit experimental data
Van der Waals EoS: P=(nRT/(V-nb))-a(n/V)2
Questions/Answers
1.1) What pressure is exerted by a sample of nitrogen gas of mass 3.055 g in a container of volume 3.00 dm3 at 32°C?
V=(3.00 dm3)((10cm/1dm)^3)=(3000cm3)(1mL/cm3)=(3000 mL)(1L/1000mL)=3L;
n=(3.055 g)(1mol/28.01g)=.109 mol;
T=32°C, C+273=K→ 32+273= 305 K;
R=8.314 Jmol*K
PV=nRT→P(3L)=(.109mol)(8.314J/mol*K)(305 K)=92.1 kPa
1.3) Much to everyone’s surprise, nitrogen monoxide (NO) has been found to act as a neurotransmitter. To prepare to study its effect, a sample was collected in a container of volume 300.0 cm3. At 14.5°C its pressure is found to be 34.5 kPa. What amount (in moles) of NO has been collected?
P=34.5 kPa;
V=300cm3(1mL/cm3)=300mL(1L/1000mL)=.3 L
R=8.314 J/mol*K
T=14.5°C, C+273=K→14.5+273= 287.5 K;
pV=nRT→(34.5kPa)(.3L)=n(8.314 J/mol*K)(287.5K)= 0.00433 mol
1.5) The effect of high pressure on organisms, including humans, is studied to gain information about deep-sea diving and anaesthesia. A sample of air occupies 1.00 dm3 at 25°C and 1.00 atm. What pressure is needed to compress it to 100 cm3 at this temperature?
P1V1/T1=P2V2/T2→1.00atm*1.00dm3/25°C=P2*(100cm3(1dm3/1000cm3)/25°C=10 atm
1.7) Until we find an economical way of extracting oxygen from sea-water or lunar rocks, we have to carry it with us to inhospitable places, and do so in compressed form in tanks. A sample of oxygen at 101 kPa is compressed at constant temperature from 7.20 dm3 to 4.21 dm3. Calculate the final pressure of the gas.
P1V1/T1=P2V2/T2→101kPa*7.20dm3/T1=P2*4.21dm3/T2;T1=T2 ; P2= 173 kPa
1.9) Hot-air balloons gain their lift from the lowering of density of air that occurs when the air in the envelope is heated. To what temperature should you heat a sample of air, initially at 315 K, to increase its volume by 25 per cent?
P1V1/T1=P2V2/T2 ;T1=315 K; P1=P2; V2=125% of V1; V1/315 K=1.25*V1/T2=393.75 K
1.11) A diving bell has an air space of 3.0 m3 when on the deck of a boat. What is the volume of the air space when the bell has been lowered to a depth of 50 m? Take the mean density of sea water to be 1.025 g cm−3 and assume that the temperature is the same as on the surface.
P2=ρ*g*h=1.025gcm31kg1000g100cm1m3*9.8ms2*50m=502.25 kPa
T1=T2; V1=S.T.P.=100kPa; P1V1/T1=P2V2/T2→100kPa*3.0m3/T1=502.25kPa*V2/T1=0.597m3
1.13) Atmospheric pollution is a problem that has received much attention. Not all pollution, however, is from industrial sources. Volcanic eruptions can be a significant source of air pollution. The Kilauea volcano in Hawaii emits 200–300 t of SO2 per day. If this gas is emitted at 800°C and 1.0 atm, what volume of gas is emitted?
P=1.0atm*101.3kPa1atm=101.3kPa;
T=800°C; C+273=K→800°C+273=1073 K;
n=200tons-300tons907185g1ton1mol64.064g=2.8326*106 to 4.248*106mol
R=8.314Jmol*K
PV=nRT→101.3kPa*V=2.8326*106 to 4.248*106mol*8.314Jmol*K*1073K=
=2.49*108 to 3.74*108L
1.15 A gas mixture being used to simulate the atmosphere of another planet consists of 320 mg of methane, 175 mg of argon, and 225 mg of nitrogen. The partial pressure of nitrogen at 300 K is 15.2 kPa. Calculate (a) the volume and (b) the total pressure of the mixture.
moles methane= .32 g*1mol16.04g= .02 mol
moles argon= .175 g*1mol39.948g= .00438 mol
moles nitrogen=.225g*1mol28.014g= .01606 mol
Pnitrogen=Xnitrogen*Ptotal→15.2kPa=.01606mol.02+.00438+.00803mol*Ptotal=61.246kPa
PV=nRT→61.246kPa*V=.02+.00438+.00803mol*8.314Jmol*K*300K=1.32L
1.17 A determination of the density of a gas or vapour can provide a quick estimate of its molar mass even though for practical work mass spectrometry is far more precise. The density of a gaseous compound was found to be 1.23 g dm−3 at 330 K and 25.5 kPa. What is the molar mass of the compound?
ρ=1.23 gdm3*1kg1000g10dm1m3=1.23 kgm3
PV=nRT→Pmρ=mMRT→MP=ρRT→M*25.5kPa=8.314Jmol*K*1.23kgm3*330K=132g/mol
1.19 A vessel of volume 22.4 dm3 contains 2.0 mol H2 and 1.0 mol N2 at 273.15 K. Calculate (a) their partial pressures and (b) the total pressure.
PV=nRT→P*22.4dm31m10dm3=2.0+1.0mol*273.15K*8.314Jmol*K=304kPa
Phydrogen=Xhydrogen*P=2/3*304kPa=202.6kPa
Pnitrogen=Xnitrogen*P=1/3*304kPa=101.3kPa
