1 Ideal Gas Law Why bother with Gases? Properties of a GasAnimation P, V, T, n Ideal Gas Law Examples of Ideal Gas Law Ideal Gas Law - #molecules version

2 Temperature and Gases Gas Properties of a GasSimplest material phase – all kinetic energy. Energy “reservoir” – convert disordered KE to useful mechanical work. Properties of a Gas Pressure Volume Temperature Quantity 4 Properties are interrelated Animation Ideal gas law

3 (sorry about the Bosnian!)Ideal Gas Animation Gas/piston animation (Java animation) (sorry about the Bosnian!) Other animations

4 Ideal Gas ObservationsObservations of P, V, T, n Pressure vs. Volume (Boyle’s Law) P proportional to 1/V Pressure vs. Temperature (Gay-Lussac’s Law) P proportional to T Volume vs. Temperature (Charle’s law) V proportional to T Pressure vs. quantity P proportional to n Volume vs. quantity V proportional to n

5 Ideal Gas Law – Primary unitsIdeal gas law combines Boyle’s, Charles, Gay-Lussac’s PV = nRT P proportional 1/V P proportional T V proportional T P proportional n V proportional n Units P in Pascals = N/m2 V in m3 T in K° (not C °) n in moles R = J/mol-K Definition of Mole n = mass/molecular mass (N2=28g/mol, O2= 32g/mol, etc.)

6 Ideal Gas Law – Alternative unitsIdeal gas law combines Boyle’s, Charles, Gay-Lussac’s PV = nRT P in atmospheres V in liters T in K° n in moles R = L-atm/mol-K Use either primary units or alternative units completely, don’t mix n’ match! Use primary units for First Law (Ch 15) calculations.

7 Celsius vs. Kelvin (again)Degree sizes same! Kelvin = Celsius with zero shifted to absolute zero. 𝑇 𝐾 =𝑇 𝐶 May use Celsius when only relative changes important. Thermal Expansion Specific and Latent Heat problems Thermal Conduction Must use Kelvin when absolute temperature important. Ideal Gas Law Kinetic Theory Thermal Radiation First and Second Law Thermodynamics Similar to Gauge vs. Absolute Pressure

8 Example 13-10 - volume 1 mole at STPSTP = 0C (273K), 1 atm (101.3 kPa) Primary units 𝑃𝑉=𝑛𝑅𝑇 𝑉= 𝑛𝑅𝑇 𝑃 = (1 𝑚𝑜𝑙)( 𝐽 𝑚𝑜𝑙 𝐾)(273 𝐾) ∙ 𝑁 𝑚 2 = 𝑁 𝑚 𝑁 𝑚 2 = 𝑚 3 =22.4 𝐿 Alternative units 𝑉= 𝑛𝑅𝑇 𝑃 = (1 𝑚𝑜𝑙)( 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝐾)(273 𝐾) 1 𝑎𝑡𝑚 =22.4 𝐿

9 Example 13-11 - helium balloonVolume of 18 cm radius sphere 𝑉= 4 3 𝜋 𝑟 3 = 4 3 𝜋 𝑚 3 =2.44∙ 10 −2 𝑚 3 Number of moles in alternative units (just being different) 𝑛= 𝑃𝑉 𝑅𝑇 = 1.05 𝑎𝑡𝑚 24.4 𝐿 ( 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝐾) 293 𝐾 =1.065 𝑚𝑜𝑙𝑒𝑠 Mass of Helium 𝑚= 𝑚𝑜𝑙 (4 𝑔 𝑚𝑜𝑙)=4.26 𝑔

10 Example 13-12 - mass of air in roomVolume of 5 x 3 x 2.5 m room 𝑉=5 𝑚 ∙3 𝑚 ∙2.5 𝑚=37.5 𝑚 3 Number of moles 𝑛= 𝑃𝑉 𝑅𝑇 = ∙ 𝑁 𝑚 2 (37.5 𝑚 3 ) 𝐽 𝑚𝑜𝑙 𝐾 (273 𝐾) =1673 𝑚𝑜𝑙𝑒𝑠 Mass of 29 g/mol (N2 28 g/mol, O2 32 g/mol) 𝑚= 1673 𝑚𝑜𝑙 (29 𝑔 𝑚𝑜𝑙)=48,536 𝑔=48.54 𝑘𝑔

11 Example 13-13 – automobile tireAn automobile tire is filled to a gauge pressure of 200 kPa at 10°C. After a long drive the temperature as risen to 40°. What is the pressure now? Strategy – Put all constant quantities on same side of Ideal Gas Law 𝑃𝑉=𝑛𝑅𝑇 → 𝑃 1 𝑇 1 = 𝑛𝑅 𝑉 = 𝑃 2 𝑇 2 Convert 200 kPa gauge pressure to 301 kPa absolute pressure Solve 𝑃 2 𝑇 2 = 𝑃 1 𝑇 → 𝑃 2 = 𝑇 2 𝑇 1 𝑃 1 = 313 𝐾 283 𝐾 301 𝑘𝑃𝑎=333 𝑘𝑃𝑎 Convert 333 kPa absolute pressure back to 232 kPa gauge pressure

12 More examples of ideal gasProblem 29 V2 = V1 (P1/P2) (T2/T1) Problem 30 T2 = T1 (P2/P1) (V2/V1) Problem ρ = 32 x 10-3/22.4 x 10-3 Problem P2 = P1 (n2/n1) Problem 33 Problem 36 n2 = n1 (P2/P1) Problem 39 P2 = P1 (V1/V2)(T2/T1)

13 Problem 29 – V change with P, TFirst move all constant quantities to one side 𝑃𝑉=𝑛𝑅𝑇 → 𝑃 1 𝑉 1 𝑇 1 =𝑛𝑅= 𝑃 2 𝑉 2 𝑇 2 Then solve for V2 𝑉 2 = 𝑃 1 𝑃 𝑇 2 𝑇 1 𝑉 1 = 1 𝑎𝑡𝑚 3.2 𝑎𝑡𝑚 𝐾 273 𝐾 3 𝑚 3 =1.07 𝑚 3 Note: Mixed units OK, since only ratios important

14 Problem 30 – T change with P, VMove all constant quantities to one side 𝑃𝑉=𝑛𝑅𝑇 → 𝑃 1 𝑉 1 𝑇 1 =𝑛𝑅= 𝑃 2 𝑉 2 𝑇 2 Solve for T2 𝑇 2 = 𝑃 2 𝑃 𝑉 2 𝑉 1 𝑇 1 = 40 𝑎𝑡𝑚 1 𝑎𝑡𝑚 𝑉 𝑜 𝑉 𝑜 293 𝐾=1302 𝐾=1029 𝐶 (Diesel car) Note: Mixed units OK, since only ratios important

15 Problem 31 – Density of O2 Volume of 1 mole at STP Mass of a mole O222.4 L = m3 (shown earlier) Mass of a mole O2 32 g = .032 kg (lookup in table) Density 𝜌= 𝑚𝑎𝑠𝑠 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝑘𝑔 𝑚 3 =1.43 𝑘𝑔 𝑚 3

16 Problem 32 – Gas substitutionCalculate mole ratio for equal mass CO2 and N2 𝑛 2 𝑛 1 = 𝑚𝑎𝑠𝑠 44 𝑔 𝑚𝑜𝑙 𝑚𝑎𝑠𝑠 28 𝑔 𝑚𝑜𝑙 = (since mass equal) Move all constant quantities to one side 𝑃𝑉=𝑛𝑅𝑇 → 𝑃 1 𝑛 1 = 𝑅𝑇 𝑉 = 𝑃 2 𝑛 2 Solve for P2 𝑃 2 = 𝑛 2 𝑛 1 𝑃 1 = 𝑎𝑡𝑚=2.32 𝑎𝑡𝑚

17 Problem 36 – Gas substitutionFirst find #moles O2 𝑛 1 = 𝑔 32 𝑔 𝑚𝑜𝑙 =812.5 𝑚𝑜𝑙𝑒𝑠 Move all constant quantities to one side 𝑃𝑉=𝑛𝑅𝑇 → 𝑃 1 𝑛 1 = 𝑅𝑇 𝑉 = 𝑃 2 𝑛 2 Solve for n2 𝑛 2 = 𝑃 2 𝑃 1 𝑛 1 = 8 𝑎𝑡𝑚 9.7 𝑎𝑡𝑚 𝑚𝑜𝑙=670 moles Then find mass 670 moles Helium 𝑚𝑎𝑠𝑠 𝐻𝑒 =670 𝑚𝑜𝑙 ∙4 𝑔 𝑚𝑜𝑙=2680 𝑔=2.68 𝑘𝑔

18 Problem 39 – P change with V, TMove all constant quantities to one side 𝑃𝑉=𝑛𝑅𝑇 → 𝑃 1 𝑉 1 𝑇 1 =𝑛𝑅= 𝑃 2 𝑉 2 𝑇 2 Solve for V2 𝑃 2 = 𝑉 1 𝑉 𝑇 2 𝑇 1 𝑃 1 = 61.5 𝐿 48.8 𝐿 𝐾 291 𝐾 2.45 𝑎𝑡𝑚=3.43 𝑎𝑡𝑚 Note: Arbitrary units OK, since only ratios important

19 Ideal Gas Law - #molecules versionIdeal Gas Law can also be written: 𝑃𝑉=𝑛𝑅𝑇 𝑃𝑉= 𝑁 𝑁 𝐴 𝑅𝑇 𝑃𝑉=𝑁𝑘𝑇 where 𝑛= 𝑁 𝑁 𝐴 (N number of molecules) 𝑘= 𝑅 𝑁 𝐴 (NA Avagadro’s number) P in Pascals (no alternative units) V in m3 T in K° Boltzman’s constant k = R/NA = 1.38e-12 J/K