CHEM 170 1st Edition Lecture 5Outline of Current Lecture I. Discussion of next week’s lab with solar cellsII. Ideal gas lawIII. Dalton’s Law or Partial PressuresCurrent LectureNext week’s laboratory experiment: Introduction to Green ChemistrySolar cells harness light energy for usage in human lives. A solar cell is made of cover glass, antireflective coating, n-type semiconductor (extra electrons, negative), p-type semiconductor (fewer electrons, positive) and back contact. Silicon’s bonding behavior issimilar to carbon, since they both have four valence electrons. The goal of the experiment is to create a solar cell using dye from blackberries and determine the current and voltage produced by the solar cell under various types of light.Chapter 11 – GasesIdeal Gas Lawex.2 NaN3 (s)→ 2 Na(l)+3 N2(g)If the pressure is 735mmHg, temperature is 26°C and there is 12.5g of NaN3 that is completely consumed, then what would be the volume of N2?12.5 g NaN3(1 mol NaN365.01 g NaN3)(3 mol NaN32 mol NaN3)=2.88 mol N3P=735 mmHg(1 atm760 mmHg)=0.967 atmT =26 °C +273=299 KPV=nRT →V =nRTPV =(2.88 mol)(0.0821L × atmmol × K)(299 K )(0.976 atm)Dalton’s Law of Partial Pressures states that the total pressure is the sum of the partial pressures of individual gases in a mixture.P=n(RTV)*V ∧T are constantTherefore P is directly related to n and (RTV) can be treated as a constant.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Ptotal=PA+PB+…Ptotal=nA(RTV)+nB(RTV)+…nA+nB+…Ptotal=(RTV)¿)Ptotal=(RTV)(ntotal)Also Ptotal is directly related to ntotal and (RTV) can be treated as a constant.PAPtotal=nARTVntotalRTVPAPtotal=nAntotal=mole fraction=XAPA=XAPtotalex. What is the pressure exerted by a mixture of 1 g H2 and 5g He in a 5.0 L container at 20°C (293K)?ntotl=1.0 g H2(1 mol2.07 g H2)+5.0 g He(1mol40 g H2)¿0 . 5 mol H2+1.25 mol He¿1.75 molP=n(RTV)P=(1.75 mol)((0.0821L× atmmol × K)(293 K)(5.0 L))=8.4 atmCHECK: PH2=nH2(RTV)=(0.5 mol)(0.0821L ×atmmol × K)(293 K )5.0 L=2.4 atmPHe=nHe(RTV)=(1.25 mol)(0.0821L ×atmmol × K)(293 K)5.0 L=6.0 atmPtotal=PH2+PHe=2.4 atm+6.0 atm=8.4 atmORPH2=nAntotalPtotal=(0.5 mol H21.75 mol He)(8.4 atm)=2.4 atmPHe=nAntotalPtotal=(0.5 mol H21.75 mol He)(8.4 atm)=6.0
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