CHEM 170 1st Edition Lecture 4Outline of Last Lecture I. Last Lecture: Boyle’s LawII. Charles’s LawIII. Avogadro’s LawIV. Ideal Gas LawOutline of Current Lecture V. Last Lecture: Ideal Gas LawVI. How To Compare a Gas Under Changing ConditionsCurrent LectureChapter 11 – GasesLast Lecture: Ideal Gas LawWhere V=volume, T=temperature, n=number of moles, P=pressure and R is the ideal gas constant.V ∝TnP → V =RTnPPV =nRTWhere R=PVnT=1 atm× 22.4 L1 mol×273.15 K=0.0821L ×atmmol × K¿1.013 ×105Pa× 22.4 ×10−3m31 mol ×273.15 K=8.314Jmol × KAnd n=massmolecular weightNote: 1.01325× 105pascals=1 atm=760 mmHg=760 torrNote: Standard Temperature and Pressure (STP) is 0°C and 1 atmex. If V=128mL, P=5×10-10torr, T=25°, then how many molecules are there?PV =nRT →n=PVRT=(5× 10−10tor)(1 atm760torr)(128 mL)(1 L1000 mL)(0.0821L× atmmol × K×(25+273 K))=4 × 10−15These 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. Both are the Ideal Gas Constant with different units4 ×10−15mol(6.02 ×1023molescules)1 mol=2 ×109moleculesex. What is the molecular weight (MW) of a gas if 1.81g of the gas occupies 1.52L at 298K and 0.970atm?PV =nRT → n=PVRT=0.970 atm× 1.52 L0.082L × atmmol × K×298 K=0.6034 moln=massmolecular weight → (MW)=massn=1.81 g0.6034 mol=30.0gmolex. What is the density of oxygen at STP? T =0 ℃ , P=1 atm , ρ=mV, PV =nRT→ V =nRTPD=mV=mPnRT=mPmass(MW )RT=(MW)PRT=32gmol×1 atm0.0821L × atmmol × K×(0+273 K )=1.43gLHow To Compare a Gas Under Changing Conditions:A gas may be placed under different conditions that causes it to change from one state to another.P1V1=n1R T1 → R=V1P1n1T1P2V2=n2R T2 → R=V2P2n2T2since R is a constant: V1P1n1T1=V2P2n2T2ex. If there is 1L of oxygen at STP and the temperature was heated to 100°C while the volume remains at 1L and the number of moles does not change, then what is the resulting pressure?V1P1n1T1=V2P2n2T2 → P2=V1P1n2T2n1T1V2¿P1T2T1¿1 atm ×(100+273 K)273 K¿1.37 atmReal life example:ex. If 6.8g of Cu was completely consumed in the following reaction at 0.970atm and 45°C, then what is the resulting volume of NO2?Since volume, number ofmoles haven’t changed those values can cancel.2+¿+2 H2O(l)−¿ yields→2 NO2(g )+Cu(aq)¿+¿+2 NO3(aq)¿Cu(s)+4 H(aq)¿6.8 g Cu(1mol Cu63.54 g Cu)(2 mol N O21 molCu)=0.214 mol N O2(g )PV =nRT →V =nRTP=(0.214 mol)(0.0821L ×atmmol × K)(318 K )0.970 atm=5.76
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