DOC PREVIEW
KU CHEM 170 - Ideal Gas Law
Type Lecture Note
Pages 3

This preview shows page 1 out of 3 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

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


View Full Document

KU CHEM 170 - Ideal Gas Law

Type: Lecture Note
Pages: 3
Download Ideal Gas Law
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Ideal Gas Law and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Ideal Gas Law 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?