Pae Behavior of Gases Chapter Gases and Their Properties Jeffrey ack California State University, Sacramento Importance of Gases Hot Air Balloons How Do They Work? Airbas fill with N as in an accident enerated by the rapid decomposition of sodim azide, NaN 3. NaN 3 (s) Na(s) + 3 N () The States of atter General Properties of Gases Gases are composed of a lare collection of particles in constant random motion. Gases occpy containers niformly and completely. Gases diffse rapidly to for homoeneos mixtres. ost of the volme of a as is empty space.
Pae Properties of Gases Gas properties are characterized by the followin qantities: volme of the as (L) T temperatre (K) n amont (moles) P pressre (atmospheres) Pressre Atmospheric pressre of air is measred with a device called a Barometer First developed by Evanelista Torricelli in 643. Pressre H(l) rises p an evacated tbe ntil force of H (down) eqals the force of atmosphere (pshin p). The pressre of the atmosphere is proportional to the heiht of the colmn and the density of the liqid. since mass (m) density volme d P since Pressre force area m A 3 olme (x ) Area (x ) P d h d A heiht (x) We conclde the pressre is proportional to the density and the heiht of a colmn of a iven sbstance. ercry has a density of 3.6 /cm 3 and water has a density of.00 /cm 3. If a colmn of mercry has a heiht of 755 mm, how hih wold a correspondin colmn of water be in feet? Soltion: We bein by settin the pressres eqal: We can write: Since P H P water P d h dh h H dwater h water ercry has a density of 3.6 /cm 3 and water has a density of.00 /cm 3. If a colmn of mercry has a heiht of 755 mm, how hih wold a correspondin colmn of water be in feet? h water 3 h d h H H water dwater 3.6 755 mm cm 33.7 ft d H h H dwater h cm 0 mm.00 cm water 3 in.54 cm This is why we don t se water in a barometer! ft in
Pae 3 Pressre easrement Units Unit: Conversion to atm: Atmosphere (atm) 760 mm H atm 760 torr atm mm H torr 0,35 Pa* atm.035 bar** atm 4.696 lb/in atm *Pa (N/m ) are often written as kpa ( kpa 0 3 Pa) **bar are often written as mbar ( mbar 0 3 bar) Boyle s Law The pressre of a system of as particles is inversely proportional to the volme of fixed nmber of moles at constant temperatre. P µ P C B P P Robert Boyle (67-69). Son of Earl of Cork, Ireland. Boyle s Law A sample of nitroen as has a pressre of 67.5 mm H in a 500.-mL flask. What is the pressre of this as sample when it is transferred to a 5-mL flask at the same temperatre? P P P P 67.5 mm H 500. ml 70. mm H 5 ml Boyle s Law A bicycle pmp is a ood example of Boyle s law. As the volme of the air trapped in the pmp is redced, its pressre oes p, and air is forced into the tire. Charles s Law The volme of a system of as particles is directly proportional to the absolte temperatre of fixed nmber of moles at constant pressre. µ T C T C T T Jacqes Charles (746-83). Isolated boron and stdied ases. Balloonist.
Pae 4 Gas: Temperatre Scales For as calclations, we need an absolte scale, one that does not take on neative vales. The conversion between C and Kelvins is: K C + 73.5 When performin calclations with absolte temperatres, on mst se the Kelvin scale. Kelvin Scale Converstions Convert 5.0 C to Kelvin: Convert 33 K to C: 5.0 C + 73.5 98. K 33 K 73.5 50. C Watch yor si. fis.!!!! Charles s Law Charles s Oriinal Balloon Balloons immersed in liqid N (at -96 C) will shrink as the air cools (and is liqefied). odern lon-distance balloon Charles s Law A 5.0-mL sample of CO as is enclosed in a astiht syrine at C. If the syrine is immersed in an ice bath (0 C), what is the new as volme, assmin that the pressre is held constant? T T æ 5.0 ml ö è ø T 73 K ç 4.6 ml T 95 K
Pae 5 General Gas Law Combinin Charles s and Boyle s Laws P constant T for a set nmber of as moles So at two sets of conditions: P P a constant T T P P T T Yo have a sample of CO in flask A with a volme of 5.0 ml. At 0.5 C, the pressre of the as is 436.5 mm H. To find the volme of another flask, B, yo move the CO to that flask and find that its pressre is now 94.3 mm H at 4.5 C. What is the volme of flask B? P P T T P T 436.5 mm H 5.0 ml 97.7 K 7 ml T P 94.3 mm H 93.7 K Avoadro s Hypothesis Eqal amonts of ases (moles) at the same temperatre (T) and pressre (P) occpy eqal volmes (). µ n A sample of ethane (C H 6 ) nderoes combstion. What volme of O (L) is reqired for complete reaction with 5. L of ethane? What volme of H O () in liters is prodced? Assme all ases are measred at the same temperatre and pressre. Write the balanced chemical eqation for the reaction. C H 6 () + 7O () 4CO () + 6H O() twice as many molecles A sample of ethane (C H 6 ) nderoes combstion. What volme of O (L) is reqired for complete reaction with 5. L of ethane? What volme of H O () in liters is prodced? Assme all ases are measred at the same temperatre and pressre. A sample of ethane (C H 6 ) nderoes combstion. What volme of O (L) is reqired for complete reaction with 5. L of ethane? What volme of H O () in liters is prodced? Assme all ases are measred at the same temperatre and pressre. C H 6 () + 7O () 4CO () + 6H O() C H 6 () + 7O () 4CO () + 6H O() 7 L O 5. L C H 8 L O 6 L CH6 6 L H O 5. L C H 6 L H O 6 L CH6
Avoadro s Hypothesis Combinin Avoadro s Hypothesis with the General Gas Law Pae 6 This in known as the Ideal Gas Law P n T constant P n R T The ases in this experiment are all measred at the same T and P. H () + O () H O() L atm R "as constant" 0.0806 mol K P nrt Rles for Ideal Gas Law Calclations. Always convert the temperatre to Kelvin (K 73.5 + C). Convert from rams to moles if necessary. 3. Be sre to convert to the nits of R (L, atm, mol & K). Types of Ideal Gas Law problems yo may enconter: Standard Temperatre & Pressre atm (760 torr or mm H & 0 C (73.5K) Calclate the standard molar volme of a as. Determination one nknown qantity of one as variable (P,, T, or n) iven the other vales directly or indirectly. Determine the new vales of P,, T, or n after a as nderoes a chane. Stoichiometry problems. Gas density and molar mass problems. P nrt 0.0806 L atm.00 mol mol K.00 atm 73.5K.4 L Standard Temperatre & Pressre In order to provide a reference point for the comparison of asses, standard temperatre and pressre are set at: atm (760 torr or mm H & 0 C (73.5K) Calclate the volme occpied by 43.7 of hydroen at STP. P nrt 43.7 H mol H 0.0806 L atm 73.5K.0 mol K 485 L.00 atm The density of a as (rams / L) can be obtained from the ideal as law. n moles mass () sbstittin "n" into the Ideal Gas Law Rearranin: m Gas Density wt æ m ö P ç RT è wt ø P wt RT ( mol) ( ) density L
Pae 7 Example Problem for findin Gas Density What is the density of oxyen at STP? mass m density olme A 0.05- sample of a as with an empirical formla of CHF is placed in a 65-mL flask. It has a pressre of 3.7 mm H at.5 C. What is the moleclar formla of the compond? as we fond before: m P wt RT The molar mass of O is 3.00 /mol m atm 3.00 mol 0.0806 L atm 73.5K mol K STP atm & 0 C.48 L A 0.05- sample of a as with an empirical formla of CHF is placed in a 65-mL flask. It has a pressre of 3.7 mm H at.5 C. What is the moleclar formla of the compond? 0.05 d 0.0758 /L 0.65 L A 0.05- sample of a as with an empirical formla of CHF is placed in a 65-mL flask. It has a pressre of 3.7 mm H at.5 C. What is the moleclar formla of the compond? 0.05 d 0.0758 /L 0.65 L atm 3.7 mm H 0.080 atm 760 mm H A 0.05- sample of a as with an empirical formla of CHF is placed in a 65-mL flask. It has a pressre of 3.7 mm H at.5 C. What is the moleclar formla of the compond? A 0.05- sample of a as with an empirical formla of CHF is placed in a 65-mL flask. It has a pressre of 3.7 mm H at.5 C. What is the moleclar formla of the compond? 0.05 d 0.0758 /L 0.65 L atm 3.7 mm H 0.080 atm 760 mm H drt 0.0758 / L 0.08057 L atm / K mol 95.7 K P 0.080 atm 0 / mol 0.05 d 0.0758 /L 0.65 L atm 3.7 mm H 0.080 atm 760 mm H drt 0.0758 / L 0.08057 L atm / K mol 95.7 K P 0.080 atm 0 / mol The moleclar formla is (CHF ) or C H F 4.
Pae 8 Gas Density Balloon filled with helim (lower density) Gas Laws & Chemical Reactions Bombardier beetle ses decomposition of hydroen peroxide to defend itself. If 0. of H O decomposes in a.50 L flask at 5 o C, what is the pressre of O & H O? Balloons filled with air If 0. of H O decomposes in a.50 L flask at 5 o C, what is the pressre of O & H O? Step : Write the balanced chemical reaction. Step : Calclate the moles of each prodct. Step 3: Find the pressre of each via P nrt If 0. of H O decomposes in a.50 L flask at 5 o C, what is the pressre of O & H O? Step : Write the balanced chemical reaction. Step : Calclate the moles of each prodct. Step 3: Find the pressre of each via P nrt H O (liq) H O() + O () Step : If 0. of H O decomposes in a.50 L flask at 5 o C, what is the pressre of O & H O? H O (liq) H O() + O () 0. H O mol H O 0.003 mol H O 34.0 If 0. of H O decomposes in a.50 L flask at 5 o C, what is the pressre of O & H O? H O (liq) H O() + O () 0. H O mol H O 0.003 mol H O 34.0 mol O 0.003 mol H O 0.006 mol O mol HO Step : Step :
Pae 9 If 0. of H O decomposes in a.50 L flask at 5 o C, what is the pressre of O & H O? nrt P O H O (liq) H O() + O () 0. H O mol H O 0.003 mol H O 34.0 mol O 0.003 mol H O 0.006 mol O mol HO 0.006 mol 0.0806 L atm/k mol 98 K.50 L 0.06 atm Step 3: If 0. of H O decomposes in a.50 L flask at 5 C, what is the pressre of O & H O? H O (liq) H O() + O () 0. H O mol H O 0.003 mol H O 34.0 mol O 0.003 mol H O 0.006 mol O mol HO nrt P HO 0.003 mol 0.0806 L atm/k mol 98 K.50 L 0.03 atm Step 3: Gas Laws & Stoichiometry A 50.0 sample of a mixtre of CO 3 and NaCl was heated and the CO () collected in a 0.0 L flask had a pressre of 755 torr at 0.0 C. How many rams of NaCl was in the oriinal sample? Gas Laws & Stoichiometry A 50.0 sample of a mixtre of CO 3 and NaCl was heated and the CO () collected in a 0.0 L flask had a pressre of 755 torr at 0.0 C. How many rams of NaCl was in the oriinal sample? Reconize that the CO () prodced is related to CO 3 (s) by CO (s) O (s) CO () 3 Gas Laws & Stoichiometry A 50.0 sample of a mixtre of CO 3 and NaCl was heated and the CO () collected in a 0.0 L flask had a pressre of 755 torr at 0.0 C. How many rams of NaCl was in the oriinal sample? Reconize that the CO () prodced is related to CO 3 (s) by CO (s) O (s) CO () 3 THERE IS NO REACTION BETWEEN THE SALTS! CO 3 (s) + NaCl(s) X NO REACTION! Gas Laws & Stoichiometry A 50.0 sample of a mixtre of CO 3 and NaCl was heated and the CO () collected in a 0.0 L flask had a pressre of 755 torr at 0.0 C. How many rams of NaCl was in the oriinal sample? olme CO moles CO mols CO 3 CO 3 se P nrt CO (s) O (s) CO () 3 mole ratio Sample CO 3 NaCl molar mass
A 50.0 sample of a mixtre of CO 3 and NaCl was heated and the CO () collected in a 0.0 L flask had a pressre of 755 torr at 0.0 C. How many rams of NaCl was in the oriinal sample? 755 torr Gas Laws & Stoichiometry atm 0.0L 760 torr 0.0806 L atm 93K mol K P n RT mol CO mol CO 3 3 84.3 CO mol CO 3 34.8 CO 3 Gas ixtres & Partial Pressres: Dalton s Law John Dalton 766-844 Pae 0 moles CO 50.0 sample 34.8 5. NaCl Dalton s Law of Partial Pressre The ole Fraction ( n ) + n + n 3... R T P tot n totrt Each as can be represented as a fraction of the total moles in the mixtre. This is called the ole Fraction. Since each as adds toether to create a total pressre (P tot ), the molar fraction that represents each of the individal ases is akin to a concentration. For a mixtre of ases,,3,4 i with moles n, n, n 3, n 4 n i moles of an individal as ole Fraction total moles of as in the mixtre X i C n n + n + n +... 3 The sm of all of the mole fractions in the mixtre is (exactly) X + X + X 3 The ole Fraction Since X i is proportional to n i, the partial pressre of as i is iven by: P P nrt i n i i n tot totrt ntot X i A sample of natral as contains 8.4 moles of CH 4, 0.4 moles of C H 6, and 0.6 moles of C 3 H 8. If the total pressre of the ases is.37 atm, what is the partial pressre of propane (C 3 H 8 )? P i X i P T X propane P T.37 atm 0.6 8.4 + 0.4 + 0.6 0.03 therefore: P i P tot X i P propane 0.03 x.37 atm 0.08 atm
The Kinetic-oleclar Theory of Gases Postlates: A as is a collection of a very lare nmber of particles that remains in constant random motion. Clasis (857) The pressre exerted by a as is de to collisions with the container walls The particles are mch smaller than the distance between them. The Kinetic-oleclar Theory of Gases The particles move in straiht lines between collisions with other particles and between collisions with the container walls. (i.e. the particles do not exert forces on one another between collisions.) Pae The averae kinetic enery (½ mv ) of a collection of as particles is proportional to its Kelvin temperatre. Gas particles collide with the walls of their container and one another withot a loss of enery. The Kinetic-oleclar Theory of Gases Gas pressre at the particle level: The Kinetic-oleclar Theory of Gases The relationship between temperatre (T) and velocity () (kinetic enery) can be fond by the followin: Ideal as law KT Settin the two eqal: nrt Nm P and P 3 solvin: The root mean sqare velocity for a as is: nrt 3nRT Nm RS Nm 3 wt wt J R 8.34 mol K æ k ö wt is in ç è mol ø Kinetic-oleclar Theory At the same T, all ases have the same averae KE. As T oes p, KE also increases and so does speed. What is the RS velocity of a nitroen molecle in miles per hr at STP? RS RS Kinetic-oleclar Theory ½ J 3 8.34 73.5K mol K N k 8.0 mol N 3 0 m/s k m recall... J s 0 cm m in.54cm ft in mile 3600s 580ft hr.03 0 3 mph pretty zippy eh?
Pae Kinetic oleclar Theory elocity of Gas Particles For a iven temperatre, heavier ases move slower than lihter ases. The velocities are described by a distribtion. Averae velocity decreases with increasin mass. Distribtion of Gas olecle Speeds Gas Diffsion & Effsion Boltzmann Distribtion Named for Ldwi Boltzmann dobted the existence of atoms. Diffsion is the process of as miration de to the random motions and collisions of as particles. It is diffsion that reslts in a as completely fillin its container. After sfficient time as mixtres become homoeneos. Gas Diffsion: Relation of mass to of Diffsion HCl and NH 3 diffse from opposite ends of tbe. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tbe. Gas Effsion EFFUSION is the movement of molecles throh a small hole into an empty container. (vacm)
Pae 3 Gas Effsion Effsion is overned by Graham s Law: The rate of effsion of a as is proportional to its RS. 0.3 ml/min at 5.0 C. Another as effses at a rate µ µ RS Thomas Graham, 805-869. Where is the molar mass of a sbstance. This implies that heavier ases will effse slower than lihter ases. 0.3 ml/min at 5.0 C. Another as effses at a rate 0.3 ml/min at 5.0 C. Another as effses at a rate µ RS µ RS µ RS µ RS Comparin the rate of effsion of CO vs. the nknown as: CO 0.3 ml/min at 5.0 C. Another as effses at a rate 0.3 ml/min at 5.0 C. Another as effses at a rate µ RS µ RS µ RS µ RS Comparin the rate of effsion of CO vs. the nknown as: Comparin the rate of effsion of CO vs. the nknown as: CO CO CO CO CO
Pae 4 0.3 ml/min at 5.0 C. Another as effses at a rate 0.3 ml/min at 5.0 C. Another as effses at a rate µ RS µ RS µ RS µ RS Comparin the rate of effsion of CO vs. the nknown as: Comparin the rate of effsion of CO vs. the nknown as: CO CO CO CO CO CO CO 0.3 ml/min at 5.0 C. Another as effses at a rate 0.3 ml/min at 5.0 C. Another as effses at a rate CO CO CO CO Solvin for the molar mass of the nknown as: æ ö ç è ø CO CO 0.3 ml/min at 5.0 C. Another as effses at a rate 0.3 ml/min at 5.0 C. Another as effses at a rate CO CO Solvin for the molar mass of the nknown as: æ ö ç è ø CO CO CO CO Solvin for the molar mass of the nknown as: æ ö ç è ø CO CO 44.0 mol
Pae 5 0.3 ml/min at 5.0 C. Another as effses at a rate 0.3 ml/min at 5.0 C. Another as effses at a rate CO CO Solvin for the molar mass of the nknown as: æ ö ç è ø CO CO CO CO Solvin for the molar mass of the nknown as: æ ö ç è ø CO CO 44.0 mol æ ml ö ç 0.3 min ç ml ç 0.363 è min ø 44.0 mol æ ml ö ç 0.3 min ç ml ç 0.363 è min ø 8.0 Water! mol Avoadro s Hypothesis, Gas Pressre, Temperatre, Boyle s Law & Kinetic-oleclar Theory P proportional to n when and T are constant P proportional to T when n and are constant P proportional to / when n and T are constant Deviations from the Ideal Gas Law Real particles have volme. Real particles interact via intermoleclar forces. At hih pressre and low temperatre, these factors become sinificant.! Deviations from the Ideal Gas Law The AN DER WAALS s EQUATION makes corrections for the volme of particles and any intermoleclar forces that exist easred P easred (ideal) æ n a ö ç P + ( - nb) nrt è ø vol. correction intermol. forces J. van der Waals, 837-93, Professor of Physics, Amsterdam. Nobel Prize 90.