Re: Chemistry Question help

Thank you guys so much!  You all saved me, and I got a good grade.  I had messed up my last chemistry assignment, and I had to bring  my grades up lest I fail the class.  It makes sense to me now, and I'm starting to enjoy chemistry.   I look forward to maybe being able to help others on this board someday.  

-Smileyranger :3

--- In [email protected], John W <JohnWW@...> wrote:
>
> That is correct, although it is based on the assumption that you are 
> dealing with theoretical "ideal gases", for which the ideal-gas 
> equation  PV = nRT applies.
> 
> 
> 
> For gases under conditions of P and T that are not far from 
> liquefaction, you would need to use a more complicated equation of state
>  that takes account the space occupied by molecules and intermolecular 
> attractions, in particular the Van Der Waals' equation, which has to be 
> solved as a cubic equation. Alternatively, one could use empirical 
> "compressibility factors", z, for a modified ideal-gas law PV = znRT. 
> These are tabulated for various gases at various temperatures and 
> pressures at ratios of their critical temperatures and pressures, and 
> solved for the other variable using a graph or gnomon. These methods are
>  described at length in Perry's Chemical Engineers' Handbook.
> 
> 
> 
> While it is true that the diffusivities of gases vary inversely as the 
> square root of their molecular (or atomic) weights, they are also 
> strongly influenced by the gases' atomicity (and hence molecular size) 
> and intermolecular attraction. For this reason, the diffusivity of Ne 
> (atomic weight 20), a monatomic inert gas with very little 
> intermolecular attraction due to lack of polarity, cannot really be 
> compared with that of CO (molecular weight 28), a diatomic gas with a 
> strongly polar multiple C=O bond and hence substantial intermolecular 
> attraction (and as the result a much higher melting/sublimation point at
>  the same pressure), simply on the basis of molecular weights. The 
> molecular size of Ne is smaller than that of CO, and in fact is smaller 
> than even that of the isoelectronic F- anion due to the greater nuclear 
> charge. The diffusivity of Ne would therefore be much higher than that 
> expected simply by comparison of its molecular weight with that of CO. 
> This is also discussed at length in Perry's Chemical Engineers' 
> Handbook, which gives a semiempirical equation for gas diffusivities.
> 
> 
> 
> John W.
> 
> --- On Thu, 17/11/11, Saim Rauf <scorpion_hellfire95@...> wrote:
> From: Saim Rauf <scorpion_hellfire95@...>
> Subject: Re: [The Chemistry Cluster] Chemistry Question help
> To: "[email protected]" <[email protected]>
> Date: Thursday, 17, November, 2011, 8:55 PM
> 
> 
> 
> 
> 
> 
> 
> 
>  
> 
> 
> 
>   
> 
> 
>     
>       
>       
>       
> 
> 
> 
> for your problem no. 1 the equation to use is of boyle's law which is
> 
> 
> 
> P.V=P'.V'   (if the temp. remains constant)
> 
> 
> 
> for problem no . 2 the equation to be used is
> 
> 
> 
> V / T=V'/T'      (where the temp. is taken in kelvins)
> 
> 
> 
> for problem no. 3 we use   boyles and charles law combined as
> 
> 
> 
> (P.V)/T = (P'.V')/T' (the temp is measure in kelvins)
> 
> 
> 
> and since the vol. is constant it can be dropped from both sides.
> 
> 
> 
> as regarding prob no. 4... as we know the volume of gas directly varies with the no. of moles so we use equation
> 
> 
> 
> V/n=V'/n' (where n'=no of moles present + added no of moles)
> 
> 
> 
> for your fifth problem ideal gas equation is to be used as
> 
> 
> 
> (P.V)/T = (P'.V')/T'
> 
> 
> 
> the sixth problem is quite easy... we have to use daltons law of partial pressures according to which
> 
> 
> 
> total pressure = pressure exerted by first gas + pressure exerted by second gas + pressure exerted by third gas + ............
> 
> 
> 
> and now your seventh problem.. the molar mass of neon is 20 where as the
>  molar mass of CO is 12+16=28 thus by grahams law of effusion... we have
> 
> 
> 
> rate of effusion of gas 1 / rate of effusion of gas 2 = sqrt(molar mass 
> of gas                                                                  
>                 1)/ sqrt(molar mass                                      
>                                                              of gas 2)
> 
> so the rate of effusion of a gas inversely varies with its molar mass 
> thus we infer the rate of effusion of neon is more than that of CO.. 
> (the same solution holds for diffusion)..
> 
> 
> 
> REGARDS
> 
> 
> 
> ________________________________
> 
> From: smileyranger_iie <smileyranger@...>
> 
> To: [email protected]
> 
> Sent: Tuesday, November 15, 2011 6:29 PM
> 
> Subject: [The Chemistry Cluster] Chemistry Question help
> 
> 
> 
>   
> 
> Cn you guys please help me with the following questions? 
> 
> 
> 
> At constant temperature, a gas occupies a volume of 840 mL at 520 mm Hg.
>  If we increase the pressure to 705 mm Hg, what will be the volume of 
> the gas?
> 
> 
> 
> At constant pressure, a gas occupies a volume of 270 mL at 320 C. What is the volume of the gas at -210 C?
> 
> 
> 
> We have 950 mL of a gas at 200	C and 350 mm Hg.	If we heat the gas to 1750	C at constant volume, what will the pressure be?
> 
> 
> 
> A balloon holding 390 mL holds 2.8 moles of argon. If we add 3.2 moles 
> of the gas at the same temperature and pressure, what will the volume of
>  the balloon be?
> 
> 
> 
> You have 540 mL of a gas at a pressure of 723 mm Hg and a temperature of
>  340 C. What is the volume if you change the pressure to 840 mm Hg and 
> the temperature to 91 0C?
> 
> 
> 
> A mixture of three gases is in a container. The mixture exerts a total 
> pressure of 456 atmospheres. If gas A exerts a pressure of 195 
> atmospheres and gas B exerts a pressure of 97 atmospheres, what is the 
> pressure of gas C?
> 
> 
> 
> Which gas will diffuse faster: Ne or carbon monoxide (CO)? Explain your answer.
> 
> 
> 
> Thanks!
> 
> 
> 
> -Smileyranger :3
> 
> 
> 
> 
> 
> [Non-text portions of this message have been removed]
>




------------------------------------

Yahoo! Groups Links

<*> To visit your group on the web, go to:
    http://groups.yahoo.com/group/thechemistrycluster/

<*> Your email settings:
    Individual Email | Traditional

<*> To change settings online go to:
    http://groups.yahoo.com/group/thechemistrycluster/join
    (Yahoo! ID required)

<*> To change settings via email:
    [email protected] 
    [email protected]

<*> To unsubscribe from this group, send an email to:
    [email protected]

<*> Your use of Yahoo! Groups is subject to:
    http://docs.yahoo.com/info/terms/