A   solution of the problem from your quiz.

You can use any set of units  but you need to convert consistently, even in dm - s ,  
    by the way 1  dm = 
      0.1  m 

One choice is to stay in SI system all the time so that you do not need to covert R to atm,  dm etc. ,  
    but you have to convert VdW  a and b constants .  
        That means that your molar volume can be calculated using R and  Pa (Pascals) for pressure
and will come in (m^3) /mole .  Try that!

You can (and some of you did try) to calculate  molecular volume iseparately in 
dm - atm units, 
    but you have to convert R (which many of you did not) .

Note that here in Mathematica  dimensional equations are solved in the same time as numericalvalues are calculated,  
    be careful how to write powers on the units

VALUES:

Speed

c = 340  m  (s^(-1))

R - constant in atm

Ra = 0.0820574 (dm^3)  atm  (mol^(-1)) (K^(-1))

w/bars

Rb = 0.0831447  (dm^3)  bar  (mol^(-1)) (K^(-1))

or  R constant in  SI units ,  
    J (mol^(-1))  (K^(-1))

Rj = 
      8.31447 kg   (m^2) (s^(-2))  (mol^(-1))  (K^(-1))

Temperature

t  = 293.15 K

Mass (not m,  
        m is akready used for meter!!!)

mm = 0.028  kg (mol^(-1))

VdW a   and b  paramters for N2 (check units,  
        textbook (s))

ARE THEY ?? ????????????????????

a =  1.408 (dm^6)  atm (mol^(-2))

b = 0.03913 (dm^3)  (mol^(-1))

OR ?? ???????????????????????????????????

ared =  1.3483 (dm^6)  atm (mol^(-2))

bred = 0.038577  (dm^3)  (mol^(-1))

this is if you use atm units ab
your PC - lab book and redbook also refers to bars

aredbar =  
      1.3661 (dm^6)  bar (mol^(-2))

or

atext = 1.408 (dm^6)  bar (mol^(-2))

btext  =  0.03913 (dm^3)   (mol^(-1))

IF you work with SI use also converted a and b constants in SI (redbook)

CONVERT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

aSI = 
      0.13661  (m^6)  kg (m^(-1)) (s^(-2)) (mol^(-2))

bSI = 0.00003857  (m^3)  (mol^(-1))

Pressure (you can stay in atm or move to Pa which is SI unit)

p = 760 tor  = 1 atm  =  ... ... ...  ..  Pa  =  kg (m^(-1)) (s^(-2))

pa = 1 atm

for 1 atm:

pbar  =  1.01325  bar

Find pPa  !!!!!!! (1 atm  =  
          101.325 k Pa)

pPa =  101.325 (10^3) kg (m^(-1)) (s^(-2))

Molar volume  v  (if you find convenient to work w/
            atm )

v =  Ra  t/ pa

if w/bars

vbar  =  Rb t /pbar

IF you work with SI this will be 24.0551 (10^(-3)) (m^3)/mol and you can use SI R

vPa  =  ((Rj t)/pPa)

IF you work with SI use also converted a and b constants in SI

Check the units on  the expression  :  (all  cancel, 
          so in SI and dm - 
            cm you have the same result)

this one is  w/
        values  a b from labtextb but in wrong units atm for a should be in bars

(1 - ((2    a)/(pa  (v^2)))   +   ((2    b)/v)   )

should be pbar

pbar =  1.01325  kg (m^(-1)) (s^(-2))

pbar =  1 bar

NOTE THAT CONVERSION IN RED BOOK IS DONE ON THE BASIS OF 1.0132 (NOT 1.01325)

aredbar/ared

(1 - ((2    aredbar)/(pbar  (vbar^2)))   +   ((2    bred)/vbar)   )

if you use redbook w /atmospheres

(1 - ((2    ared)/(pa  (v^2)))   +   ((2    bred)/v)   )

So the last two are a bit different!!!!

THIS IS WHAT YOU SHOULD HAVE  -  all in SI

(1 - ((2    aSI)/(pPa  (vPa^2)))   +   ((2    bSI)/vPa)   )

in SI

ccccccccccccccccccccccccccccccccccccccccccccc

Check SI

((2 bSI)/vPa)

((2    aSI)/(pPa  (vPa^2)))

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

Now calculate the main expression:

in redbook bar

g=(( mm (c^2))/(Rj   t     (1 - ((2    aredbar)/(pbar (v^2)))   +   ((2    bred)/v)   )))

(this one is close to SI only discrepancy is becose of Pa - conversion in redbook)

...... ...  ...  ...  ...  ...  ...  ...  ...  ...  ...  ...  ...

NOW in texbook bar  *this should have been your result

g=(( mm (c^2))/(Rj   t     (1 - ((2    atext)/(pbar (v^2)))   +   ((2    btext)/v)   )))

...... ...  ...  ...  ...  ...  ...  ...  ...  ..

in redbook atmosphere

g=(( mm (c^2))/(Rj   t     (1 - ((2    ared)/(pa (v^2)))   +   ((2    bred)/v)   )))

this is ok ,  same as SI here our conversion

IN SI

g=(( mm (c^2))/(Rj   t     (1 - ((2    aSI)/(pPa  (vPa^2)))   +   ((2    bSI)/vPa)   )))

OK no units

note  :  
      THE RESULT VALUE can be very close and you would get no points!!!!!

example calculation

g=(( 0.028   (((340))^2))/(8.31447   293.15     (1 - ((2    1.408)/(1  (((((8.314 293.15)/1)))^2)))   +   ((2    0.03913)/(((8.314 293.15)/1)))   )))