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Over [174]
3 years ago
14

The molar volume of oxygen,O2, is 3.90 dm3 mol-1 at 10.0 bar and 200 degree centigrade. Assuming that the expansion may be trunc

ated after the second term, calculate the second virial coefficient of oxygen at this temperature.
Chemistry
1 answer:
schepotkina [342]3 years ago
4 0

Answer:

B = - 0.0326 dm³/mol

Explanation:

virial eq until second term:

  • PVm = RT [ 1 + B/Vm ]

∴ P = 10 bar * (atm/ 1.01325 bar) = 9.869 atm

∴ T = 200°C = 473 K

∴ Vm = 3.90 dm³/mol

∴ R = 0.08206 dm³.atm/K.mol

⇒ PVm / RT = 1 + B/Vm

⇒ ((9.869 atm)*(3.90 dm³/mol)) / ((0.08206 dm³.atm/mol.K)*(473K)) = 1 + B/Vm

⇒ 0.99164 = 1 + B/Vm

⇒ B/Vm = - 8.357 E-3

⇒ B = (3.90 dm³/mol)*( - 8.357 E-3 )

⇒ B = - 0.0326 dm³/mol

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svp [43]
Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958   = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass 
                                      = 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water 
                                        = 4.7 / 0.88 = 5.34 moles 
∴ moles of the solution = total moles in solu - moles of water 
                                       = 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
                                                    = 53.8 / 0.64 = 84 g/mole

Q: C

moles of urea (NH2)2 CO = mass weight / molar mass
                                           = 4.49 g / 60 g /mol
                                           = 0.07 mol
moles of methanol = mass weight / molar mass 
                                 = 39.9  g / 32  g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg 
∴Pv(solu) = 84.55 mmHg


 
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Answer: Id say C

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