(Please either write in latex or on a piece of paper so I can follow your method

Question

(Please either write in latex or on a piece of paper so I can follow your method. Test is tomorrow)

Methanol, C3OH, has specific heat of 0.328 cal/g °C for its vapor and its heat of vaporization is 35.21 kJ/mole at its boiling point of 64.60 °C. Calculate the specific heat in J/g °C of liquid methanol if 122.1 kJ of heat is required to convert 100.0 g of liquid methanol at 32.60 °C to vapor at 95.60 °C. At wts:

C=12.00 g/mol, H=1.008 g/mol and O= 16.00 g/mol and

1cal=4.1840 J.

Ans = _______________

c) (6 points) Use the pertinent data from part b above and the Clausius-Clapeyron equation

the vapor pressure in Torr at 32.60 C. R=8.314

J/mol.K Hint: You need to use 273.2

Explanation / Answer

Solution :-

a) The attractive forces between molecules of a substance and their container is called Intermolecular forces.

The temperature where all three phases of a substance are is equilibrium is called the Triple point.

Raoults law involves a relationship between the mole fraction and the vapor pressure of the gas in a solution.

The abbreviation for the concentration unit of the ratio of solute mass to solution mass times one billion is ppb.

When the temperature increases most gases become less soluble in solution.

b) Methanol, C3OH, has specific heat of 0.328 cal/g °C for its vapor and its heat of vaporization is 35.21 kJ/mole at its boiling point of 64.60 °C. Calculate the specific heat in J/g °C of liquid methanol if 122.1 kJ of heat is required to convert 100.0 g of liquid methanol at 32.60 °C to vapor at 95.60 °C. At wts:

C=12.00 g/mol, H=1.008 g/mol and O= 16.00 g/mol and

1cal=4.1840 J.

Solution :- specific heat of methanol gas =0.328 cal * 4.184 J =1.3723 J per g C

Delta H vap = 35.21 kJ per mol * 1000 J / 1 kJ = 35210 J per mol

Q = 122.1 kJ * 1000 J / 1kJ = 122100 J

Initial temperature = 32.60 C boiling point = 64.60 C and final temperature = 95.60 c

Specific heat of liquid methanol = ?

q= q liquid methanol + q methanol gas

q= (m*c*delta T) + (m/MM*delta H vap )(m*c*delta T)

122100 J = (100 g *c*(64.60-32.60))+[(100g/32.04 g per mol)*35210 J per mol] +[100g*1.3723 J per g C*(95.60 – 64.06 C)

122100 J =3200 *c + 109893 + 4254.13

122100 J = 3200 C + 114147

(122100 J – 114147)/3200 = c

2.46 J per g C= c

So the specific heat of the methanol liquid is 2.46 J per g C

Use the pertinent data from part b above and the Clausius-Clapeyron equation

to find the vapor pressure in Torr at 32.60 C. R=8.314

J/mol.K Hint: You need to use 273.2

Solution :-

T1 = 64.6 +273.2 = 337.8 K

P1 = 760 torr

T2 = 32.60 C +273 = 305.6 K

P2 = ?

ln[P2/P1] = Delta H vap / R * [(1/T1)-(1/T2)]

ln[P2/760] = 35210 J per mol /8.314 J per mol K * [ (1/337.8)-(1/305.6)]

ln[P2/760] = -1.321

p2 / 760 = anti ln [-1.321]

p2/760 = 0.227

p2 = 0.227*760 torr

P2 = 173 torr

Therefore the vapor pressure at 32.6 C is 173 torr

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