The vapour pressure of methyl alcohol is 40 mmHg at 5°C. Use this value and the calsius-clapeyron equation to estimate the normal boiling point of mythl alcohol.

Clausius*

The boiling point is the temperature at which the vapor pressure is 1 atm = 760 mm Hg, which is 19 times higher.

You will need the heat of vaporization of methyl alcohol to apply the Clausius-Clapeyron equestion. It is 35278 kJ/kmol. Call it Hv. You will also need the molar gas constant R, which is 8.314 kJ/(mole*K)

For an explanation of how to apply the C-C equation, see
http://www.science.uwaterloo.ca/~cchieh/cact/c123/clausius.html

P1 DHvap 1 1

ln (---) = ---- (--- - ---)
P2 R T2 T1

ln1/0.53= 38/8.31(1/T1 - 1/T2)
2.93 =4.57(1/278 - 1/T2)
.641 = 1/278 - 1/ T2
.6407 = -1/T2
T2 = -1.56

that answer is suppose to be T=339 K
which part is wrong?

I gave you the wrong heat of vaporization by a factor of 1000, but you did catch that. It should be 35.278 kJ/mole. You used 38. I don't see a ln(19) that should be in your calculation. That is the natural log of the pressure ratio. Done propertly, you cannot end up with a negative absolute temperature

To estimate the normal boiling point of methyl alcohol using the Clausius-Clapeyron equation, we need to make use of the given data point and the known constants. The Clausius-Clapeyron equation relates the vapor pressure of a substance to its boiling point temperature. It can be written as:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively.
ΔHvap is the enthalpy of vaporization.
R is the ideal gas constant (8.314 J/(mol*K)).
T1 and T2 are the corresponding temperatures in Kelvin.

In this case, we are given P1 = 40 mmHg at T1 = 5°C = 278K.

To estimate the normal boiling point (T2), we assume the vapor pressure at the boiling point is equal to atmospheric pressure, which is 760 mmHg. So, P2 = 760 mmHg.

First, we need to convert the given temperatures from Celsius to Kelvin:
T1 = 5°C + 273.15 = 278.15K

Next, we rearrange the equation to solve for T2:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

Let's calculate the natural logarithm of the ratio of pressures:

ln(760/40) = -ΔHvap/R * (1/T2 - 1/278.15)

To isolate T2, we rearrange the equation further:

1/T2 = ((760/40) - 1) * (-R/ΔHvap) + 1/278.15

Now, we can substitute the values and solve for T2:

1/T2 = (19 - 1) * (-8.314 J/(mol*K)/ΔHvap) + 1/278.15

Simplifying:

1/T2 = 18 * (-8.314 J/(mol*K)/ΔHvap) + 1/278.15

Next, we invert both sides of the equation:

T2 = 1 / (18 * (-8.314 J/(mol*K)/ΔHvap) + 1/278.15)

Finally, substituting the given value for P1 and rearranging:

T2 = 1 / (18 * (-8.314 J/(mol*K)/ΔHvap) + 1/278.15)

Now, we have the equation to estimate the normal boiling point of methyl alcohol. To obtain the final answer, we need to know the value of ΔHvap, the enthalpy of vaporization for methyl alcohol. Once we know that value, we can plug it into the equation and calculate the estimated normal boiling point.