A liquid is heated from 33.96°C to 104.32°C. The vapor pressure increases from 23.54torr to 984.35torr. What is the heat of vaporization of the liquid in kJ/mol?
The change in temperature is 70.36°C.
Change in vapor pressure = 960.76 torr.
What is the equation to find the heat of vaporization given that we know the change in pressure, and change in temperature?
Isn't that the Clausius-Clapeyron equation?
To find the heat of vaporization, you can use the Clausius-Clapeyron equation, which relates the change in vapor pressure to the heat of vaporization:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1),
where:
P2 and P1 are the final and initial vapor pressures respectively,
ΔHvap is the heat of vaporization,
R is the gas constant (8.314 J/(mol·K)),
T2 and T1 are the final and initial temperatures respectively (in Kelvin).
In this case, we are given the change in vapor pressure (P2 - P1) and the change in temperature (T2 - T1). So, let's rearrange the equation:
ln((P2 - P1)/P1) = (ΔHvap/R) * (1/T1 - 1/T2).
Now, let's plug in the values we have:
ln((960.76 torr)/23.54 torr) = (ΔHvap/8.314 J/(mol·K)) * (1/(33.96 + 273.15 K) - 1/(104.32 + 273.15 K)).
Simplifying further:
ln(40.838) = (ΔHvap/8.314) * (0.003696 - 0.002853).
We can solve this equation to find the value of ΔHvap in J/mol. Divide it by 1000 to convert to kJ/mol if required.