Question

The enthalpy of vaporization of a certain liquid is found to be 14.4 kJ mol-1 at 180 K, its normal boiling point. The molar volumes of the liquid and the vapour at the boiling point are 115 cm3 mol-1 and 14.5 dm3 mol-1, respectively. (a) Estimate dp/dT from the Clapeyron equation and (b) the percentage error in its value if the Clausius–Clapeyron equation is used instead.

Answer 1

Answer :

(a) The value of $\frac{dP}{dT}$ is $5.56\times 10^3Pa/K$

(b) The percentage error will be 2.5 %

Explanation :

(a) First we have to calculate the $V_{vap}$.

$V_{vap}=V_2-V_1$

$V_1$ = volume of liquid = $115cm^3mol^{1-}=115\times 10^{-6}m^3mol^{1-}$

$V_1$ = volume of vapor = $14.5dm^3mol^{1-}=14.5\times 10^{-3}m^3mol^{1-}$

$V_{vap}=V_2-V_1$

$V_{vap}=(14.5\times 10^{-3}m^3mol^{1-})-(115\times 10^{-6}m^3mol^{1-})$

$V_{vap}=1.44\times 10^{-2}m^3mol^{1-}$

Now we have to calculate the value of $\frac{dP}{dT}$

The Clausius- Clapeyron equation is :

$\frac{dP}{dT}=\frac{\Delta H_{vap}}{T\Delta V_{vap}}$

where,

T = temperature = 180 K

$\Delta H_{vap}$ = heat of vaporization = 14.4 kJ/mole = 14400 J/mole

$V_{vap}=1.44\times 10^{-2}m^3mol^{1-}$

Now put all the given values in the above formula, we get:

$\frac{dP}{dT}=\frac{(14400J/mole)}{(180K)\times (1.44\times 10^{-2}m^3mol^{1-})}\times \frac{1Pa}{1J/m^3}$

$\frac{dP}{dT}=5.56\times 10^3Pa/K$

(b) Now we have to calculate the percentage error.

Now we have to calculate the value of $\frac{dP}{dT}$ at normal boiling point.

The Clausius- Clapeyron equation is :

$\frac{dP}{dT}=\frac{\Delta H_{vap}}{T\Delta V_{vap}}$

As we know that : PV = nRT

So,

$\frac{dP}{dT}=\frac{P\Delta H_{vap}}{RT^2}$

where,

R = gas constant = 8.314 J/K.mol

T = temperature = 180 K

$\Delta H_{vap}$ = heat of vaporization = 14.4 kJ/mole = 14400 J/mole

P = pressure at normal boiling point = 101325 Pa

Now put all the given values in the above formula, we get:

$\frac{dP}{dT}=\frac{(101325Pa}\times (14400J/mole)}{(8.314J/K.mol)\times (180K)^2}$

$\frac{dP}{dT}=5.42\times 10^3Pa/K$

Now we have to determine percentage error.

$\%\text{ error}=\frac{(5.56\times 10^3Pa/K)-(5.42\times 10^3Pa/K)}{5.56\times 10^3Pa/K}\times 100$

$\%\text{ error}=2.5\%$

Therefore, the percentage error will be 2.5 %