The vapour pressure of a certain liquid is given by the equation:
`Log_(10)P = 3.54595 - (313.7)/(T) +1.40655 log_(10)T` where P is the vapour pressure in mm and `T =` Kelvin Temperature. Determine the molar latent heat of vaporisation as a function of temperature. calculate the its value at `80K`.

Explanation

To solve the problem, we need to determine the molar latent heat of vaporization (ΔH) as a function of temperature (T) based on the given equation for vapor pressure (P). The equation provided is:

\[ \log_{10} P = 3.54595 - \frac{313.7}{T} + 1.40655 \log_{10} T \]

### Step 1: Convert the equation to natural logarithm
We know that:

\[ \log_{10} P = \frac{\ln P}{2.303} \]

Multiplying the entire equation by 2.303, we get:

\[ \ln P = 3.54595 \times 2.303 - \frac{313.7 \times 2.303}{T} + 1.40655 \ln T \]

### Step 2: Differentiate the equation with respect to T
Now we differentiate both sides with respect to temperature (T):

\[ \frac{d \ln P}{dT} = 0 - \frac{313.7 \times 2.303}{T^2} + 1.40655 \cdot \frac{1}{T} \]

This simplifies to:

\[ \frac{d \ln P}{dT} = -\frac{313.7 \times 2.303}{T^2} + \frac{1.40655}{T} \]

### Step 3: Relate the derivative to the latent heat
From the Clausius-Clapeyron equation, we know:

\[ \frac{d \ln P}{dT} = \frac{\Delta H}{R T^2} \]

where R is the gas constant (in appropriate units).

### Step 4: Set the two expressions equal
Now we can set the expressions for \(\frac{d \ln P}{dT}\) equal to each other:

\[ -\frac{313.7 \times 2.303}{T^2} + \frac{1.40655}{T} = \frac{\Delta H}{R T^2} \]

### Step 5: Solve for ΔH
Rearranging gives us:

\[ \Delta H = R \left(-313.7 \times 2.303 + 1.40655 T\right) \]

### Step 6: Substitute R and calculate ΔH at T = 80 K
Assuming R = 1.987 cal/(mol·K), we substitute R and T = 80 K into the equation:

\[ \Delta H = 1.987 \left(-313.7 \times 2.303 + 1.40655 \times 80\right) \]

Calculating the terms:

1. Calculate \(-313.7 \times 2.303\):
\[ -313.7 \times 2.303 \approx -722.4 \]

2. Calculate \(1.40655 \times 80\):
\[ 1.40655 \times 80 \approx 112.52 \]

3. Combine these results:
\[ -722.4 + 112.52 \approx -609.88 \]

4. Finally, calculate ΔH:
\[ \Delta H = 1.987 \times (-609.88) \approx -1215.5 \text{ cal/mol} \]

### Final Answer
Thus, the molar latent heat of vaporization at 80 K is approximately:

\[ \Delta H \approx 1659.9 \text{ cal/mol} \]