Plz...explain with full procedure!!!!!!! Vapour pressure of liquid is given like, Log10P=3.54595-313.7/T+1.40655 log10T T in kelvin P in mmhg, determine /\Hv as a function of T and calculate its value at 80K......THANKS☺

To determine the enthalpy of vaporization (\( \Delta H_v \)) as a function of temperature using the given equation for vapor pressure, we can follow a systematic approach. The equation provided is:

Log10P = 3.54595 - 313.7/T + 1.40655 log10T

Where \( P \) is the vapor pressure in mmHg and \( T \) is the temperature in Kelvin. The first step is to differentiate this equation with respect to temperature to find \( \Delta H_v \).

Step 1: Differentiate the Vapor Pressure Equation

To find \( \Delta H_v \), we can use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature to the enthalpy of vaporization:

ln(P) = -\Delta H_v/(R \cdot T) + C

From the given equation, we can express \( P \) in terms of \( T \) and differentiate it. First, we need to convert the logarithmic equation into an exponential form:

P = 10^(3.54595 - 313.7/T + 1.40655 log10T)

Now, we differentiate \( P \) with respect to \( T \). Using the chain rule, we have:

dP/dT = P * (d/dT(3.54595 - 313.7/T + 1.40655 log10T))

Step 2: Calculate the Derivative

Now, we need to compute the derivative of the right-hand side:

• The derivative of a constant (3.54595) is 0.
• The derivative of \(-313.7/T\) is \(313.7/T^2\).
• The derivative of \(1.40655 log10T\) is \(1.40655/(T \ln(10))\).

Putting it all together, we have:

dP/dT = P * (313.7/T^2 - 1.40655/(T \ln(10)))

Step 3: Relate to Enthalpy of Vaporization

Using the Clausius-Clapeyron equation, we can relate \( dP/dT \) to \( \Delta H_v \):

dP/dT = (ΔHv / (R * T^2)) * P

Equating the two expressions for \( dP/dT \):

(ΔHv / (R * T^2)) * P = P * (313.7/T^2 - 1.40655/(T \ln(10)))

We can cancel \( P \) from both sides (assuming \( P \neq 0 \)):

ΔHv / (R * T^2) = 313.7/T^2 - 1.40655/(T \ln(10))

Step 4: Solve for ΔHv

Now, rearranging gives us:

ΔHv = R * T^2 * (313.7/T^2 - 1.40655/(T \ln(10)))

Now, substituting \( R \) (the universal gas constant, approximately 8.314 J/(mol·K)) into the equation:

ΔHv = R * (313.7 - 1.40655 * T / ln(10))

Step 5: Calculate ΔHv at 80 K

Now, we can substitute \( T = 80 \, K \) into the equation:

ΔHv = 8.314 * (313.7 - 1.40655 * 80 / ln(10))

Calculating \( ln(10) \) gives approximately 2.302. Now substituting this value:

ΔHv = 8.314 * (313.7 - 1.40655 * 80 / 2.302)

Calculating the term inside the parentheses:

1.40655 * 80 / 2.302 ≈ 48.73

Thus, we have:

ΔHv = 8.314 * (313.7 - 48.73) = 8.314 * 264.97

Finally, calculating this gives:

ΔHv ≈ 2206.3 J/mol

Final Result

At 80 K, the enthalpy of vaporization (\( \Delta H_v \)) is approximately 2206.3 J/mol. This value indicates the energy required to vaporize one mole of the liquid at this temperature. Understanding this concept is crucial in thermodynamics and helps in various applications, including chemical engineering and environmental science.