Question

What Is The Vapor Pressure Of A Liquid (In MmHg) At 298.15 K If Its AHvap = 28.9 KJ/Mol And Its Normal Boiling Point Is 341.88 K?

Answer 1

The vapor pressure of a liquid can be calculated using the Clausius-Clapeyron equation. By substituting the given values into the equation, the vapor pressure of the liquid at a temperature of 298.15 K is approximately 73.1 mm Hg.

Step-by-step explanation:

The vapor pressure of a liquid is the pressure exerted by its vapor in equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The Clausius-Clapeyron equation can be used to calculate the vapor pressure at a different temperature using the enthalpy of vaporization. The equation is:

`ln(P2/P1) = (-ΔHvap/R)((1/T2) - (1/T1))`

where P₁ is the vapor pressure at temperature T₁, P₂ is the desired vapor pressure at temperature T₂, ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant. Rearranging the equation gives:

`P2 = P1 * exp((-ΔHvap/R) * ((1/T2) - (1/T1)))`

Substituting the given values into the equation, we have:

P₂ = 101.3
`kPa * exp((-28.9 kJ/mol) / (8.314 J/mol·K) * ((1/298.15 K) - (1/341.88 K)))`

P₂ ≈ 73.1 mm Hg

Answer 2

the vapor pressure of the liquid at
`T_(1)` =298.15K is approximately 760.00034mmHg.

The vapor pressure of a liquid at a given temperature can be calculated using the Clausius–Clapeyron equation:

`\ln \left((P_2)/(P_1)\right)=-\frac{\Delta H_{\text {vap }}}{R}\left((1)/(T_2)-(1)/(T_1)\right)`

where:

`P_(1)` is the vapor pressure at temperature
`T_(1)`

`P_(2)` is the vapor pressure at temperature
`T_(2)`

`\Delta H_{\text {vap }}` is the enthalpy of vaporization,

R is the ideal gas constant (8.314 J/(mol·K)),

`T_(1)` is the initial temperature,

`T_(2)` is the final temperature.

Given that
`\Delta H_{\text {vap }}=28.9 \mathrm{~kJ} / \mathrm{mol}` and the normal boiling point is
`T_2=341.88 \mathrm{~K}` , we want to find the vapor pressure at
`T_1=298.15 \mathrm{~K}`.

Let's first convert
`\Delta H_{\text {vap }}` to joules:

`\begin{aligned}& \Delta H_{\text {vap }}=28.9 \mathrm{~kJ} / \mathrm{mol} * 1000 \mathrm{~J} / 1 \mathrm{~kJ} \\& \Delta H_{\text {vap }}=28,900 \mathrm{~J} / \mathrm{mol}\end{aligned}`

The vapor pressure of a liquid at a given temperature can be calculated using the Clausius–Clapeyron equation:

`\ln \left((P_2)/(P_1)\right)=-\frac{\Delta H_{\text {vap }}}{R}\left((1)/(T_2)-(1)/(T_1)\right)`

We'll use
`P_(1)` as the vapor pressure at
`T_(1)` (which is what we're trying to find), and
`P_(2)` as the vapor pressure at
`T_(2)`.

`\ln \left((P_2)/(P_1)\right)=-(28,900)/(8.314)\left((1)/(341.88)-(1)/(298.15)\right)`

Now, solve for
`P_(1)`:

`\begin{aligned}& (P_2)/(P_1)=e^{-(28,900)/(8.314)\left((1)/(341.88)-(1)/(298.15)\right)} \\& P_1=\frac{P_2}{e^{-(28,900)/(8.314)\left((1)/(341.88)-(1)/(298.15)\right)}}\end{aligned}`

Let's substitute the given values into the equation and calculate the result:

`\begin{aligned}& P_1=\frac{P_2}{e^{-(28.900)/(8.314)\left((1)/(341.88)-(1)/(298.15)\right)}} \\& P_1=\frac{P_2}{e^{-(28.900)/(8.314)\left((1)/(341.88)-(1)/(298.15)\right)}} \\& P_1 \approx \frac{P_2}{e^{(-3.478) * 10^(-7)}} \\& P_1 \approx (P_2)/(0.999999652)\end{aligned}`

Now, if we assume
`P_(2)` is the vapor pressure at the normal boiling point, it would be 1 atmosphere (760 mmHg). Therefore:

`\begin{aligned}& P_1 \approx (760)/(0.999999652) \\& P_1 \approx 760.00034\end{aligned}`