Question

Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32.1 kj/mol.

Answer 1

The normal boiling point of the substance is approximately -272.83 °C.

Step-by-step explanation:

The normal boiling point of a substance is the temperature at which its vapor pressure equals atmospheric pressure at sea level. To determine the normal boiling point, we need to use the Clausius-Clapeyron equation. First, we can calculate the temperature at which the vapor pressure is 83.4 kPa using the given information (T₁ = 35 °C, P₁ = 55.1 mm Hg, ΔHvap = 32.1 kJ/mol). Then, we can convert the temperature to Kelvin and Celsius to obtain the normal boiling point.

Using the Clausius-Clapeyron equation:

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

Where:

• P₁ = vapor pressure 1 (55.1 mm Hg)
• T₁ = temperature 1 in Kelvin (35 °C + 273.15 = 308.15 K)
• P₂ = vapor pressure 2 (83.4 kPa)
• T₂ = temperature 2 in Kelvin (unknown)
• ΔHvap = enthalpy of vaporization (32.1 kJ/mol)
• R = ideal gas constant (8.314 J/(K * mol))

By substituting the values into the equation and solving for T₂, we can find the normal boiling point.

ln(83.4 kPa/55.1 mm Hg) = -(32.1 kJ/mol / (8.314 J/(K * mol))) * (1/T₂ - 1/308.15 K)

0.508 = -3.865 * (1/T₂ - 1/308.15 K)

Simplifying the equation:

0.508 = -3.865/T₂ + 12.537

3.865/T₂ = 12.037

T₂ = 3.865/12.037 = 0.321 K

Converting back to Celsius:

T₂ = 0.321 K - 273.15 = -272.83 °C

Therefore, the normal boiling point of the substance is approximately -272.83 °C.

Answer 2

389.78681 K

Step-by-step explanation:

`P_1` = Initial pressure = 55.1 mmHg

`P_2` = Final pressure = 1 atm = 760 mmHg

`T_2` = Boiling point

`T_1` = Initial temperature = 35°C

`\Delta H_(vap)` = Heat of vaporization = 32.1 kJ/mol

From the Clausius-Claperyon equation

`ln(P_2)/(P_1)=(-(\Delta H_(vap))/(R))((1)/(T_2)-(1)/(T_1))\\\Rightarrow (1)/(T_2)=-ln(P_2)/(P_1)(R)/(\Delta H_(vap))+(1)/(T_1)\\\Rightarrow (1)/(T_2)=-ln(760)/(55.1)(8.314)/(32.1* 10^(3))+(1)/(273.15+35)\\\Rightarrow T_2=\left(-ln\left((760)/(55.1)\right)(8.314)/(32.1* \:10^3)+(1)/(273.15+35)\right)^(-1)\\\Rightarrow T_2=389.78681\ K`

The normal boiling point of the substance is 389.78681 K