at which temperature will water boil when the external pressure is 30 kPa?

Well, water boils when it just can't take the heat anymore and decides to throw in the towel. But when the external pressure is 30 kPa, water actually gets a bit of relief. It says, "Hey, I don't have to work as hard to reach boiling point!" So, when the pressure is low like that, water gets a bit lazy and decides to start boiling at a lower temperature of around 48.3 degrees Celsius. It's like water's version of a mini vacation. Ah, the perks of being H2O!

The boiling point of water varies with the external pressure. At sea level (standard atmospheric pressure), water boils at 100 degrees Celsius or 212 degrees Fahrenheit. However, when the external pressure is reduced, the boiling point decreases.

To find the boiling point of water at a specific pressure, you can refer to a phase diagram of water. At 30 kPa, the boiling point of water is approximately 85 degrees Celsius or 185 degrees Fahrenheit.

To determine the temperature at which water boils under a specific external pressure, we need to consult a phase diagram for water. A phase diagram is a graphical representation of the relationship between temperature, pressure, and the states of a substance.

In this case, we are interested in the boiling point of water at a pressure of 30 kPa. We can use the Clausius-Clapeyron equation to determine this value.

The Clausius-Clapeyron equation is given as:

ln(P₁/P₂) = ΔH_vap/R * (1/T₂ - 1/T₁)

Where:
- P₁ and P₂ are the initial and final pressures respectively
- ΔH_vap is the molar enthalpy of vaporization
- R is the ideal gas constant
- T₁ and T₂ are the initial and final temperatures respectively

To solve for the boiling temperature (T₂) at 30 kPa, we can assume the initial pressure (P₁) to be ambient pressure (1 atm) and use the known boiling point of water at that pressure. The molar enthalpy of vaporization for water is approximately 40.79 kJ/mol. The ideal gas constant (R) is 8.314 J/(mol⋅K).

ln((30 kPa)/(1 atm)) = (40.79 kJ/mol)/(8.314 J/(mol⋅K)) * (1/T₂ - 1/373.15 K)

Simplifying the equation, we get:

ln(30) = 4.92 * (1/T₂ - 1/373.15)

Now, we can rearrange the equation to solve for T₂:

1/T₂ = (ln(30)/4.92) + 1/373.15

Finally, we can calculate T₂ by taking the reciprocal of the right-hand side of the equation:

T₂ = 1 / ((ln(30)/4.92) + 1/373.15)

Using the above equation, we can determine the boiling temperature of water when the external pressure is 30 kPa.

Do you have a graph to read this from? If so that is how you get the answer.

If no graph of vapor pressure vs temperature, then I assume you are to use the Arrhenius equation. Use the following data in that equation.
p1 = 30 kPa
T1 = ?K
P2 = 101.3
T2 = 373 K