Question

A certain substance has a heat of vaporization of 61.92 kj/mol at what kelvin temp will vapor pressure be 6.00 times higher then it was at 297k

Answer 1

To find the temperature at which the vapor pressure is 6 times higher than at 297K, we can use the Clausius-Clapeyron equation. Substituting the given values, the temperature is approximately 2685K.

Step-by-step explanation:

To find the temperature at which the vapor pressure will be 6 times higher than at 297K, we can use the Clausius-Clapeyron equation. Rearranging the equation, we have: T₂ = T₁ * (P₂/P₁) * (△Hvap/R), where T₁ is the initial temperature (297K), P₁ is the initial vapor pressure, P₂ is the desired vapor pressure (6 times higher), △Hvap is the heat of vaporization (61.4 kJ/mol), and R is the gas constant (8.314 J/mol·K).

Let's substitute the given values into the equation: T₂ = 297K * (6/1) * (61.4 kJ/mol / 8.314 J/mol·K).

By calculation, T₂ ≈ 2685K. Therefore, at approximately 2685K, the vapor pressure will be 6 times higher than at 297K.