Suppose the vapor pressure of a substance is measured at two d...

Question

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius–Clapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, P1 and P2, and the absolute temperatures at which they were measured, T1 and T2: (b) Gasoline is a mixture of hydrocarbons, a component of which is octane (CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₃). Octane has a vapor pressure of 13.95 torr at 25 °C and a vapor pressure of 144.78 torr at 75 °C. Use these data and the equation in part (a) to calculate the heat of vaporization of octane. (c) By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.81. (d) Calculate the vapor pressure of octane at - 30 °C.

Accepted Answer

1. The Clausius-Clapeyron equation is given by: $$\ln\left(\frac{P_2$}{P_1$}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2$} - \frac{1}{T_1$}\right)$$, where $$P_1$$ and $$P_2$$ are the vapor pressures at temperatures $$T_1$$ and $$T_2$$ respectively, $$\Delta H_{vap} is $$the heat of vaporization, and $$R is $$the ideal gas constant. This equation allows us to relate the change in vapor pressure of a substance with the change in temperature.. 2. To calculate the heat of vaporization of octane, we can rearrange the Clausius-Clapeyron equation to solve for $$\Delta H_{vap}$$: $$\Delta H_{vap} = -R\left(\frac{1}{T_2$} - \frac{1}{T_1$}\right)\ln\left(\frac{P_2$}{P_1$}\right). We $$can then substitute the given values for $$P_1$$, $$P_2$$, $$T_1$$, and $$T_2$$ into this equation. Remember to convert the temperatures from Celsius to Kelvin by adding 273.15.. 3. To calculate the normal boiling point of octane, we need to find the temperature at which the vapor pressure of octane is equal to 1 atm or 760 torr. We can rearrange the Clausius-Clapeyron equation to solve for $$T_2$$: $$T_2$$ = \frac{1}{\frac{1}{$$T_1$$} - \frac{R}{\Delta H_{vap}}\ln\left(\frac{$$P_2$$}{$$P_1$$}\right)}$$. We can then substitute the given values for P_1$, P_2$, T_1$, and the calculated value for $$\Delta H_{vap}$$ into this equation.. 4. To calculate the vapor pressure of octane at -30 °C, we can use the Clausius-Clapeyron equation again. This time, we rearrange the equation to solve for P_2$: P_2$ = P_1$\exp\left(-\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2$} - \frac{1}{T_1$}\right)\right). We $$can then substitute the given values for $$P_1$$, $$T_1$$, the calculated value for $$\Delta H_{vap}$$, and the desired temperature $$T_2 ($$converted to Kelvin) into this equation.. 5. After performing these calculations, you should have the heat of vaporization of octane, the normal boiling point of octane, and the vapor pressure of octane at -30 °C. Remember to check your answers for reasonableness based on what you know about the properties of octane and other similar substances.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Clausius–Clapeyron Equation

Heat of Vaporization

Normal Boiling Point