Answer:
The temperature at which the vapor pressure would be 0.350 atm is 201.37°C
Explanation:
The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.
Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.
The equation for Clausius- Clapeyron Equation can be expressed as:
![\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }](https://tex.z-dn.net/?f=%5Cmathtt%7BIn%20%5Cdfrac%7BP_2%7D%7BP_1%7D%3D%20%5Cdfrac%7B%5CDelta%20%5C%20H%20_%7Bvap%7D%7D%7BR%7D%20%5Cbegin%20%7Bpmatrix%7D%20%5Cdfrac%7BT_2%20-T_1%7D%7BT_2%20%5C%20T_1%7D%20%5Cend%20%7Bpmatrix%7D%20%7D)
where ;
is the vapor pressure at temperature 1
is the vapor pressure at temperature 2
∆Hvap = enthalpy of vaporization
R = universal gas constant
Given that:
= 1 atm
= 0.350 atm
∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol
= 282 °C = (282 + 273) K = 555 K
R = 8.314 J/mol/k
Substituting the above values into the Clausius - Clapeyron equation, we have:
![\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }](https://tex.z-dn.net/?f=%5Cmathtt%7BIn%20%5Cdfrac%7BP_2%7D%7BP_1%7D%3D%20%5Cdfrac%7B%5CDelta%20%5C%20H%20_%7Bvap%7D%7D%7BR%7D%20%5Cbegin%20%7Bpmatrix%7D%20%5Cdfrac%7BT_2%20-T_1%7D%7BT_2%20%5C%20T_1%7D%20%5Cend%20%7Bpmatrix%7D%20%7D)
![\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }](https://tex.z-dn.net/?f=%5Cmathtt%7BIn%20%20%5Cbegin%20%7Bpmatrix%7D%20%20%5Cdfrac%7B0.350%7D%7B1%7D%20%5Cend%20%7Bpmatrix%7D%20%7D%20%3D%20%5Cdfrac%7B28.5%20%5Ctimes%2010%5E3%20%7D%7B%208.314%20%7D%20%5Cbegin%20%7Bpmatrix%7D%20%5Cdfrac%7BT_2%20-%20555%7D%7B555T_2%7D%20%5Cend%20%7Bpmatrix%7D%20%7D)
![\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }](https://tex.z-dn.net/?f=%5Cmathtt%7BIn%20%20%5Cbegin%20%7Bpmatrix%7D%20%20%5Cdfrac%7B0.350%7D%7B1%7D%20%5Cend%20%7Bpmatrix%7D%20%7D%20%3D%20%5Cdfrac%7B28.5%20%5Ctimes%2010%5E3%20%7D%7B%208.314%20%7D%20%5Cbegin%20%7Bpmatrix%7D%20%5Cdfrac%7B1%7D%7B555%7D-%20%5Cdfrac%7B1%7D%7BT_2%7D%20%5Cend%20%7Bpmatrix%7D%20%7D)
![- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }](https://tex.z-dn.net/?f=-%201.0498%3D%203427.953%20%20%20%5Cbegin%20%7Bpmatrix%7D%20%5Cdfrac%7B1%7D%7B555%7D-%20%5Cdfrac%7B1%7D%7BT_2%7D%20%5Cend%20%7Bpmatrix%7D%20%7D)
![\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }](https://tex.z-dn.net/?f=%5Cdfrac%7B-%201.0498%7D%7B3427.953%7D%3D%20%20%20%20%5Cbegin%20%7Bpmatrix%7D%20%5Cdfrac%7B1%7D%7B555%7D-%20%5Cdfrac%7B1%7D%7BT_2%7D%20%5Cend%20%7Bpmatrix%7D%20%7D)
![- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }](https://tex.z-dn.net/?f=-%203.06246906%20%5Ctimes%2010%5E%7B-4%7D%3D%20%20%20%20%5Cbegin%20%7Bpmatrix%7D%20%5Cdfrac%7B1%7D%7B555%7D-%20%5Cdfrac%7B1%7D%7BT_2%7D%20%5Cend%20%7Bpmatrix%7D%20%7D)
![\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7BT_2%7D%20%3D%20%20%20%20%5Cbegin%20%7Bpmatrix%7D%20%5Cdfrac%7B1%7D%7B555%7D%2B%20%283.06246906%20%5Ctimes%2010%5E%7B-4%7D%20%29%20%5Cend%20%7Bpmatrix%7D%20%7D)
![\dfrac{1}{T_2} = 0.002108048708](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7BT_2%7D%20%3D%20%20%200.002108048708)
![T_2 = \dfrac{1}{0.002108048708}](https://tex.z-dn.net/?f=T_2%20%3D%20%5Cdfrac%7B1%7D%7B0.002108048708%7D)
= 474.37 K
To °C ; we have
= (474.37 - 273)°C
= 201.37 °C
Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C