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3 years ago

Question 1 (1 point)

Chemistry

1 answer:

lilavasa [31]3 years ago

4 0

Answer:

I believe the answer is C. Stopping an object from moving

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3 years ago

The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.350 atm? (∆Hvap = 28.5

victus00 [196]

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:

$\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }$

where ;

$P_1$ is the vapor pressure at temperature 1

$P_ 2$ is the vapor pressure at temperature 2

∆Hvap = enthalpy of vaporization

R = universal gas constant

Given that:

$P_1$ = 1 atm

$P_ 2$ = 0.350 atm

∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol

$T_1$ = 282 °C = (282 + 273) K = 555 K

R = 8.314 J/mol/k

Substituting the above values into the Clausius - Clapeyron equation, we have:

$\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }$

$\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }$

$\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }$

$\dfrac{1}{T_2} = 0.002108048708$

$T_2 = \dfrac{1}{0.002108048708}$

$\mathbf{T_2 }$ = 474.37 K

To °C ; we have $\mathbf{T_2 }$ = (474.37 - 273)°C

$\mathbf{T_2 }$ = 201.37 °C

Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C

3 0

3 years ago