Chapter 20, Problem 33P

(a)

To determine

To Calculate:The temperature at which water boils on mountain.

Answer to Problem 33P

  $\text{90°C}$

Explanation of Solution

Given:Atmospheric pressure $\text{70}\text{.0 kPa}$ .

Formula used:

Clausius-Clapeyron equation is

  $\text{ln}\left(\frac{\text{P2}}{\text{P1}}\right)\text{=}\left(\frac{\text{ΔHvap}}{\text{R}}\right)\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)$

Here, $P_2$ is the final pressure, $P_1$ is the initial pressure, $\Delta H_{vap}$ is the enthalpy of vaporization ofwater, R is the universal gas constant, T is the boiling temperature of the water, and $\Delta T$ is thevariable temperature.

Calculation:

Rewrite the equation,

$\text{ln}\left(\frac{\text{P2}}{\text{P1}}\right)\text{=}\left(\frac{\text{ΔHvap}}{\text{R}}\right)\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)$ for $T_2$

  $\begin{array}{l}\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{=ln}\left(\frac{\text{P2}}{\text{P1}}\right)\left(\frac{\text{R}}{\text{ΔHvap}}\right)\\ \frac{\text{1}}{{\text{T}}_{\text{2}}}\text{=}\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-ln}\left(\frac{\text{P2}}{\text{P1}}\right)\left(\frac{\text{R}}{\text{ΔHvap}}\right)\\ \text{T2=}\frac{\text{1}}{\left[\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\left(\ln \left(\frac{\text{P2}}{\text{P1}}\right)\left(\frac{\text{R}}{\text{ΔHvap}}\right)\right)\right]}\end{array}$

The boiling point of the water is $\text{100°C}$ and the atmospheric pressure is $\text{1atm}$ .

Covert Celsius temperature to Kelvin temperature as follows:

  $\begin{array}{l}T1=\left(100+273\right)\text{K}\\ T1=373\text{K}\end{array}$

Substitute $\text{373 K = T1, 70}\text{.0 kPa =P2, 1 atm = P1, 8}\text{.314J/mol}\text{.K= R, 40}{\text{.66×10}}^{\text{9}}{\text{J/mol=ΔH}}_{\text{vap}}$ in the equation

  $\text{T2=}\frac{\text{1}}{\left[\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\left(\ln \left(\frac{\text{P2}}{\text{P1}}\right)\left(\frac{\text{R}}{\text{ΔHvap}}\right)\right)\right]}$

P, R and solve for T

$\text{T2=}\frac{\text{1}}{\left[\frac{\text{1}}{\text{373K}}\text{-}\left(\ln \left(\frac{\text{70}\text{.0kPa}}{\text{1atm}\left(\frac{\text{101}\text{.325kPa}}{\text{1atm}}\right)}\right)\left(\frac{\text{8}\text{.314J/mol×K}}{\text{40}{\text{.66×10}}^{\text{3}}\text{J/mol}}\right)\right)\right]}$${\text{T}}_{\text{2}}\text{= 363}\text{.0 K}$

Covert Kelvin temperature to Celsius temperature as follows:

  $\begin{array}{l}\text{T2 = }\left(\text{363– 273}\right)\text{°C}\\ \text{T2 = 90°C}\end{array}$

Conclusion: Therefore, the temperature at which the water boils on the mountain is $\text{90°C}$ .

(b)

To determine

To Calculate:The temperature at which water boils in a container.

Answer to Problem 33P

  $\text{ 81°C}$

Explanation of Solution

Given: pressure $\text{0}\text{.500 atm}$

Calculation: The water boils inside the container and the pressure inside the container is $\text{0}\text{.500 atm}$ .

The temperature, where the water boils in a container is,

$\text{T2=}\frac{\text{1}}{\left[\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\left(\ln \left(\frac{\text{P2}}{\text{P1}}\right)\left(\frac{\text{R}}{\text{ΔHvap}}\right)\right)\right]}$

Substitute ${\text{373 K =T}}_{\text{1}}\text{, 0}{\text{.500 atm = P}}_{\text{2}}{\text{, 1 atm = P}}_{\text{1}}\text{, 8}{\text{.314J/mol K = R, and T}}_{\text{2}}\text{40}{\text{.66x10J/mol = ΔH}}_{\text{vap}}$ in the equation

$\text{T2=}\frac{\text{1}}{\left[\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\left(\ln \left(\frac{\text{P2}}{\text{P1}}\right)\left(\frac{\text{R}}{\text{ΔHvap}}\right)\right)\right]}$ to solve T₂

$\begin{array}{l}\text{T2=}\frac{\text{1}}{\left[\frac{\text{1}}{{\text{373K}}_{}}\text{-}\left(\text{ln}\left(\frac{\text{0}\text{.500atm}}{\text{1atm}}\right)\left(\frac{\text{8}\text{.314J/mol}\text{.K}}{\text{40}{\text{.66×10}}^{\text{3}}\text{J/mol}}\right)\right)\right]}\\ {\text{T}}_{\text{2}}\text{, = 355K}\end{array}$

Covert Kelvin temperature to Celsius temperature as follows:

  $\begin{array}{l}\text{T2 =}\left(\text{354– 273}\right)\text{°C}\\ \text{T2 = 81°C}\end{array}$

Therefore, the temperature at which the water boils in the container is $\text{ 81°C}$

Conclusion: By usingClausius-Clapeyron equation,the temperature at which the water boils in the container can be calculated.

(c)

To determine

To Calculate:The pressure at water boils at $\text{115°C}$ is to be calculated.

Answer to Problem 33P

  $\text{170 kPa}$

Explanation of Solution

Given: Temperature $\text{115°C}$ .

Calculation:The Clausius-Clapeyron equation is

  $\ln \left(\frac{\text{P2}}{\text{P1}}\right)\text{=}\left(\frac{\text{ΔHvap}}{\text{R}}\right)\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)$

Apply anti-logarithm to the above equation on both sides

  $\begin{array}{l}\left(\frac{\text{P2}}{\text{P1}}\right){\text{=e}}^{\left(\frac{\text{ΔHvap}}{\text{R}}\right)\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)}\\ {\text{P2=P1e}}^{\left(\frac{\text{ΔHvap}}{\text{R}}\right)\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)}\end{array}$

Covert Celsius temperature to Kelvin temperature as follows:

${\text{P}}_{\text{2}}{\text{=P}}_{\text{1}}{\text{e}}^{\left(\frac{\text{ΔHvap}}{\text{R}}\right)\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)}$

  $\begin{array}{l}\text{T1 = }\left(\text{115 + 273}\right)\text{K}\\ \text{T1 = 388K}\end{array}$

Substitute

  ${\text{ 373 K =T}}_{\text{1}}{\text{, 388 K = T}}_{\text{2}}{\text{,1 atm = P}}_{\text{1}}\text{, 8}\text{.314J/mol}\text{. K = R,40}\text{.66}\times {\text{10}}^3{\text{ J/mol = ΔH}}_{\text{vap}}$ in the equation and solve for P₂.

  $\begin{array}{l}\text{P2=}\left(\text{1atm}\right){\text{e}}^{\left(\frac{\text{40}{\text{.66×10}}^3\text{J/mol}}{\text{8}\text{.314J/molK}}\right)\left(\frac{\text{1}}{\text{373K}}\text{-}\frac{\text{1}}{\text{388K}}\right)}\\ \text{P2=}\left(\text{1atm}\left(\frac{\text{101}\text{.33kPa}}{\text{1atm}}\right)\right){\text{e}}^{\left(\frac{\text{40}{\text{.66×10}}^3\text{J/mol}}{\text{8}\text{.314J/molK}}\right)\left(\frac{\text{1}}{\text{373K}}\text{-}\frac{\text{1}}{\text{388K}}\right)}\end{array}$ $\begin{array}{l}{\text{P}}_{\text{2}}\text{ = 168}\text{.23 kPa}\\ {\text{P}}_2\approx \text{170 kPa}\end{array}$

Therefore, the pressure at which the water boils at $\text{115°C}$ temperature is $\text{170 kPa}\text{.}$

Conclusion:Bythe Clausius-Clapeyron equation, the pressure at which the water boils at $\text{115°C}$ temperature can be calculated.