Question

What is the pressure of water vapor at a temperature of ${130^ \circ }C$ in a steam steril-lithographic chamber of an autoclave with a capacity of \[V = 221\] , if the water evaporates spent \[30 KJ\] ? Specific heat of evaporation = \[2.23{\text{ }}MJ/Kg\]

Answer 1

Hint: The heat transferred formula is to apply here. This is for specified temperatures that we get the unknown temperature and then by applying the Clausius-Clapeyron equation we get the pressure of the water vapor in a steam steril-lithographic chamber at a specified temperature by knowing its enthalpy of vaporization and universal Gas constant as applicable in the equation.

Complete step by step answer:
 Given ${h_{fg}}$ = specific heat of evaporation \[ = 2.23{\text{ }}MJ/Kg\]\[ = 2.23{\text{ }} \times {10^6}J/Kg\]
\[V = \] Water Capacity \[ = 221\]
\[Q = \] heat transferred \[ = 30{\text{ }}KJ\]
\[{T_2} = \] The temperature of water vapor $ = {130^ \circ }C$\[ = 130 + 273 = 403K\]
\[R = 8.314{\text{ }}J{\text{ }}mo{l^{ - 1}}{K^{ - 1}}\]
Therefore applying the heat transferred formula
 $ \Rightarrow Q = {h_{fg}}\left( {{T_2} - {\text{ }}{T_1}} \right)$ given ${h_{fg}}$ = specific heat of evaporation = \[2.23 \times {10^6}J/Kg\]
\[ \Rightarrow 30 \times {10^3} = 2.23{\text{ }} \times {10^6}\left( {403 - {T_1}} \right)\]
\[{T_1} = \;402.98{\text{ }}K\]
Hence applying the Clausius-Clapeyron equation we get
$\ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right) = \dfrac{{\Delta {h_{fg}}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$
 $ \Rightarrow \ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right) = \dfrac{{2.23{\text{ }} \times {\text{ }}{{10}^6}}}{{8.314}}\left( {\dfrac{1}{{402.98}} - \dfrac{1}{{403}}} \right)$
\[ \Rightarrow \ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right) = 0.033\]
$ \Rightarrow \dfrac{1}{{{P_2}}} = {e^{ - 0.033}}$
$ \Rightarrow \dfrac{1}{{{P_2}}} = 0.9675$
Hence ${P_2} = 1.04atm$ pressure.

Note: The Clausius-Clapeyron equation gives us the quantitative relation between a body’s vapor pressure P and its temperature Tandit also predicts the rate at which vapor pressure increases per unit increase in temperature.
$P = A\dfrac{{\Delta {h_{fg}}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$ Where \[{\mathbf{\Delta }}{h_{{\mathbf{fg}}}}\;\] = the enthalpy of vaporization for the liquid, \[R\] is the gas constant and \[A\;\] is a constant whose value depends on the chemical identity of the body. Temperature $T$ must be in Kelvin Scale in this equation. However, since the relationship between vapor pressure and temperature is not linear, the equation is rearranged into logarithmic form to obtain the linear equation. We can obtain the pressure of water at a certain temperature by using the Clausius-Clapeyron equation by knowing its enthalpy of vaporization.