A vessel containing a reaction mass under going an exothermic reaction is under vacuum, with temperature and pressure constant. When the source of vacuum is closed, the reactor begins to increase pressure thus increasing temperature. This will continue until a pressure relief device is activated, equilibrium will not be achieved. If the source of vacuum is opened again and vapor pressure begins to decrease the temperature follows. Can someone explain the principle of why the reaction mass cooled when pressure is decreased using vacuum.
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The underlying principle why the reaction mass is cooled when the pressure is decreased is simply evaporation cooling. The relevant quantity is the enthalpy of vaporization $\Delta H_\text{vap}$ of the considered substance. In equilibrium, the observed quantity can also be the equilibrium temperature that corresponds to the equilibrium vapour pressure. Note that the vapour pressure increases non-linearly with temperature according to the Clausius–Clapeyron relation.
For example, assuming the relevant substance (a reaction product or solvent) in the reactor is water and the initial pressure maintained by a vacuum pump is constant at $p_0=0.030\ \mathrm{bar}$, the corresponding equilibrium temperature is $T_0=24\ \mathrm{^\circ C}$. The reaction mass is cooled by the specific enthalpy of vaporization $\Delta h_{\text{vap},0}=2443.9\ \mathrm{kJ\ kg^{-1}}$. If the power of the reactor increases, the resulting evaporation rate will increase, if the power of the reactor decreases, the resulting evaporation rate will decrease, thus keeping the temperature approximately constant as long as the vacuum system can keep the pressure approximately constant.
When the vacuum system is closed, the pressure will no longer be kept constant. The pressure will increase due to the evaporating water. Accordingly, the equilibrium temperature will increase, too. For example, when reaching a pressure of $p_1=3.000\ \mathrm{bar}$, the temperature has increased to $T_1=133.5\ \mathrm{^\circ C}$. The reaction mass is still cooled by the specific enthalpy of vaporization $\Delta h_{\text{vap},1}=2163.5\ \mathrm{kJ\ kg^{-1}}$; however, no heat is transferred from the reactor vessel since the steam cannot escape.
Assuming that after reaching a pressure of $p_1=3.000\ \mathrm{bar}$, a steam relief valve is opened so that steam can escape from the reactor to the environment, and assuming this valve is big enough, the steam pressure will quickly drop to the ambient pressure of $p_2=1.000\ \mathrm{bar}$. Additional water will evaporate until the temperature has reached the equilibrium temperature of $T_2=99.6\ \mathrm{^\circ C}$. After that, water will continue to evaporate with the specific enthalpy of vaporization $\Delta h_{\text{vap},2}=2257.4\ \mathrm{kJ\ kg^{-1}}$.
If the vacuum system is opened again (and the steam relief valve is closed), the pressure will decrease to the initial value of $p_3=p_0=0.030\ \mathrm{bar}$. Additional water will evaporate until the temperature has reached the equilibrium temperature of $T_3=T_0=24\ \mathrm{^\circ C}$ again.
