Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering

Third law of thermodynamics tells about entropy it says that entropy becomes constant at absolute zero temperature. There are some deviations observed from this law for some substances.

As first law of thermodynamics failed to tell about the direction of energy transfer and spontaneity of a process the second law of thermodynamics is introduced it tells about the direction of energy transfer and tells about spontaneity of a process etc.

For an ideal solution the heat of mixing or enthalpy change and volume change due to mixing is zero.
Since volume change mixing $$ = {V^t}\left( {T,P} \right) - \sum {} {x_i}{V_i}\left( {T,P} \right).$$
Where $${V^t} = $$   total molar volume of the solution at temperature $$T$$, pressure $$P$$.
$${V_i} = $$   molar volume of species when existed as pure species at same $$T$$, $$P$$
From summability : \[{{V}^{t}}=\sum{{}}{{x}_{i}}{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{V}}}_{l}}\]     where \[{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{V}}}_{l}}=\]  partial molar volume since for ideal solution
\[\sum{{}}{{x}_{i}}{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{V}}}_{l}}=\sum{{}}{{x}_{i}}{{V}_{i}}\left( T,P \right)\]       $$ \Rightarrow $$ Volume change of mixing $$= 0$$
Similarly, enthalpy change of mixing or heat of mixing $$= 0.$$

Since, internal energy $$(U)$$ is a point function the change over cyclic process will be equal to zero.
$$dU=0.$$

Consider an elementary reaction $$A \Leftrightarrow B$$   and let equilibrium concentrations of $$A$$ and $$B$$ be $${{C_a}}$$ and $${{C_b}},$$
So equilibrium constant $$\left( K \right)$$ is $${K_1} = \frac{{{C_b}}}{{{C_a}}}$$
When the concentrations of both the species doubled equilibrium constant becomes $$K = \frac{{2{C_b}}}{{2{C_a}}} \Rightarrow K = {K_1}.$$     Hence remains constant.

Since in case of free expansion external pressure $$\left( {{P_{ext}} = 0} \right)$$   the expansion work done will be zero as $$W = \int {{P_{ext}}dV.} $$

Critical solution temperature is that temperature at which the complete miscibility or a homogeneous mixture is formed. Sometimes the temperature has to be raised from the existing temperature or sometimes it should be decreased from the existing temperature to reach this critical solution temperature.

Since $${K_p} = {K_c}R{T^{\Delta n}}$$   out of the given reactions the reaction $${N_2}{O_4} \Leftrightarrow 2N{O_2}$$    has $$\Delta n > 0 \Rightarrow {K_p} = {K_c}.$$

Reduced properties are defined as the ratio of properties at the existing point to the properties at critical point.
So, reduced pressure $$ = \frac{{{\text{pressure}}}}{{{\text{critical point pressure}}}}$$