Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering

Translational kinetic energy of an ideal gas is given by $$ = \frac{3}{2}KT.$$

An ideal solution is that where the components behaves like pure components.
For an ideal solution the heat of mixing or enthalpy change and volume change due to mixing is zero.
Since, volume change of 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 $$= 0$$

Joule Thomson coefficient is nothing but the slope of temperature and pressure plot drawn during throttling so, it is dependent on both temperature and pressure.

During the joule-thomson plug experiment fluid is allowed to pass through valve or porous plug and observations are required for this experiment if we write steady flow energy equation we will get to know that enthalpy remains constant.

Area under $$p-v$$  diagram is minimum when work is done on a system under isothermal conditions hence $$n=1.$$

Heat of combustion is the change in heat content of the compound when one mole of that compound is burnt in the presence of oxygen. This type of reaction is where common in petroleum refinery where coke is combusted and removed.

Under reversible conditions if we see the area under $$P-V$$  plot the area is maximum in case of isothermal conditions hence the adiabatic work is less than isothermal work.

From second law of thermodynamics
$$\eqalign{ & TdS \geqslant \delta Q \cr & \Rightarrow TdS \geqslant dU + \delta W \cr} $$
For an irreversible process $$TdS - dU - \delta W > 0$$
And for a reversible process $$TdS - dU - \delta W = 0$$
For any spontaneous process there should be finite changes so, we can consider it as an irreversible process and we know for irreversible process from second law of thermodynamics by above discussion: $$TdS - dU - PdV > 0$$
Under constant temperature and volume process $$ - dF > 0 \Rightarrow dF < 0$$
Similarly for an constant temperature and pressure process $$d\left( {TS - U - PV} \right) > 0 \Rightarrow dG < 0$$

We can say by Le Chatelier's principle on increasing the pressure the system will react in an way to minimize the effect( the volume decreases) so, generally substances tend to form solid but in case of water the as the liquid occupies less volume than solid the high pressure tends to form liquid (melting).

High viscosity is obviously an undesirable property of good refrigerant as it increases the losses of energy converts the useful energy in to non-conservative frictional energy.