Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering

The unit of equilibrium constant depends upon the change in number of moles of products and reactants and general units will be $${\left( {mol\,d{m^{ - 3}}} \right)^{\Delta n}}$$
Where, $$\Delta n =\,$$ Different in number of moles of products and reactants.

Since all the gases at same reduced pressure and temperature all the gases behaves similarly the compressibility factor will be same for all the gases.

We know the change in entropy $$dS = \frac{{\delta Q}}{T}$$   and for expansion since the heat has to be added to the system and by our convention heat added to the system isothermally is positive so change in entropy is positive.

When a gas expands in a vessel, its molecules do work on the surroundings by pushing against external pressure.

Internal energy of a gas is directly related to its temperature. If the internal energy decreases during expansion, it means the gas is losing energy.

If the system is not allowed to exchange heat with the surroundings (Q = 0), the energy decrease can only be due to the work done by the gas. This condition defines an adiabatic process.

In contrast, an isothermal process maintains constant internal energy, and reversible/irreversible refers to the manner of expansion, not the energy change.

Hence, the correct answer is Adiabatic.

All the given quantities are effected by temperature and the relations are like this :
Free energy change $$dG = VdP - SdT$$
Fugacity $$dG = RTd\left( {lnf} \right)$$
Activity coefficient $$\gamma = \frac{f}{{{f^{id}}}}$$

Here as the reaction is exothermic we can say by Le Chatelier's principle the decrease in temperature favours the forward reaction but since here the number of moles of both the reactants and products are same the pressure has no effect by Le Chatelier's principle.

The liquefication involves both adiabatic expansion and joule-thomson effect of change in pressure as we do the observations at adiabatic conditions during the flow of gas through orifice (joule Thomson effect).

For a cyclic process by first law of thermodynamics we can say $$\delta Q - \delta W = 0 \Rightarrow \frac{{\delta Q}}{{\delta W}} = 1$$
So, a fixed ratio always exists and we can also the difference between the heat and work is a point function.

Definition of Boyle temperature: the temperature at which a real gas obeys the ideal gas laws over wide range of pressures that particular temperature is called as Boyle temperature.