Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering

After throttling, the temperature of the gas may increase or decrease because here there are two opposing effects:
1. As the pressure energy is converted to intermolecular energy the temperature of the gas should increase.
2. By throttling since the pressure decreases or expansion will takes place by equation of state we can tell the temperature also decreases.
So, it all depends on what is initial state of the gas and nature of the gas.

Since at or above critical state there is no strong demarcation between the vapor and liquid that means the properties doesn’t show any abrupt changes the gas can’t be liquefied.

We know that the equation which relates the easily measurable and sensible properties (basic properties of a substance) nothing but the temperature, pressure and volume is called equation of state.

Since for an spontaneous process $$TdS > dU + PdV$$
For, constant temperature and constant pressure process the above equation can be written as
$$\eqalign{ & d\left( {TS - U} \right) > PdV \cr & \Rightarrow d\left( { - A} \right) > PdV \cr & \Rightarrow {\left( {dG} \right)_{T,P}} < 0 \cr} $$
For spontaneous process.

The free energy change for a chemical reaction in terms of equilibrium constant is given by.
$$\Delta {G^ \circ } = - RTlnK$$
Since at equilibrium $$\Delta G$$ of the reaction $$=0$$

Since all the gases except $${H_2}$$ $$He$$ etc, have maximum inversion point greater than atmospheric (room) temperature cooling can take place by throttling if the initial conditions lie in a region where joule thomson coefficient is positive. But for $${H_2}$$ the maximum inversion point is very low so usually heating takes place by throttling. But if we do the decrease the temperature of $${H_2}$$ from room temperature to below the maximum inversion point temperature by some primary cooling techniques than there are chances where cooling of $${H_{2\,gas}}$$ can be done by throttling.

Since in case of reversible or equilibrium process $$\delta Q = TdS$$   and for isolated system $$\delta Q = 0$$
So, $$dS=0$$

According to Le Chatlier's principle the system will react in such a way that it minimizes the effect of deviation from equilibrium. Since, ice has the more volume than water Equilibrium shifts towards water (less volume) side thereby, minimizing the effect of pressure and regaining a new equilibrium. So, we can say the melting point is reached fast or decreased.

The given is statement of third law of thermodynamics.

An isentropic process is that process where entropy remains constant. The reversible adiabatic process is considered as isentropic process.