Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering

For Thermodynamic equilibrium to exist between the phases the phases should have same pressure (so mechanical equilibrium exists) and to satisfy the thermal equilibrium the temperature between the phases should be same similarly, for chemical equilibrium to exist the chemical potential of a constituent between all the phases should be the same.

By first law of thermodynamics we can write under adiabatic conditions $$dU = \delta W.$$   Since the work done is a path function to bring a certain internal energy obviously the work done depends upon path specified. this can be understood by taking the example of climbing up an hill in different ways the initial and end points are same but when he goes to climb the hill by using an rope he needs different amount of energy and if he goes by steps he needs different amount of energy (just for understanding). So, to bring a certain internal energy change one requires different amount of work when different paths are involved.

The heat of formation of an element in standard state is considered as zero. Since, it is a natural form thereby, doesn’t require any energy to obtain this form.

The variation is given by Clapeyron equation and the equation is:
$$\frac{{dp}}{{dT}} = \frac{{\Delta H}}{{T\Delta V}}$$

The given number of component $$= 1$$
Given number of phases $$= 2$$
We know, degree of freedom (number of independent variable) $$ = C - \emptyset + 2.$$
(Assuming no reaction taking place)
$$ \Rightarrow f = 1 - 2 + 2 = 1.$$

Actually the internal energy ($$U$$) of a substance is a function of $$dU = CvdT - \left[ {P + T\left( {\frac{{\left( {\frac{{\partial V}}{{\partial T}}} \right)P}}{{\left( {\frac{{\partial V}}{{\partial P}}} \right)T}}} \right)dV} \right]$$

For an ideal gas, $$PV = RT$$

So, $$\left( {\frac{{\partial V}}{{\partial T}}} \right)P = \frac{R}{P}$$   and  $$\left( {\frac{{\partial V}}{{\partial P}}} \right)T = - \frac{{RT}}{{{P^2}}}$$
Hence, $$dU = CvdT$$
So, Internal energy of an ideal gas is purely a function of temperature. (Here the $${C_V}$$ is considered only as a function of temperature which is satisfied by many of the substances.)

Enthalpy of a substance is defined as the internal energy of the substance $$+$$ pressure energy per unit mass of the substance.
The reason for defining such kind of relationship is that under constant pressure system the change of this property (enthalpy) denotes the heat added or removed from the system.

Since entropy is obviously a point or state function by its definition itself.
If we see the following steps on how we arrived to entropy it will be clear that entropy is a state function:
For, reversible heat engines working under same thermal reservoirs since,
$$\eqalign{ & \oint {} \frac{{\partial Q}}{T} = 0 \cr & \Rightarrow \frac{{\delta Q}}{T} = dS \cr} $$
Since, $$\frac{{\delta Q}}{T}$$   is a path independent quantity we designated it as a property which is nothing but entropy.

Here the $$F$$ is Gibbs free energy and $$A$$ is Helmholtz free energy.
So, $$F=H-TS$$   and $$A=U-TS$$
$$\eqalign{ & \Rightarrow F = H + A - U \cr & \Rightarrow F = A + PV \cr} $$

Mollier chart is a graph between enthalpy and entropy.