Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering

By T-Ds Equations at constant entropy
$$\eqalign{ & {C_p}dT = T\frac{{\partial V}}{{\partial {T_P}}}dP{\text{ and}} \cr & {C_v} = - T{\left( {\frac{{\partial P}}{{\partial T}}} \right)_P}{\left( {\frac{{\partial V}}{{\partial T}}} \right)_S} \cr & \Rightarrow \frac{{{C_P}}}{{{C_V}}} = \frac{{\left( {\frac{{\partial P}}{{\partial V}}} \right)S}}{{\left( {\frac{{\partial P}}{{\partial V}}} \right)T}} \cr} $$
Since, $${C_P}$$  is always greater than $${C_V}$$  the ratio of isothermal compressibility and isentropic (reversible adiabatic) process is always greater than $$1 \Rightarrow $$   the difference is greater than zero.

If we see the $$P-V$$  plots for isobaric, adiabatic and isothermal process the area under the graph is more in case of isobaric process hence the work done in isobaric process is maximum.

The joule Thomson coefficient is given as $${\mu _i} = {\left( {\frac{{\partial T}}{{\partial P}}} \right)_H},$$    And since for an ideal gas enthalpy is strictly only function of temperature which implies constant temperature and hence the joule thomson coefficient becomes zero. Physically both the contervining effects of throttling are balancing each other.

All the given three equations are models to find activity co-efficient.

Given, $$\frac{{{G^E}}}{{RT}} = A{x_1}{x_2} \Rightarrow \frac{{n{G^E}}}{{RT}} = \frac{{A{n_1}{n_2}}}{n}$$
On partially differentiating it $$\left( {\frac{{\partial \frac{{n{G^E}}}{{RT}}}}{{\partial {n_1}}}} \right)T,P,{n_2} = \frac{{A{n_1}{n_2}}}{n} = A{x_2}^2$$
As $$\left( {\frac{{\partial \frac{{n{G^E}}}{{RT}}}}{{\partial {n_1}}}} \right)T,P,{n_2} = \frac{{\overline {n{G^E}_l} }}{{RT}} = ln{\gamma _1}$$
Hence $$ln{\gamma _1} = A{x_2}^2$$

Since activity co-efficient is defined as
$$\eqalign{ & {\gamma _i} = \frac{{{\text{Fugacity in real solution}}}}{{{\text{Fugacity in ideal solution}}}}, \cr & ln{\gamma _1} = \frac{{\overline {n{G^E}_l} }}{{RT}} = 0\left( {{\text{for ideal solution}}} \right), \cr} $$
Since there will be no excess gibbs free energy for ideal solution.
So, in the question it will be better if replace ideal gas by ideal solution.

A combination of separating and throttling is normally used for measuring the dryness fraction.

Entropy is the measure of randomness or disorderness in the system by third law it s given that at absolute zero temperature the entropy is zero, for a perfect crystal.

The third option is nothing but third law of thermodynamics and we know that entropy is a function of both temperature and pressure.

The inversion curve is drawn by joining all the inversion points (which are nothing but maximum points for particular conditions) as the slope (joule thomson coefficient) is zero for a maximum the joule Thomson coefficient is zero.