Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering

The quantitaive relationship between equilibrium constant (equilibrium) and temperature is given by vant hoff equation .whereas the arrhenius quation gives an quantitative relationship between temperature and rate constant .and the lechalters principle gives the quilitative relationship between temperature and chemical equilibrium.
Vant hoff equation : $$lnK = - \frac{{\Delta H}}{{RT}} + \frac{{\Delta S}}{R}$$

The gibbs duhem equation is an equation that relates the partial modal properties of different species in the solution with composition at constant temperature and pressure and the equation is given by :
Writing the gibbs duhem equation in terms of partial gibbs free energy :

$$\sum\nolimits_{i\, = \,1}^n {} {x_i}d{\overrightarrow G _i} = 0$$

Since $$d\stackrel{\rightharpoonup }{G_l}=RTdln\left({\stackrel{⌢}{f}}_i\right)$$

We can write \[\sum\nolimits_{i\,=\,1}^{n}{{}}{{x}_{i}}d{{\left( ln\overset\frown{f} \right)}_{i}}=0\]
So, by gibbs duhem equation the composition and fugacity are related.

Because the enthalpy of an real gas is given by
$$dH = {C_p}dT + \left[ {V - T\left( {\frac{{\partial V}}{{\partial T}}} \right)}p \right]dp$$
For ideal gas since $$\left( {\frac{{\partial V}}{{\partial T}}} \right)p = \frac{R}{P} \Rightarrow dH = {c_p}dT$$
Hence only for an ideal gas enthalpy is a function of temperature.

The inversion point is an point where an sign change occurs so we can define this point as point of inflation. So here the slope (joule thomson cofiicient) is zero hence $${\left( {\frac{{\partial T}}{{\partial P}}} \right)_H} = 0$$

As the real gas can deviate positively or negatively from ideal behavior it can have
$$Z$$ (compressibility factor) > 1 → positive deviation.
$$Z$$ < 1→ negative deviation.

Conside piston cylinder arrangement when the expansion occurring in vacuum there is no work has to be done by the system. But for an real gas on striking the piston losses some of its momentum and hence velocity of molecules decreases and therefore cooling takes place. But for an ideal gas since its velocity doesn’t changes the temperature remains same.

Since always losses are inherint with irreversible process.
eg :some of the work done will be lost as either friction loss or intermolecular energy etc.
whereas an reversible process takes place in infinitesimal steps so we can neglect losses or consider as no losses so work obtained will be high.

During phase transition since the system will be at same temperature and pressure we can say the system is in thermodynamic equilibrium and hence during melting and freezing the system will be in equilibrium.