Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering

High viscosity is obviously an undesirable property of good refrigerant as it increases the losses of energy converts the useful energy in to non-conservative frictional energy.

By energy balance $$m{c_p}\left( {{T_i} - {T_f}} \right){\text{ of block}} = m{c_p}\left( {{T_f} - {T_i}} \right){\text{ of water}}$$
So, we will get $${T_f} = 303.16\,k$$
We know,
$$\eqalign{ & \Delta {S_{total}} = \Delta {S_{BLOCK}} + \Delta {S_{WATER}} \cr & \Rightarrow \Delta s = m{c_p}ln\frac{{{T_f}}}{{{T_I}}} + m{c_p}ln\frac{{{T_f}}}{{{T_i}}} \cr} $$
$$ \Rightarrow $$ on substitution $$w$$ will get $$\Delta s = - 1.869 + 3.118 = 1.26$$

Since there will be a linear relationship between enthalpy and concentration at constant temperature the isotherm will be straight line.

If we see the typical $$p-T$$  plot any substance below the triple point we have only sublimation zone that means no liquefaction is not possible if we still decrease the pressure hence always the pressure should be greater than triple point pressure and the value of temperature depends on the pressure at which the substance is present.

For a spontaneous process of a system the change in entropy should be always greater than zero and for an equilibrium (more or less reversible) process the entropy must be equal to zero.

The change in enthalpy is given as enthalpy of the products - enthalpy of the reactants since, the given reaction is opposite of given reaction only the sign changes.

We know COP of a refrigerator
$$\eqalign{ & = \frac{{{\text{Cooling effect}}}}{{{\text{Work taken}}}} \cr & = \frac{{{Q_L}}}{{{W_{net}}}} \cr & = \frac{{{Q_2}}}{{{Q_1} - {Q_2}}} \cr} $$
For a reversible refrigerator COP $$ = \frac{{{T_2}}}{{{T_1} - {T_2}}}$$
Since $${T_1} - {T_2} < {T_2}$$
So COP is always greater than zero.

Compressor the process equipment is usually not used during the vapour compression refrigeration system and absorption compression refrigeration system.

For an ideal gas $$PV = nRT$$
$$\eqalign{ & So,{\left( {\frac{{\partial P}}{{\partial V}}} \right)_T} = \frac{{ - nRT}}{{{V^2}}} \cr & \frac{{\partial V}}{{\partial {T_P}}} = \frac{{nR}}{P} \cr & \frac{{\partial P}}{{\partial {V_V}}} = \frac{{nR}}{V}{\left( {\frac{{\partial T}}{{\partial V}}} \right)_V} \cr} $$

Heat evolved or absorbed during allotropic change is called heat of transition.