Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering

At the critical temperature we can say there will be no demarcation between the liquid and vapor but the intermolecular forces still exist so, the volume occupied by the substance there will still be less than the ideal gas.

An isolated system is that kind of system which will be in no contact with surroundings in terms of both mass and energy. Generally this kind of system is called as universe that means the system and the interacting surroundings are taken in to single system and analysis is done.

COP of refrigerating system $$ = \frac{{{\text{heat taken from the low temperature reserviour}}}}{{{\text{work supplied}}}}$$

So, we can say there are no units for COP because the heat taken and work both are energy in transits and therefore have same units.

Domestic refrigeration cycle works on the principle of absorption cycle which basically states heat is removed from the food items by passing refrigerant around them with the help of compressor, condenser, throttling etc.

Since, most of the gases exhibit a linear relationship between temperature and volume.
So, we can say generally, When the temperature is 0K the volume also become zero.

The melting and boiling points of all the substances increase with increase in pressure which can be explained by Le Chatelier's principle or molecular behavior.
Molecular behavior: when the pressure increases the intermolecular or attractive forces between the molecules increases so, more thermal energy is required for conversion of solid to liquid. Hence, melting point increases.
There are some exceptions for this behavior also eg: water.
So, for paraffin wax with increase in pressure the melting point increases.

For a pure substance in a closed system has a degree of freedom = 2
So, entropy(s) can be expressed as a function of two variables (properties) $$S = fn(T,V)$$

\[\begin{align} & \Rightarrow dS={{\left( \frac{\partial S}{\partial T} \right)}_{V}}dT+{{\left( \frac{\partial S}{\partial v} \right)}_{T}}dT \\ & \Rightarrow dS=\frac{{{C}_{V}}dT}{T}+{{\left( \frac{\partial P}{\partial T} \right)}_{V}}dV\left( \text{for one mole of ideal gas}{{\left( \frac{\partial P}{\partial T} \right)}_{V}}=\frac{R}{V} \right) \\ & \Rightarrow dS=\frac{{{C}_{V}}}{T}dT+\frac{nR}{V}dV \\ \end{align}\]
So, on integrating this equation from temperature T₂ to T₁ expanding volume from V₁ to V₂ GIVES the equation

$$\Delta S = nRln\left( {\frac{{{V_2}}}{{{V_1}}}} \right)\, + nCvln\left( {\frac{{{T_2}}}{{{T_1}}}} \right)$$

for an ideal gas of n moles.

For an adiabatic process → $$P{V^y}$$ = constant
For an isothermal process → $$PV$$ = constant
So, slope for adiabatic process in $$PV$$ plane is $$\frac{{dp}}{{dv}} = - y\frac{p}{v}$$
Slope for isothermal process is $$\frac{{dp}}{{dv}} = - \frac{p}{v}$$
Hence, $${\left( {\frac{{\partial p}}{{\partial v}}} \right)_S} = - y{\left( {\frac{{\partial p}}{{\partial v}}} \right)_T}$$

Troutons rule states that for most of the liquid the entropy of vaporisation is constant at boiling points and the entropy of vaporisation is given by ratio of enthalpy of vaporisation to normal boiling temperature $$\left( {{T_b}} \right)$$ nothing but troutons ratio.
$$\Delta {S_{vap}} = \frac{{\Delta {H_{vap}}}}{{{T_b}}}$$

Isentropic process is the process where entropy(s) remains constant. So, $$ds = 0$$
Generally, $$ds$$ $$ = \frac{{\delta Q}}{T}$$   for reversible process, so for reversible adiabatic process $$ds = 0$$
Hence, reversible adiabatic process is called isentropic process.