Chemical Engineering Thermodynamics — Part 1 MCQs with Explanations

Sublimation means the conversion of solid phase directly in to vapor phase (without involvement of phase liquid). Actually for every substance this direct phase conversion occurs when the substance is below its triple point pressure and corresponding temperature. Actually one of the point where the dry ice sublimates is 1 atm pressure and corresponding temperature is $$ - {78^ \circ }C.$$

The given equation is a thermodynamic equation of state which is true for every substance since it is relating point variables.

Henrys law is closely obeyed by a gas when its solubility is very low in an solution.

For an ideal solution both the volume of mixing and enthalpy of mixing is zero since an ideal solution behaves like a pure solvent.

Given $$\Delta H = - 22.4\,kcal$$ so it represents an exothermic reaction as heat is released out during the process.

As, we know the relation $$\Delta G = \Delta {G^ \circ } + RTlnQ$$ where $$Q =$$ reaction quotient, higher the value of gibbs free energy change $$ \Rightarrow $$ higher is the value of $$Q \Rightarrow $$ faster is the rate of the reaction.

For a given refrigeration effect, the system with the minimum value of Coefficient of Performance (COP) is the least efficient in terms of energy usage. Option A: Reverse Carnot cycle. The Carnot cycle represents the maximum possible efficiency for a given temperature difference. Therefore, the reverse Carnot cycle would have the highest COP among the options. Option B: Ordinary vapour-compression cycle. While not as efficient as the Carnot cycle, the vapour-compression cycle is commonly used in refrigeration systems and typically has a higher COP compared to other cycles. Option C: Vapour-compression process with a reversible expansion engine. This would have a higher COP than the air refrigeration cycle because it involves a vapor-compression process. Option D: Air refrigeration cycle. Air refrigeration cycles typically have the lowest COP among the options due to the lower efficiency of air as a refrigerant compared to vapor-compression cycles. Therefore, the correct answer is Option D: Air refrigeration cycle .

Here work means expansion work since the expansion work ($$W$$) and we know expansion work $$W = \int {} {P_{ext}}dV$$ and since for isochoric process there will be no volume change i..e.., volume remains constant and hence $$dV = 0.$$ So, work done $$= 0$$

Since, the area under $$P-V$$ plot for isothermal is less than the area under $$p-v$$ plot for adiabatic process during compression. We can say adiabatic work done is more than isothermal work done.

Since for an adiabatic process \[T{V^{\gamma - 1}} = {\rm{CONSTANT}}\] and on integration of this equation we get \[ln\left( {\frac{{{T_2}}}{{{T_1}}}} \right) = - \left( {\left( {\gamma - 1} \right)ln\frac{{{V_2}}}{{{V_1}}}} \right)\] since for expansion process \[{V_2} > {V_1}\] Hence \[ \Rightarrow {T_2} < T_1\] So, cooling takes place.

All the three will be effected by temperature and the relation between the temperature and fugacity is given by : \[{\left( {\frac{{\partial ln{f_i}}}{{\partial T}}} \right)_{P,n}} = \frac{{ - \overline {{H_l}} + {{\overline {{H_l}} }^{^ \circ }}}}{{R{T^2}}}\] Activity co-efficient and temperature given by : \[\frac{{\partial lna}}{{\partial {T_P}}} = \frac{{{H^{^ \circ }} - H}}{{R{T^2}}}\] Free energy and temperature is given by : \[{\left( {\frac{{\partial G}}{{\partial T}}} \right)_P} = S\]

A closed system is that in which there will be no mass transfer or material transfer but there will be energy transfer here the identity of system also remains same. Since in the given options these conditions are satisfied by liquid cooling system where our attention is on energy transfer rather than mass transfer.

Entropy of an ideal gas depends upon : \[\Delta S = \int_{{T_i}}^{{T_f}} {\frac{{{C_P}dT}}{T} - nRln\left( {\frac{{{P_f}}}{{{P_i}}}} \right)} \] \[ \Rightarrow \] it depends upon both temperature and pressure.

The chemical potential and fugacity of an real gas is related by : \[\mu = {\mu ^ \circ } + RT\frac{{lnf}}{{{f^ \circ }}}\] Where \[{f^ \circ } = \] standard fugacity given $$=1$$

Partial molar properties are defined as in non ideal systems the molar properties of an component in an mixture is not equal to molar properties of the same component when present as pure component so as to designate the properties of individual components in the mixture we used partial molar properties. These partial molar properties and the total mixture property is related as \[{M^t} = \sum {{x_i}\overline {{M_l}} } \]

Group contribution method is used to measure the critical properties.

The efficiency of an heat engine is given by \[\mathop \eta \limits^\iota = \frac{{{Q_1} - {Q_2}}}{{{Q_1}}}\] For heat pump it is given by \[\mathop \eta \limits^\iota = \frac{{{Q_1}}}{{{Q_1} - {Q_2}}}\] Where \[{{Q_1}}\] and \[{{Q_2}}\] are given by heat taken and heat given to the thermal reservoirs respectively. \[ \Rightarrow {\left( {\mathop \eta \limits^\iota } \right)_{heat\,\,engine}} = \frac{1}{{{{\left( {\mathop \eta \limits^\iota } \right)}_{heat\,\,pump}}}}\]

The room temperature is considered as standard stat at which internal energy is considered or taken as zero basically it is considered as Datum for calculations of internal energy.

A critical point is a point where the physical distinction between the liquid and vapor vanishes since there will be no further difference in properties like surface tension, volume, density etc. Between the liquid and vapor. So we call the state of substance at and above the critical point as gas.

The slope of isothermal process in $$p-v$$ plane is \[\frac{{dP}}{{dV}} = \frac{{ - P}}{V}\] The slope of adiabatic process in $$P-V$$ plane is \[\frac{{dP}}{{dV}} = - \gamma \frac{P}{V}\] \[ \Rightarrow \] Slopes become equal when \[\gamma = 1\]

The Carnot cycle consists of $$1.$$ Reversible isothermal expansion so, \[T = {\rm{CONSTANT}}\] $$2.$$ Reversible adiabatic expansion so, \[S\left( {{\rm{entropy}}} \right) = {\rm{CONSTANT}}\] $$3.$$ Reversible isothermal compression so, \[T = {\rm{CONSTANT}}\] $$4.$$ Reversible adiabatic compression so, \[S\left( {{\rm{entropy}}} \right) = {\rm{CONSTANT}}\]

Given, vapor-liquid equilibrium : Number of components \[\left( {{\rm{water}}} \right) = 1\] Number of phases involved \[\left( {{\rm{vapor - liquid}}} \right) = 2\] So, degree of freedom : \[\begin{array}{l} \left( f \right) = C - \emptyset + 2\\ 1 - 2 + 2 = 1 \end{array}\]

We know heat and work are path functions as they are dependent on path specified. But according to first law the difference between heat and work is an point function \[\delta Q - \delta W = dU\]

Helmholtz free energy : \[\begin{array}{l} A = U - TS\\ dA = - {W_R} \end{array}\] So, the decrease in Helmholtz energy denotes maximum work done in an reversible process under isothermal conditions.

Gibbs free energy : \[\begin{array}{l} G = H - TS\\ \Rightarrow - dG = {W_R} - PdV \end{array}\] So, it represents maximum useful work in a reversible process.

According to second law of thermodynamics, the entropy \[{\left( {dS} \right)_{universe}} > 0,\] for an irreversible or spontaneous process and \[d{S_{universe}} = 0\] for a reversible process or almost equilibrium reached process. so for an natural process since there will be irreversibility always the system entropy goes on increasing until equilibrium is reached where the system entropy becomes equal to zero. Hence a maximum exists for a system.

The enthalpy of formation of an element in its pure state is zero .since it doesn’t require any energy to obtain this state ($$form$$) it is naturally occurring state of the element. \[\begin{array}{l} {O_2}\left( g \right) \to {O_2}\left( g \right)\\ \Rightarrow \Delta H = 0 \end{array}\]

Helmholtz function : \[\begin{array}{l} A = U - TS\\ \Rightarrow dA = - PdV - SdT \end{array}\] So, we can derive the Maxwell function : \[{\left( {\frac{{\partial P}}{{\partial T}}} \right)_V} = {\left( {\frac{{\partial S}}{{\partial V}}} \right)_T}\]

A reversible process means the process in which both system and surroundings both should be restored to their original conditions when the system is bought to initial conditions. Since in case of free expansion of gas it is not possible to bring back the surroundings to initial state when system is brought back to initial conditions we call it as an irreversible process.

In both Rankine and Brayton cycle the heat is added at constant pressure.

Heat is added at constant temperature in an ideal Stirling cycle.

The standard state of an gas is taken as ideal gas state where the fugacity is equal to pressure and the pressure is usually taken as $$1,$$ similarly the activity under ideal state is equal to $$1$$ since the activity is the ratio of fugacity at conditions of our interest to the fugacity at standard state and we are taking the standard state as pressure $$= 1$$ So, activity = fugacity = 1.

As ethylene has flammable characters it is not commonly used as a refrigerant.

Since the entropy of mixing is positive we can say the mixing process or solution formation process is an irreversible process.

By second law of thermodynamics the \[TdS \ge \delta Q\] For an isolated system \[\delta Q = 0\] \[ \Rightarrow \] for an irreversible (nothing but spontaneous process) \[dS > 0\] and for an reversible (or the system at equilibrium) \[dS = 0\]

The Isobaric, isothermal and adiabatic process represented in $$P – V$$ as the following. Since, area under $$P-V$$ plot for isobaric process is more than the isothermal and adiabatic process the work done in isobaric process is highest.

Given, number of components $$= 2$$ Number of phases $$= 1$$ So, degree of freedom \[\begin{array}{l} f = C - \emptyset + 2\\ \Rightarrow 2 - 1 + 2\\ \Rightarrow 3\\ F = 3 \end{array}\]

Since here heat is taken from the system \[Q = - ve\] From second law of thermodynamics \[TdS \ge \delta Q\] Since given reversible process \[TdS = \delta Q\] $$ \Rightarrow dS = - ve$$ From first law of thermodynamics $$\eqalign{ & \delta Q = dU \cr & \Rightarrow dU = - ve \cr} $$

We know $$S$$ as a function of $$T$$ and $$P$$ as : $$dS = \frac{{{C_P}dT}}{T} - {\left( {\frac{{\partial V}}{{\partial T}}} \right)_P}dP$$ For isothermal change $$dT = 0$$ for ideal gas $$\frac{{\partial V}}{{\partial {T_P}}} = \frac{{nR}}{P}$$ $${\text{So, }}dS = - nRln\left( {\frac{{{P_2}}}{{{P_1}}}} \right)$$ For isothermal volume change of an ideal gas for both reversible and irreversible process.

The degree of freedom means the minimum number of intensive variables which are needed to specify the system and its minimum value is zero. Triple point of a substance is one of the state for which the degree of freedom $$= 0.$$