Chemical Engineering Thermodynamics — Part 2 MCQs with Explanations

The equilibrium constant $$\left( {{K_P}} \right)$$ for the reaction below can be written as : $${K_P} = {\left( {atm} \right)^{\Delta x}}$$ Where $$\Delta x = $$ change in number of moles (difference in number of moles of products $$-$$ reactants).

The equilibrium constant $$\left( {{K_C}} \right)$$ is dependent only on temperature and nature of reactants and products involved and the relation is given as : $${K_C} = A{e^{ - \frac{{{E_A}}}{{RT}}}}$$

According to the gas law we can write $$\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}$$ So by reducing the pressure to $$\frac{P}{4}$$ at constant temperature the volume increases to $$4$$ times the initial volume.

Given, number of components $$= 1$$ Number of phases $$= 2$$ So, degree of freedom $$f = 1 - 2 + 2 = 1$$

The distribution laws says that for two non-miscible solvents will be in equilibrium if the chemical potential of solute is same in both the phases the concentration need not be equal but the ratio of concentrations will be remains constant once equilibrium is reached.

Given, number of components $$= 2$$ Number of phases $$= 2$$ $$\eqalign{ & {\text{So, }}f = 2 - 2 + 2 \cr & \Rightarrow f = 2 \cr} $$ Assuming no reaction or reaction taking place very slowly.

The first law of thermodynamics is another form of conservation of energy which states that energy can’t be destroyed or created only it can be converted from one form to another. Here we relate the two energy in transits (heat and work).

Since, the degree of randomness decreases when the water is converted to ice the entropy decreases and also in an similar way when the gas is expanding from high pressure to low pressure the entropy increases.

One of the corollaries of second law of thermodynamics states that heat can’t be continuously transferred from low temperature body to high temperature body without taking any external work. This is known as clausius statement of second law of thermodynamics.

For a triple point the degree of freedom is zero. So, we can say any parameter can’t change its state. It is a fixed state point.

The isentropic process is nothing but reversible adiabatic process since between the isothermal and adiabatic process the isothermal process has maximum area under $$P-V$$ diagram the isothermal process will have maximum work done during expansion. Since the slope of $$p-v$$ plot for adiabatic process is greater than isothermal process.

During fusion process the heat is absorbed by the solid to convert to liquid leading to increase in enthalpy.

A phase indicates physically how a matter is existing. So, minimum it should be $$=1.$$

Extensive properties are those which depend upon the extent ($$mass\,$$) of the system. Typical extensive properties are volume, internal energy, Gibbs free energy etc..,

Van Laar equation is an activity based model developed by Van Laar during 1910-1913. Which will be very useful during vapor-liquid equilibrium calculations and the equations for a binary mixture are given as : $$\eqalign{ & ln{\gamma _1} = \frac{{AX_2^2}}{{{{\left( {\frac{{A{X_1}}}{B} + {X_2}} \right)}^2}}} \cr & ln{\gamma _2} = \frac{{BX_1^2}}{{{{\left( {\frac{{B{X_2}}}{A} + {X_1}} \right)}^2}}} \cr} $$

Insulated cylinder $$ \Rightarrow $$ adiabatic system $$\eqalign{ & \Rightarrow P{V^\gamma } = {\text{CONSTANT}} \cr & \Rightarrow T{V^{\gamma - 1}} = {\text{CONSTANT}} \cr & \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \frac{{{V_2}^{\gamma - 1}}}{{{V_1}}} \cr & \Rightarrow \frac{{500}}{{{T_2}}} = {\left( 5 \right)^{\gamma - 1}} \cr & {C_P} - {C_V} = R\left( {{\text{For ideal gas}}} \right) \cr & \Rightarrow \gamma = 1.65 \cr & {\text{So, }}{T_2} = 176.05\,K \cr} $$

According to Le Chatelier's principle the system will react in such a way that it minimizes the effect of deviation from equilibrium. Since, ice has the more volume than water equilibrium shifts towards water (less volume) side thereby, minimizing the effect of pressure and regaining a new equilibrium.

Hess law : if different process are used to bring the same chemical reaction than the enthalpy change is same for all of them. ( Note : enthalpy is a point function doesn’t depends on the path between.)

It is the point where the air becomes saturated and gives its first liquid bubble of more volatile substance in an solution this point corresponding temperature and pressure is called dew temperature and dew pressure.

The First Law of Thermodynamics is a statement of the conservation of energy . It expresses that the total energy of an isolated system remains constant — energy can be transferred or transformed, but it cannot be created or destroyed. Option A: Direction of energy transfer This option is incorrect . While the first law deals with the amount of energy transferred , it does not determine the direction of energy flow. The Second Law of Thermodynamics addresses the direction of energy transfer — for example, heat naturally flows from a hotter body to a cooler one. Option B: Reversible processes only This option is incorrect . The First Law applies to all types of processes — both reversible and irreversible . It only accounts for the total energy balance and is not concerned with the nature of the process. Option C: Irreversible processes only This is also incorrect . As with Option B, the law is universal and does not distinguish between reversible and irreversible processes. It applies equally to both. Option D: None of these This is the correct answer . None of the listed options correctly describe what the First Law of Thermodynamics actually deals with. The first law is about energy conservation , not about direction (Option A), nor is it limited to any specific process type (Options B and C). Therefore, the correct answer is: None of these.

The $${C_P}$$ and $${C_V}$$ are the amount of heat required to raise the temperature of a gas by an $${1^ \circ }C$$ under constant pressure and constant volume respectively. Since in case of constant volume the entire heat is used for raising the temperature of the gas only no work is done but in case of constant pressure process some of the heat given is used for doing work also hence more amount of heat is required in case of constant pressure process to obtain the $${1^ \circ }C$$ change in temperature of the gas. But in case of solids and liquids since volume expansivity is small the $${C_P}$$ and $${C_V}$$ values remains same.

$$\frac{{1kw}}{{ton}} = \frac{{3.516}}{{c.o.p}}$$

For an azeotropic mixture since the composition of a constituent remains same in vapour and liquid one of the variable to define the system reduces. So, $$f = C - \emptyset + 2$$ Given number of components $$= 2,$$ number of phases $$= 2$$ So, degree of freedom $$= 1$$

We know : $$G = {G^ \circ } + RTln\frac{f}{{{P^ \circ }}}$$ $$ \Rightarrow $$ where $${G^ \circ }$$ and $${P^ \circ }$$ are at standard states So, we can say when $${P^ \circ } \to 0, \Rightarrow {G^ \circ } \to \infty $$

Air cycle has minimum coefficient of performance.

Definition of heat capacity: the amount of heat required to raise the temperature of a substance or material by a small amount is called heat capacity. $$\mathop {\lim }\limits_{dT \to 0} \,dQ = C\,\left( {{\text{heat capacity}}} \right)$$ Every substance has its own molecules and every molecule in a substance has its own properties and structure depending on the nature of the substance so different molecules exhibit different heat capacity at the same temperature. But every substance molecules at 0K will come to rest and to raise the temperature of a substance by a negligible amount doesn’t depend upon the nature of the substance but once the substance reaches $$dT$$ amount of temperature different substance or materials exhibit different heat capacity.

We know, the property relation: $$dG = vdp - sdT$$ Which is valid for both reversible and irreversible process since it is a property relation. when the system undergoes isothermal change $$dT = 0$$ $$\eqalign{ & {\text{So, }}dG = vd \cr & \Rightarrow dG = \frac{{nRT}}{P}dP\left( {{\text{for, ideal gas}}} \right) \cr} $$ On, integration $$ \Rightarrow \Delta G = nRT.ln\frac{{{P_2}}}{{{P_1}}}$$ So, for a system containing ideal gas and undergoing isothermal change of volume or pressure this expression is valid.

Since $$P{V^Y} = $$ constant is valid only for reversible process but since as free expansion is irreversible because it is working on a cycle by taking heat from a single reservoir and producing net expansion work when considered reversible and thus violating Kelvin-planck statement when considered reversible hence we can conclude free expansion is an irreversible process. And hence $$P{V^Y} = $$ constant can’t be valid.

$$A.$$ solid ⟹ liquid → melting $$B.$$ liquid ⟹ vapor → vaporization $$C.$$ solid ⟹ vapor → sublimation So, enthalpy change for the reaction $${H_2}O\left( {liq} \right) \Rightarrow {H_2}O\left( {gas} \right)$$ is enthalpy change of vaporization. In the question entropy should be replaced by enthalpy.

During phase change suppose consider liquid to vapor the entropy increases because randomness increases and the liquid converts to vapor by taking enthalpy so, enthalpy changes and generally the enthalpy of vapor will be greater than liquid. We may think internal energy remains constant during phase change since temperature remains constant but here the potential energy changes thereby, changing internal energy since in vapor the distance between molecules is greater than the distance between the molecules in liquid hence the work made is different. So internal energy is different. But we know Gibbs free energy is a function of pressure and temperature and during the phase change since pressure and temperature remains constant so, Gibbs free energy remains constant, but not zero.

Activity coefficient measures the extent to which a real gas deviates from ideality. Hence, for ideal gas activity coefficient = 1.

The internal energy ($$U$$) is a function of $$dU = CvdT - \left[ {P + T\left\{ {\frac{{\left( {\frac{{\partial V}}{{\partial T}}} \right)p}}{{\left( {\frac{{\partial V}}{{\partial P}}} \right)T}}} \right\}dV} \right]$$ For an ideal gas, $$PV = RT$$ So, $$\left( {\frac{{\partial V}}{{\partial T}}} \right)p = \frac{R}{P}{\text{ and}}\left( {\frac{{\partial V}}{{\partial T}}} \right)T = \frac{{ - RT}}{{{P^2}}}$$ Hence, $$dU = CvdT$$ So, for an ideal gas if it undergoing isothermal change $$\left( {dT = 0} \right) \Rightarrow dU = 0$$ So, the questioned should be changed and should be mentioned for an ideal gas.

According to law of corresponding states if two gases have same reduced temperature and reduced pressure than they will have same reduced volume. Law of corresponding state: $$\left[ {Pr + \frac{3}{{{v_r}^2}}} \right]\left[ {3Vr - 1} \right] = 8R{T_r}$$ So, the answer is reduced volume.

Gibbs phase rule is, $$G = C - \in + 2$$ Where : $$C$$ = Number of components $$ \in $$ = Number of phases $$G$$ = Degree of freedom Clearly we can see Gibbs phase rule is more useful when two or more phases are involved in the given options the condensation is the only process where more than one phase is involved so, the Gibbs phase rule finds its application there.

The phase change occurs at constant temperature and pressure under normal situations. But the volume changes because between two phases always there is a change in intermolecular forces except at critical condition. Eg: when liquid is converting to vapor the vapor formed will have more volume than liquid.

According to Henrys law the solubility increases with increase in pressure at a particular temperature and the relation is $$S = H * P$$ Where : $$S$$ = solubility $$H$$ = henrys constant $$P$$ = pressure.

Chemical engineering thermodynamics is more concerned with the equilibrium states behavior. How much energy or mass is present at equilibrium like what is the equilibrium concentration, temperature etc. or we can say by basic thermodynamics the amount of energy transferred from one form to another by the use of first law of thermodynamics. But, basically the chemical engineering thermodynamics deals with equilibrium states.

Only below and at the critical point there is strong demarcation in properties of liquids and gases above the critical point we can’t recognize the difference. Hence for a gas which is under or at the critical point can be liquefied. For a gas above critical point can also undergo a pressure change but we can’t notice the difference in properties.

Closed system is also called as control mass system where there is no exchange of mass between system and surroundings but energy transfer can takes place. In closed system there shouldn`t be change in identity of mass (molecules) of system even.

clausis-claperyon equation can be applied for any equilibrium system consisting of two phases the two phases can be solid and liquid during melting or solid and vapor during sublimation and the relation is given as. $$lnP = \frac{{ - \Delta {H_{phase\,\,change}}}}{{RT}} + C$$