Chemical Thermodynamics MCQ Quiz - Objective Question with Answer for Chemical Thermodynamics - Download Free PDF

The Answer is  Degrees of freedom

Concept:-

• Degrees of Freedom (DOF): Degrees of freedom refer to the number of variables that can be independently varied without changing the number of phases in equilibrium. In the context of the phase rule, DOF indicates the system's flexibility to adjust certain parameters.
• Components: Components are the minimum number of independently variable chemical entities needed to describe the composition of each phase/system. Understanding components is crucial for applying the phase rule to different systems.
• Phase Equilibrium: Phase equilibrium involves the coexistence of different phases in a system under specific conditions. The phase rule helps analyze and predict the behavior of phases in equilibrium based on the given components and phases.
• Phase Diagrams: Phase diagrams visually represent the equilibrium conditions of a system, showing how different phases coexist at varying temperatures and pressures. The phase rule is often applied in the interpretation of phase diagrams.

Explanation:-

The phase rule equation, F = C - P + 2, represents the number of degrees of freedom (F) in a system at equilibrium. 

• F (Degrees of Freedom): This is the number of intensive variables (e.g., temperature, pressure, concentration) that can be independently varied without altering the number of phases in equilibrium.
• C (Number of Components): This is the minimum number of independently variable chemical entities needed to describe the composition of each phase/system.
• P (Number of Phases): This is the number of distinct, homogenous, and mechanically separable parts in the system that coexist under certain conditions.

CONCEPT:

Adiabatic Reversible Compression of Ideal Gas

• In adiabatic reversible processes for an ideal gas:

  T₂ = T₁ × (V₁/V₂)^γ-1

• γ (gamma) = C_p/C_v is the adiabatic index:
  • For monoatomic gases, γ = 5/3
• T₁ can be calculated using the ideal gas equation:

  P × V = nRT ⇒ T₁ = (P × V₁) / (nR)

EXPLANATION:

• Given:
  • n = 1 mol
  • P = 1 atm = 1.013 × 10⁵ Pa
  • V₁ = 49.2 L = 0.0492 m³
  • V₂ = 24.6 L = 0.0246 m³
  • R = 8.314 J/mol·K
• Calculate initial temperature T₁:

  T₁ = (P × V₁) / (nR) = (1.013 × 10⁵ × 0.0492) / 8.314 ≈ 599 K

• Use adiabatic relation:

  T₂ = T₁ × (V₁/V₂)^γ-1

  γ = 5/3 ⇒ γ - 1 = 2/3

  T₂ = 599 × (49.2 / 24.6)^2/3 = 599 × 2^2/3 ≈ 599 × 1.587 ≈ 950.7 K

• Closest value: 952 K

Therefore, the final temperature of the gas is closest to Option 1: 952 K.

CONCEPT:

Partial Molar Volume (V̄) from Total Volume Expression

• Partial molar volume of a solute in solution is defined as the change in total volume with respect to the number of moles of solute, keeping solvent mass constant.
• Given total volume V as a function of molality (m), and water mass (W) is fixed, we use:

  V̄_solute = (dV/dn_solute)_{W constant}

• Since molality m = n_solute / (mass of solvent in kg), we write V in terms of m and compute:

  V̄_solute = dV/dm × dm/dn = (dV/dm) × (1/W)

EXPLANATION:

• Use given volume expression:

  V (cm³) = 1000 + 60(m/m°) − 0.5(m/m°)²

  where m° = 1 mol/kg.
• Given:
  • Mass of ethanol = 460 g → n = 460 / 46 = 10 mol
  • Mass of water = 2 kg → m = 10 mol / 2 kg = 5 mol/kg
• Let x = m/m° = 5. Then:

  V = 1000 + 60x − 0.5x²

  dV/dx = 60 − x

  At x = 5 ⇒ dV/dx = 60 − 5 = 55

• V̄ = (dV/dx) × (dx/dm) = 55 × (1/m°) = 55 cm³/mol

Therefore, the partial molar volume of ethanol is 55 cm³/mol. The correct answer is Option 3.

CONCEPT:

Clausius-Clapeyron Equation for Phase Transition

• For solid-liquid equilibrium, the Clapeyron equation is:

  dT/dP = ΔV / ΔS

• Where:
  • ΔV = Change in molar volume during transition (in m³ mol^-1)
  • ΔS = Change in molar entropy during transition (in J K^-1 mol^-1)
• This equation tells how the transition temperature (like freezing point) changes with pressure.

EXPLANATION:

• Given:
  • ΔV = -1.6 cm³ mol⁻¹ = -1.6 × 10^-6 m³ mol⁻¹
  • ΔS = 22 J K⁻¹ mol⁻¹
  • ΔP = 100 bar = 100 × 10⁵ Pa = 10⁷ Pa
• From Clapeyron equation:

  dT = (ΔV / ΔS) × ΔP

  = (-1.6 × 10^-6 / 22) × 10⁷

  = -0.0727 × 10⁷

  = -0.727 K

• Negative sign indicates decrease in temperature.

Therefore, the lowering in freezing point is approximately 0.73 K is correct.

CONCEPT:

Residual Entropy at T = 0 K

• According to the Third Law of Thermodynamics, the entropy of a perfect crystal at 0 K is 0 J·K⁻¹·mol⁻¹.
• However, for some substances like carbon monoxide (CO), residual entropy exists at 0 K due to orientational disorder.
• In CO, the molecule can be arranged in two indistinguishable orientations: C≡O or O≡C. These are energetically identical but distinguishable quantum mechanically.
• This leads to a nonzero entropy even at 0 K, given by:

  S = R ln(W), where W is the number of microstates.

EXPLANATION:

• For 1 mole of CO molecules with two possible orientations each:

  W = 2^N, where N = Avogadro’s number.

  So, S = R ln(2)

• For 1 mole:

  S = (8.314 J·mol⁻¹·K⁻¹) × ln(2) ≈ 5.76 J·mol⁻¹·K⁻¹

• For 2 moles:

  S = 2 × 5.76 = 11.53 J·K⁻¹

Therefore, the entropy of 2 moles of CO at 0 K is closest to 11.53 J·K⁻¹.

Concept:

• Carnot engine: The theoretical engine which works on the Carnot cycle is called a Carnot engine.
  • It gives the maximum possible efficiency among all types of heat engines.

• Heat source: The part of the Carnot engine which provides heat to the engine is called a heat source.
  • The temperature of the source is maximum among all the parts.

• Heat sink: The part of the Carnot engine in which an extra amount of heat is rejected by the engine is called a heat sink.

• The amount of work that is done by the engine is called work done.

The efficiency (η)of a Carnot engine is given by:

\(η = 1 - \frac{{{T_C}}}{{{T_H}}} = \;\frac{{Work\;done\left( W \right)}}{{{Q_{in}}}} = \;\frac{{{Q_{in}} - \;{Q_R}}}{{{Q_{in}}}}\)

Where TC is the temperature of the sink, TH is the temperature of the source, W is work done by the engine, Qin is the heat given to the engine/heat input and QR is heat rejected.

Calculation:

Given:
The temperature of the source T_H = 500K

The temperature of the sink T_c = 400K

The efficiency is given by:

\(η = 1 - \frac{T_{2}}{T_{1}}\)

Substituting the values in the above equation, we get:
\(η = 1 - \frac{400}{500}\)

or, η = .2, converting it in percentage, we get: .2 × 100 = 20%

Hence the efficiency is 20%.

Concept:

Law of Mass action and Equilibrium constant:

• At a constant temperature, the rate of a chemical reaction is directly proportional to the product of molar concentrations of reactants present at any given time. This is the law of mass action.
• The equilibrium constant for the reversible reaction of type \(aA + bB \rightleftharpoons cC + dD\) is represented as

\({K_{\rm{c}}}{\rm{ = }}\frac{{{{{\rm{[C]}}}^{\rm{c}}}{{{\rm{[D]}}}^{\rm{d}}}}}{{{{{\rm{[A]}}}^{\rm{a}}}{{{\rm{[B]}}}^{\rm{b}}}}}\)

Characteristics of an equilibrium constant:

• It is a definite value for all chemical reactions.
• ​The constants Kp and Kc are both equilibrium constants.
• Kp is used when the concentration terms are given in partial pressures i.e, in gaseous reactions.
• Kc is used when the reaction terms are expressed in molarities.
• The relation between Kp and Kc is given by:

\({K_p} = {K_c} \times {\left( {RT} \right)^{\Delta n}}\) where R = Universal gas constant, T = Temperature, and \(\triangle n\) = change in moles of gases in the reaction.

Calculation:

Given:

The reaction is:  CO (g) + Cl2 (g) ⇌ COCl2

The partial pressure of CO = .7bar

The partial pressure of CO₂ = 1 bar

The partial pressure of COCl2 (g) at equilibrium = 0.15 bar

• For the given reaction,

• The partial pressure of CO at equilibrium = .7 - .15 = .55
• The partial pressure of CO2 at equilobrium= 1 - .15 = .85
• The partial pressure of COCl2 (g) at equilibrium = 0.15 bar
• The Kp for the reaction is:

\({K_{\rm{p}}}{\rm{ = }}{ p_{COCl_2}\over p_{CO}\times p_{Cl_2}}\)

\(k_p= .{15\over .55 \times .85} = .32\)

Hence, the equilibrium constant for this reaction at 500 °C is .32.

Concept:

• Phase of a system is defined as the physical state that can be separated from other part of system mechanically.
• Components are distinct constituents that are independently variable.
• In a system, minimum number of independently variable factors that can be used to define the system gives degree of freedom (F). The factors can be pressure, temperature , or composition/concentration.

Degree of Freedom(F) = C- P+ 2

C is the number of components of system

and P is the number of phase

Explanation:

For given system, number of components, C = 2

number of phase, P = 3

Degree of freedom, F should be

\(F= C-P+2 \)

\(F=2-3+2 \)

\(F=1 \)

Conclusion:

Hence, the degree of freedom of a two component system with its  3 phases in simultaneous equilibrium is 1

Concept:-

For second-order phase transition, volume, entropy, and enthalpy do not change at the transition

Explanation:-

Yellow phosphorus -> Red phosphorus

given , Δ V= 0
           Δ S = 0
so Phase transition = second order
That option 2 is correct

Concslusion:-

So, The order of this phase transition is most likely to be second order

Concept:

Gibbs Free Energy

In thermodynamics, Gibbs free energy is used to calculate maximum reversible work performed by a thermodynamic system at a given temperature and pressure. 

The Gibbs free energy of a system can be calculated as shown below,

G = H - TS 

where, 

G = Gibbs free energy 

H = Enthalpy of the system 

T = Temperature in kelvin

S = Entropy of the system

• At constant temperature, the change in Gibbs free energy can be written as 

\(\Delta G\, = \,\Delta H - T\Delta S\)

• Under standard conditions 

\(\Delta \mathop G\nolimits^0 \, = \,\Delta \mathop H\nolimits^0 - T\Delta \mathop S\nolimits^0 \)

Explanation:

Given, at  298 K 

\(\begin{array}{l} \Delta \mathop G\nolimits_f^0 \, = \,370.7\,KJmo{l^{ - 1}}\\ \Delta \mathop H\nolimits_f^0 \, = \,416.3\,KJmo{l^{ - 1}} \end{array}\)

we a calculated from the equation,

\(\begin{array}{c} \Delta \mathop G\nolimits_f^0 = \,\Delta \mathop H\nolimits_f^0 - T\Delta \mathop S\nolimits_f^0 \\ 370.7\,KJmo{l^{ - 1}}\, = \,416.3\,KJmo{l^{ - 1}}\, - \left( {298K \times \Delta \mathop S\nolimits_f^0 } \right)\\ \Delta \,\mathop S\nolimits_f^0 = \,0.1530\,KJmo{l^{ - 1}}\,{K^{ - 1}} \end{array}\)

Further, it is given that, \(\Delta{H}_f^0\)  is constant in the interval 250 K to 375K. i.e., \(\Delta{H}_f^0\)= 416.3 KJmol^-1 

at 375K , 

\(\begin{array}{c} \Delta \mathop G\nolimits_f^0 \,\, = \,\,\Delta \mathop H\nolimits_f^0 \, - \,\,T\Delta \mathop S\nolimits_f^0 \\ \Delta \mathop G\nolimits_f^0 = 416.3\,KJmo{l^{ - 1}} - \left( {375K \times 01530KJmo{l^{ - 1}}{K^{ - 1}}} \right)\\ = 358.92\,KJmo{l^{ - 1}} \end{array}\)

\(\Delta{G}_f^0\)  for Fe(g) at 375 K is,  \(\Delta \mathop G\nolimits_f^0 \,\, = \,\,359\,KJmo{l^{ - 1}}\)

Concept:

Enthalpy of formation (Heat of formation):

• It is defined as the enthalpy change when one mole of a compound is formed from its elements.
• It is designated as ΔHf.

​∆Hf = ΔHf (products) - ΔHf (reactants)  

Where, ΔHf (products) = Enthalpy of formation of the product, ΔHf (reactants) = Enthalpy of formation of reactant

Standard enthalpy of formation:

• The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation is called Standard Molar Enthalpy of Formation.
• Its symbol is ΔHfo.
• The enthalpy of formation of all free elements in their standard states is regarded as zero.

i.e., ΔHfo = ΔHfo (products) - ΔHfo  (reactants) 

Where, ΔHfo = ​Standard enthalpy of a formation, ΔHfo (products) = Standard enthalpies of formation of all products, ΔHfo  (reactants) = Standard enthalpies of formation of all reactants.

Calculation:

Given:

• The reaction taking place is: 2 P (s) + 3 Br2 (l) → 2 PBr3 (l) 
• The enthalpy change of the reaction is -243kJ.
• Atomic Wt. of P = 31

In the above reaction, we see that two moles of Phosphorus are reacting, which gives us the energy of -243kJ.

• The number of moles pf Phosphorus in 3.1 g =

3.1 / 32 = .1mole

So, if 2 moles of P liberates -243kJ, .1 mole of P will liberate =

\({.1 × 243\over2 }{=-12.15kJ}\)

Hence, if the amount of phosphorus consumed is 3.1 g, the change in enthalpy is -12.15kJoule.

Concept:

• According to the first law of thermodynamics,

\(dq = du + dw\). where, \(dq,\; du\; \)and \(dw\) are the amount of heat change, internal energy change and work done by the system.

• When the expansion of the gas is carried out under isothermal conditions, the temperature of the system remains constant (dT=0). For a perfect gas, the pressure (P) of the system changes with the volume (V) according to the equation

PV = K, where K is a constant. 

• When the expansion of the gas is carried out under adiabatic conditions, no heat change takes place between the system and surroundings (dq=0). For a perfect gas, the pressure (P) of the system changes with the volume (V) according to the equation

​\(P{V^\gamma } = K\), where K is a constant.

• For a monoatomic gas, the value of \(\gamma \) is

​\({5 \over 3}\) (\(\gamma = {{{C_p}} \over {{C_v}}}\)).   

Explanation:

• For an adiabatic process, the pressure (P) of the system changes with the volume (V) according to the equation 

​\(P{V^\gamma } = K\)

• Under adiabatic conditions, the pressure will change with volume as, 

​\(P = {K \over {{V^\gamma }}} \) 

or, \(P \propto {1 \over {{V^\gamma }}} \)

• Now, for or a monoatomic gas, the value of \(\gamma \) is 5/3. 

So, \(P \propto {1 \over {{V^{{5 \over 3}}}}}\) 

• Thus, the pressure of the gas will fall more rapidly under adiabatic conditions because 

 \(P \propto {1 \over {{V^{{5 \over 3}}}}}\)

​Conclusion:

Hence, the pressure of the gas will fall more rapidly under adiabatic conditions because  p ∝ \(\rm \frac{1}{v^{5/3}}\) 

Concept:

According to Gibb's Phase rule,

\(F=C-P+2\)

Where,

F is the the number of degrees of freedom, C is the number of chemically independent constituents of the system, and P is the number of phases.

Explanation:

We know that, \(F=C-P+2 \)

\(P=C+2-F \)

Given,

For one component system, \(C=1\)

Now we get, \(P=1+2-F=3-F\) 

For P to be maximum, F should be equal to zero.

Hence, \(P=3-F=3-0=3\)

Conclusion: -

The maximum number of phases that can be simultaneously in equilibrium for a one component system is 3. So the option 3 is correct.

Concept: 

The Clausius–Clapeyron relation characterizes a discontinuous phase transition between two phases of matter of a single constituent. On a pressure-temperature (P–T) diagram, the line separating the two phases is known as the coexistence curve. The Clausius–Clapeyron relation gives the slope of the tangents to this curve. The Clausius–Clapeyron relation can be used to find the relationship between pressure and temperature along the phase.

Explanation: 

The given phase diagram is as follows: 

We know that the melting point is the temperature at which liquid changes into solid or vice-versa.

As the diagram depicts by the dotted line between solid and liquid as the pressure increases the melting point is decreasing. 

Thus, based on the given diagram the statement: 'Melting point increases with pressure ', is wrong.

Hence, the correct option is (1).

Conclusion:-

So, the INCORRECT statement is Melting point increases with pressure

Explanation:-

• 25 mg Compound contains (25-16) g of Water = 9 g of water,

Using,

No. of moles = Given Mass / Molecular Mass

No. of moles = 250 g of the compound will contain 90 g of water; and

250 g of  Water will contain = 90 / 18 moles of Water = 5 moles.