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

CONCEPT:

Standard Heat of Formation

• The standard heat of formation of a substance is the amount of heat absorbed or evolved when one mole of the substance is formed from its constituent elements in their standard states under standard conditions (usually 298 K and 1 atm).
• The standard heat of formation of ions in aqueous solution can be calculated using thermodynamic data such as heats of crystallization and formation of compounds containing these ions.

EXPLANATION:

Standard heat of formation of BaSO₄(s) = Standard heat of formation of Ba²⁺ (aq) + Standard heat of formation of SO₄²⁻ (aq) + Standard heat of crystallization of BaSO₄(s)

• Rearranging the equation:

  Standard heat of formation of Ba²⁺ (aq) = Standard heat of formation of BaSO₄(s) – Standard heat of formation of SO₄²⁻ (aq) – Standard heat of crystallization of BaSO₄(s)

• Substituting the values:
  • Standard heat of formation of Ba²⁺ (aq) = –349 kcal/mol – (–216 kcal/mol) – (–4.5 kcal/mol)
  • = –349 + 216 + 4.5
  • = –128.5 kcal/mol

Therefore, the standard heat of formation of Ba²⁺ (aq) is –128.5 kcal/mol.

CONCEPT:

Enthalpy Change (ΔH) and Energy Diagram

• The enthalpy change (ΔH) of a reaction represents the difference in energy between the products and reactants.
• When ΔH is negative, the reaction is exothermic, meaning energy is released and the products have lower energy than the reactants.
• Energy diagrams for exothermic reactions show the reactants starting at a higher energy level and the products at a lower energy level, with an energy "drop" between them.

EXPLANATION:

• In the given reaction:

  C(s) + 2H₂(g) → CH₄(g); ΔH = -74.8 kJ mol^-1

• ΔH is negative, indicating that the reaction is exothermic.
• In an energy diagram for this reaction:
  • The reactants (C and H₂) start at a higher energy level.
  • The products (CH₄) are at a lower energy level, showing that energy is released during the reaction.
  • The difference in energy between the reactants and products corresponds to the magnitude of ΔH (74.8 kJ mol^-1).

Therefore, the correct energy diagram will show the reactants at a higher energy level than the products, with an energy drop of 74.8 kJ mol^-1.

CONCEPT:

• Enthalpy Change at Constant Pressure: It is defined as the heat exchanged at constant pressure, and is represented by the formula: ΔH = ΔU + PΔV
• First Law of Thermodynamics: This law states that energy can neither be created nor destroyed, only transformed. It is expressed as: ΔU = q + w, where:
  • ΔU = change in internal energy
  • q = heat added to the system
  • w = work done on the system
• Enthalpy Change at Constant Volume: At constant volume, no work is done (w = 0), so the heat added is equal to the change in internal energy. This also relates to ideal gases as: q_p = ΔU + Δn_gRT
• Change in Degree of Randomness: It is associated with entropy (ΔS). A reversible process has entropy change defined as: ΔS = q_rev / T_initial

EXPLANATION:

•  A - (r): Enthalpy change at constant pressure is given by ΔH = ΔU + PΔV. So, A matches with (r).
•  B - (p): The First Law of Thermodynamics is ΔU = q + w. So, B matches with (p).
•  - (q): At constant volume and for ideal gases, the relation q_p = ΔU + Δn_gRT holds true. So, C matches with (q).
•  D - (s): Change in randomness or entropy is defined by ΔS = q_rev / T_initial. So, D matches with (s).

Therefore, the correct answer is Option (a): A-(r), B-(p), C-(q), D-(s).

Concept:

• Change in Internal Energy
  • When a gas absorbs heat and performs work by expanding against an external pressure, the change in its internal energy can be determined using the first law of thermodynamics.

Explanation:

The first law of thermodynamics is given by:
\(\Delta U = Q - W\)
where:
\(\Delta U\) is the change in internal energy,
Q is the heat absorbed by the system,
W is the work done by the system.

Given:
\(Q = 200 \text{ J}\)
The gas expands by\( \Delta V = 500 \text{ cm}^3 = 500 \times 10^{-6} \text{ m}^3\)
The external pressure \(P = 2 \times 10^5 \text{ N m}^{-2}\)

The work done by the gas when it expands against the external pressure is given by:
\(W = P \Delta V\)
Substituting the given values:
\(W = 2 \times 10^5 \text{ N m}^{-2} \times 500 \times 10^{-6} \text{ m}^3\)
\(W = 2 \times 10^5 \times 0.0005\)
\(W = 100 \text{ J}\)

Using the first law of thermodynamics:
\(\Delta U = 200 \text{ J} - 100 \text{ J}\)
\(\Delta U = 100 \text{ J}\)

Conclusion:

Therefore, The change in internal energy is 100 J.

Concept:

Standard Enthalpy of Formation (Δ_fH°)

• The standard enthalpy of formation (Δ_fH°) of a compound is the change in enthalpy when 1 mole of the compound is formed from its elements in their standard states.

Calculation:

First, we need to calculate the molar mass of ethane (C₂H₆):

• C: 12.01 g/mol
• H: 1.01 g/mol

Molar mass of C₂H₆ = 2(12.01 g/mol) + 6(1.01 g/mol) = 24.02 g/mol + 6.06 g/mol = 30.08 g/mol

Given 3 g of ethane, we can calculate the amount in moles:

n = (mass) / (molar mass) = 3 g / 30.08 g/mol ≈ 0.0997 mol

Heat liberated for formation of 0.0997 mol of ethane: 8.84 kJ

Since this amount of heat corresponds to 0.0997 mol, we need to scale this up to find the heat liberated for the formation of 1 mole:

Δ_fH° = (8.84 kJ) / (0.0997 mol) ≈ 88.63 kJ/mol

Since heat is liberated, the Δ_fH° is negative:

Δ_fH° ≈ -88.63 kJ/mol

The closest value to this in the options provided is -84.4 kJ

CONCEPT:

• Latent heat of vaporization: The heat energy required to change the state of a material from liquid state to its gaseous state is called the latent heat of vaporization.
  • It is also the amount of heat energy released when vapour changes into a liquid.
• Latent heat of fusion: The heat energy required to change the state of a material from solid-state to its liquid state is called the latent heat of fusion.
  • It is also the amount of heat energy released when at melting point liquid changes to solid.
• Specific heat or specific heat capacity (C) of a body is the amount of heat required for a unit mass of the body to raise the temperature by 1 degree Celsius.
• Thermal capacity (S): The amount of heat required for a mass of a material to produce a unit change in its temperature is called a Thermal capacity/heat capacity of that material.

EXPLANATION:

• The heat given to the substance during the phase change is called latent heat. It can be either the latent heat of fusion or the latent heat of vaporization depending upon the phase change. So option 2 is correct.

EXTRA POINTS:

These are some important terms used related to latent heat

• Liquid to solid: Latent heat of solidification
• Liquid to vapour: Latent heat of evaporation/vaporization (CD)
• Vapour to liquid: Latent heat of condensation

Explanation:

Joule – Thomson coefficient:- When the gas in steady flow passes through a constriction, e.g. in an orifice or valve, it normally experiences a change in temperature. From the first law of thermodynamics, such a process is isenthalpic and one can usefully define a Joule – Thomson coefficient as

\(\mu = {\left( {\frac{{\partial T}}{{\partial P}}} \right)_H}\)

As a measure of the change in temperature which results from a drop in pressure across the construction.

• For an ideal gas, μ = 0, because ideal gases neither warm not cool upon being expanded at constant enthalpy.
• If μ is +ve, then the temperature will fall during throttling.
• If μ is -ve, then the temperature will rise during throttling.

Explanation:

Critical point:

• The point at which the saturated liquid line and saturated vapour line of a pure substance meet is called the critical point.
• At a critical point, the liquid is directly converted into vapour without having a two-phase transition. So, enthalpy of vaporization at a critical point is zero i.e. At the critical point, saturated liquid and saturated vapour phases are identical.
• The figure below represents the P-V diagram for a pure substance (water).

Below critical point, latent heat is required to convert water into vapour but at this point, there is no need for latent heat i.e. at this point latent heat equals zero.

For water critical point parameters are:

Pressure (P_c) = 22 MPa, Temperature (T_c) = 374°C

Triple Point

• The Triple point is a point on the P-T diagram where all the three phases solid, liquid and gases exist in equilibrium.
• At a pressure below the triple point line, the substance cannot exist in the liquid phase and the substance when heated, transforms from solid to vapour by absorbing the latent heat of sublimation from the surroundings.
• The triple point is merely the point of intersection of the sublimation and vaporization curves.
• It has been found that on a ‘P-T’ diagram the triple point is represented by a point and on a ‘P-V’ diagram it is a line, and on a ‘U-V’ diagram it is a triangle. In the case of ordinary water, the triple point is at a pressure of 4.58 mm Hg and a temperature of 0.01°C.

dh = C_pdT, so specific enthalpy is a function of temperature only.

The ideal gas is defined as a gas which obeys the following equation of state: Pv = RT

The internal energy of an ideal gas is a function of temperature only. That is, u = u(T)

Using the definition of enthalpy and the equation of state of an ideal gas, h = u + Pv = u + RT

Since R is a constant and u = u(T), it follows that the enthalpy of an ideal gas is also a function of temperature only.

h = h(T)

Since u and h depend only on the temperature for an ideal gas, the constant volume and constant pressure specific heats c_v and c_p also depend on the temperature only. c_v = c_v (T), c_P = c_P (T) 

For all ideal gases:

• The specific heat at constant volume (c_v) is a function of T only.
• The specific heat at constant pressure (cp) is a function of T only.
• A relation that connects the specific heats cp, cv , and the gas constant is cp - cv = R
• The specific heat ratio, γ = cp/cv, is a function of T only and is greater than unity.

Concept:

• The first law of thermodynamics establishes equivalence between the quantity of heat used and the mechanical work but does not specify the conditions under which conversion of heat into work is possible, nor the direction in which heat transfer can take place.
• The second law of thermodynamics tells that heat will flow naturally from one energy reservoir to another at a lower temperature, but not in the opposite direction without assistance. This is very important because a heat engine operates between two energy reservoirs at different temperatures.

The correct option is None of the above

Concept:

• The rating of SI engine fuels is based on their antilock property.
• Rating of a particular fuel is accomplished by comparing its performance with that of a standard reference fuel which is usually a combination of iso-octane (C8H18) and normal heptane (C7H18) plus tetraethyl lead.
• Iso-octane being a very good anti-knock fuel is arbitrarily assigned a rating of 100 octane number.

Explanation:

• A common way to gauge an engine's performance or the quality of aviation gasoline is to use the octane number or octane rating.
• Iso-octane has an octane number of 100.
• Neo-octane is not a standard reference compound for the octane scale.
• n-Octane (normal octane) has a low octane number but it is not zero.
• The octane rating scale's zero point, n-octane, is what causes autoignition or combustion to occur more quickly.

Concept:

Both Heat and Work are boundary phenomena. Both are observed only at the boundaries of the system and both represent energy crossing the boundary of the system.

Point function:

• The thermodynamic properties which depend on the end state only (independent of the path followed) are known as point function like temperature, pressure, density, volume, enthalpy, entropy etc.

Path function: 

• The thermodynamic properties which depend on the end states, as well as the path followed, are known as path function like heat and work.

Only thermodynamic properties are exact differential and properties are point function. As both heat and work are path functions so, they are not thermodynamic properties. So, they are inexact differentials.

Explanation:

Entropy:

• The Second Law states that spontaneous process leads to an increase of the total entropy
• The total entropy increase will continue until it reaches the maximum possible value under the given conditions
• Thus, the state of equilibrium is the state of the maximum total entropy
• The Second Law may thus be stated as: "the total entropy tends toward a maximum"
• The third law states that the entropy of a substance is zero at the absolute zero of temperature and it represents the maximum degree of order
• Entropy is the degree of randomness in a system
• The entropies of gases are much larger than those of liquids, which are larger than those of solids
• So, the substance in the solid phase has the least entropy
• The total entropy change is zero for a reversible process (entropy conservation takes place) and positive for an irreversible process

Explanation:

Second Law of thermodynamics: 

The entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium.

The 2nd law of thermodynamics provides the feasibility or the spontaneity or the direction of possible energy conversion through the concept of entropy and states that every spontaneous process is an irreversible process.

Clausius's statement of the Second Law of Thermodynamics:

It is impossible to construct a system that will operate in a cycle, and transfers heat from the low-temperature reservoir (or object) to the high-temperature reservoir (or object) without any external effect or work interaction with the surrounding.

• It is impossible to construct a device that operates on a cycle and produces no effect other than the transfer of heat from a lower temperature body to a higher temperature body.
• The Clausius statement is related to refrigerators as well as heat pumps.

Perpetual motion machine of the second kind (PMM2): 

A fictitious machine that produces net-work in a complete cycle by exchanging heat with only one reservoir is called the PMM2.

It violates the Kelvin plank statement.

If Q2 = 0

i.e. Wnet = Q1

or Efficiency = 100%

The heat engine will produce network in a complete cycle by exchanging heat with only signal reservoir thus violating the Kelvin-Plank statement.

Thus, PMM2 violates the second law of thermodynamics and It is a hypothetical machine.

Thus a perpetual motion machine is a hypothetical machine whose operation would violate the law of thermodynamics.

Third law of thermodynamics: 

As the temperature approaches absolute zero, the entropy of a system/crystal approaches a constant minimum.

ΔST at 0 K = 0

where ΔS = change in entropy.

Explanation:

The second law of thermodynamics:

• The entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium.

ΔSTotal = ΔSSystem + ΔSsurrounding

• The second law of thermodynamic actually deals with the quality of the energy. As per the second law of thermodynamic whenever low-grade energy transfer from one state to another state some randomness get associated with the energy and the quality of energy get degraded.
• The second law of thermodynamics introduces the concept of entropy.