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# First Law of Thermodynamics Questions

The branch of science that deals with the transfer of energy from one form to another, as well as the relationship between heat and temperature, energy, and work done, is known as thermodynamics. The laws of thermodynamics are used to transfer heat energy linked with chemical reactions into different useful forms.

Energy transformation is based on the fact that energy can only be transformed from one form to another and used in many industries. The laws of thermodynamics are concerned with energy changes that occur during a reaction, not with the rate at which the reaction occurs.

 Definition: The first law of thermodynamics is a thermodynamic version of the concept of conservation of energy, differentiating three types of energy transfer: heat, thermodynamic work, and energy associated with matter transfer, and linking these to an internal energy function of a body.

## First Law of Thermodynamics Chemistry Questions with Solutions

Q1: ΔU = 0 means that the process is isothermic. True or False?

a) True

b) False

Explanation: Temperature is the only factor that affects internal energy. As a result, if ΔU = 0, the process will have a constant temperature or be isothermic.

Q2: Temperature is developed during a fluid flow as a result of________.

a) Increase in density

b) Change in pressure

c) Fluid level

d) Translational Kinetic Energy

Explanation: When a high rate of fluid flow is present, the molecules are more likely to clash. The average translational kinetic energy of the particles increases in this state. The temperature established due to this is called Kinetic temperature.

Q3: When enthalpy increases, the _________ rises as well.

a) Internal energy

b) Volume

c) Pressure

d) Mass

Explanation: The number of molecular interactions increases as the temperature increases. According to the equation from the first rule of thermodynamics, internal energy increases with temperature. As a result, as enthalpy rises, internal energy rises as well.

Q4: Which of these reactions isn’t exothermic?

a) Combustion reaction

b) Neutralization reaction

c) Evaporating liquid water

d) Thermite reaction

Explanation: An exothermic reaction releases energy in the form of light or heat. Endothermic reactions are the polar opposite of exothermic processes. Evaporating liquid water is an endothermic reaction in this situation. Endothermic reactions take place when a system consumes heat from its environment.

Q5: During photosynthesis, what reaction occurs?

a) Endothermic reaction

b) Exothermic reaction

c) Redox reaction

d) Combustion reaction

Explanation: Photosynthesis involves the absorption of heat and energy from the environment. The reaction that occurs during photosynthesis is an endothermic reaction because it is a response in which the system absorbs heat from its environment.

Q6: What is the entropy equation?

a) Ratio of absolute temperature to reversible heat transfer

b) Ratio of reversible transfer of heat to absolute temperature

c) Ratio of adiabatic heat to macroscopic variables

d) Ratio of macroscopic variables to adiabatic heat

Answer: b) Ratio of reversible transfer of heat to absolute temperature

Explanation: Rudolf Clausius established this equation by defining entropy as the ratio of reversible heat transfer to absolute temperature. This concept of entropy is also known as the macroscopic definition.

Q7: Calculate the final volume of one mole of an ideal gas at 0°C and 1 atm pressure. During reversible isothermal expansion, it absorbs 2000 calories of heat.

The gas is at standard temperature and pressure, i.e. it is at STP.

As a result, V1 = 22.4 dm3 and V2 must be determined.

Assuming that the expansion is isothermal and reversible, ∆U = 0.

We know that ∆U = q + w

But ∆U = 0

Thus, q = – w = 2000 cal

= 2000 x 4.184 J = 8368 J

w = −nRT ln (V2/ V1) is the work done in reversible isothermal expansion.

∴ nRT ln (V2/ V1) = – w = 8368 J

⇒ (I mol) x (8.314 J K mol) x (273 K) ln (V2/ 22.4 dm3) = 8368 J

V2 = 242.50 dm3

As a solution, one mole of an ideal gas at 0°C and 1 atm pressure has a final volume of 242.50 dm3.

Q8: Give the limitations of the First Law of Thermodynamics.

Limitations of the first law of thermodynamics are as follows:

• There are no constraints on the flow of heat in either direction: the first law provides a direct relationship between the heat received and the work done by a system. The first law does not state whether or not heat can travel from a cold to a hot end. For example, chilling ice to a low temperature will not allow us to extract heat from it. There is some external work that remains to be done.
• Does not indicate if the reaction is possible: the first law does not specify whether a process is feasible or not. For example, when a rod is heated at one end, equilibrium must be achieved, which is only attainable with some energy expenditure.
• It is not possible to convert heat energy into an equivalent quantity of work in practise.

Another law, known as the second law of thermodynamics, is required to overcome these limitations.

Q9: In an isothermal process, the first law of thermodynamics states that-

(a) the internal energy of the system fluctuates and there exists an imbalance between work and heat

(b) a constant amount of heat is added to the system

(c) temperature is not affected by the internal energy of the system, work, or heat

(d) it can be implied that heat added to the system is equal to work done by the system

(e) temperature is not affected by the pressure or volume changes of the system

Answer: (d) it can be implied that heat added to the system is equal to work done by the system

Explanation: The difference between the heat energy added to the system and the work done by the system, according to the first law of thermodynamics, determines the change in internal energy of a closed system:

△U = Q − W

Temperature remains constant in an isothermal process. The temperature of a system is a measure of its internal energy. There is no variation in internal energy if the temperature is constant.

0J = Q − W

It can be assumed that the amount of heat added to the system equals the amount of work done by the system.

Q = −W

Q10: Compare Isothermal Expansion and Adiabatic Expansion.

Assume that isothermal and adiabatic expansions of an ideal gas have the same initial volume Vi and pressure Pi, leading to the same end volume Vf. If the final pressures are Piso and Padia, then:

For the purpose of isothermal expansion, Pi Vi = Piso Vf

As a result, Vf/Vi = Pi/Piso

As for expansion (Vf) > (Vi) and also for all gases γ > 1 , hence

{(Vf/Vi)γ} > {Vf/Vi}

Q11: Define the enthalpies of vaporisation and fusion.

Enthalpy of Vaporisations: Some heat is absorbed from the surroundings during the evaporation of a liquid. As a result, an increase in enthalpy is required for liquid evaporation.

While evaporating one mole of water at 25°C, the increase in enthalpy is 43.93 kJ.

This result can be expressed in the form of thermo-chemical equation which is given below:

H2O(l) → H2O(g); ∆H = + 43.93 kJ

The evolution of heat occurs when vapours condense into liquid state. As a result, a drop in enthalpy causes liquid to condense.

The enthalpy decreases by 43.93 kJ when one mole of water vapour is condensed at 25°C. Therefore,

H2O(g) → H2O(l); ∆H = – 43.93 kJ

Enthalpy of Vaporization is the change in enthalpy (∆H) when a liquid transforms into a vapour and a vapour transforms into a liquid.

Enthalpy of Fusion: Liquid is formed when melting of solid takes place. The absorption of heat occurs during this time.

The system’s enthalpy rises as a result of this. While fusing one mole of water at 0°C, the increase in enthalpy is 6.02 kJ.

It is as follows:

H2O(s) → H2O(l); ∆H = + 6.02 kJ

When a liquid freezes and transforms into a solid state, heat is emitted. As a result, the enthalpy of the process drops, i.e.

H2O(l) → H2O (s); ∆H = – 6.02 kJ

Enthalpy of Fusion is defined as the change in enthalpy, i.e. ∆H, when a solid becomes a liquid and a liquid becomes a solid.

Q12: What are the various forms of Energy?

The various forms of Energy are:

• Heat Energy

A transfer (flow) of thermal energy across a boundary (for example, from a hot body to a cold body via the area of their contact) is defined as heat. Heat is defined as energy that transfers from a high-temperature item to a lower-temperature object. The energy in an object is referred to as its internal energy rather than “heat.” The action of heating might increase this internal energy.

A temperature difference causes heat energy to be transferred. A warm body with a higher temperature transfers energy as heat to a cold body with a lower temperature.

Heat flow is the transmission of energy that occurs only as a result of a temperature differential. The SI unit of power, the Watt, can be defined as 1 J/s of heat flow. Calorie is another unit used to measure heat energy.

• Chemical Energy

Chemical energy is the energy derived from the interactions of atoms in molecules and other types of matter aggregation. It is defined as the work done by electric forces during the aggregation process when mutual positions of electric charges, electrons and protons, are rearranged. So, the electrostatic potential energy of electric charges is basically what it is. When a system’s chemical energy is reduced during a chemical reaction, the difference is passed to the environment in some way (often heat or light). If, on the other hand, a system’s chemical energy increases as a result of a chemical reaction, the difference is filled by the environment (usually again in form of heat or light).

• Mechanical Energy

Mechanical energy comes in a variety of forms, but it may be divided into two categories: elastic potential energy and kinetic energy. Potential energy, on the other hand, is a very broad phrase that encompasses all force fields, including gravitation, electrostatics, and magnetic fields. The energy that any object receives as a result of its position in a force field is referred to as potential energy.

• Electric Energy

The electric potential energy of a provided configuration of charges is characterised as the work which must be done against the Coulomb force to rearrange charges from infinite separation to this configuration.

Q13: C2H2(g), C (graphite), and H2(g) have molar heats of combustion of 210.62 Kcal, 84.05 Kcal, and 69.32 Kcal, respectively. Calculate the standard heat of formation of C2H2(g).

The required reaction is: 2C + H2 → C2H5

Standard heat of formation equation, i.e. ΔH =?

Creating a thermochemical equation based on the information provided:

(i) C2H2 + 5/2 O2 → 2CO2 + H2O, ΔH = –210.62 kcal (given)

(ii) C + O2 → CO2, ΔH = –84.05 kcal (given)

(iii) H2 + 1/2 O2 → H2O, ΔH = –69.32 kcal (given)

Now, multiplying (ii) eq. by 2 and then adding (iii) eq. and subtracting (i) eq. i.e.

i.e. (ii) 2C + 2O2 → 2CO2, ΔH = –168.1 kcal

As a result, we get the following equation after adding (iii) eq. and subtracting (i) eq.

2C + H2 → C2H2, ΔH = 26.80 kcal

As a result, ΔH = 26.80 kcal is the standard heat for the formation of C2H2.

Q14: Define Gibbs Free energy. Also, derive the Relationship between Equilibrium Constant and Gibbs free energy and hence Calculate ΔG° for conversion of oxygen to ozone 3/2 O2 (g) → O3 (g) at 300 K, if Kp for this conversion is 9.47×10-29.

Gibbs free energy: The Gibbs free energy (or Gibbs energy) is a thermodynamic potential that can be used to calculate the maximum reversible work that a thermodynamic system can accomplish at a constant temperature and pressure in thermodynamics. It is symbolised by ΔG°.

Relationship between Free Energy and Equilibrium Constant: When equilibrium is not achieved, the reaction’s free energy change in any state is denoted as ΔG, which is related to the reaction’s standard free energy change, ΔG°.

ΔG = ΔG° + RT InQ, where Q is the reaction quotient

When equilibrium is achieved, there is no further change in free energy, i.e. ΔG = 0 and the reaction quotient Q equals the equilibrium constant. As a result, the equation above will appear like this:

ΔG° = –RT In Keq

Or, ΔG° = –2.303 RT log Keq

Galvanic cells are an example of this. The electrical work done by the cell is related to Gibbs energy change ΔG.

ΔG = -nFEcell, where, n = no. of moles of electrons involved

E = emf of the cell

When both the reactants and the products are in their standard states,

ΔG° = nFE°cell

3/2 O2(g) → O3(g)

Now, ΔG° = –2.303 RT log Kp

Where R = 8.314 J/K mol, Kp = 9.47×10-29, T = 300K

Thus, Gibbs energy, ΔG° = 160971 J/mol = 160.971 KJ/mol

Q15: One mole of an ideal gas is heated at constant pressure from 0°C to 200°C.

(a) Calculate the amount of work completed.

(b) What must be the final pressure if the gas is expanded isothermally and reversibly at 0°C from 1 atm to some other pressure Pt if the maximum work is equal to the work involved (a).

(a) Work done:

Work done during the heating of gas from 0 to 200°C.

T1 = 0°C = 273 K

T2 = 200°C = 473 K

Now

W = – PΔV = -P (V2 – V1)

= – P [(nRT2/P) − (nRT1/P)]

= – nR (T2–T1) = –1 × 1.987 × (473 – 273)

= – 397.4 cal

(b) If 397.4 cal of work is applied to a gas at 0°C to cause isothermal expansion from 1 atm to pressure Pt, the work done in reversible isothermal expansion is:

w = – 2.303nRTlog (P1/P2)

–397.4 = – 2.303 × 1.987 × 273log (1/Pt)

Therefore, Pt = 1.272 atm

Hence, final pressure is 0.694 atm.

## Practise Questions on First Law of Thermodynamics

Q1: First law of thermodynamics is based on?

a) Conservation of energy

b) Conservation of mass

c) Conservation of momentum

d) Conservation of work

Q2: The equation for the average kinetic energy is_________

a) 0.5 KT

b) 2.5 KT

c) 1.5 KT

d) 3.5 KT

Q3: How is the first law of thermodynamics applied in chemistry?

Q4: The first law of thermodynamics is represented by which of the following?

(a) While melting, an ice cube remains at the same temperature

(b) If a refrigerator is unplugged, eventually, everything inside of it will return to room temperature

(c) The specific heat of an object explains how easily it changes temperatures

(d) Two systems in equilibrium with a third system are in equilibrium with each other

(e) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy

Q5: Show that ∆U = 0 and ∆H = 0 in an Isothermal Expansion of an Ideal Gas.

Click the PDF to check the answers for Practice Questions.

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