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(a)If both (A) and (R)are true and (R) is the true explanation of (A).

(a)If both (A) and (R)are true and (R) is the true explanation of (A).

(a) ΔS_total  > 0, the process is spontaneous 

(b) ΔS_total < 0, the process is non-spontaneous 

(c) ΔS_total = 0, the process is at equilibrium.

System: The part of the Universe which is under investigation. Surroundings: The remaining part of the Universe other than system. The system and surroundings are separated by well defined boundary. 

Types of a system: 

(i) Open system 

(ii) Closed system 

(iii) Isolated system

Energy can neither be created nor destroyed but can be transformed from one form to another. 

 ΔU = q + w 

ΔU = q + ω

Where ∆U Change in internal energy

q = heat

ω = Work done on the system

(1) Isothermal process : This is a process which is carried out at constant temperature. Since internal energy, U of the system depends on temperature there is no change in the internal energy U of the system. 

Hence,

ΔU = 0.

By first law of thermodynamics,

ΔU = Q +W 

∴ 0 = Q + W 

∴ Q = -W or W = -Q.

• Hence in expansion, the heat absorbed by the system is entirely converted into work on the surroundings.
• In compression, the work done on the system is converted into heat which is transferred to the surroundings by the system, keeping temperature constant.

(2) Isobaric process : In this, throughout the process pressure remains constant. Hence the system performs the work of expansion due to volume change ΔV.

W= -P_ext × ΔV

Let Q_P be the heat absorbed by the system at constant pressure.

By first law of thermodynamics,

ΔU = Q_P + W.

∴ ΔU = Q_P – P_exΔV 

or Q_P = ΔU + P_exΔV

In this process, the heat absorbed Q_P is used to increase the internal energy (ΔU) of the system.

(3) Isochoric process : In this process the volume of the system remains constant. 

Hence, 

ΔV = 0. 

Therefore, 

The system does not perform mechanical work.

∴ W = -PΔV = -P × (0) = 0

Let Q_V be the heat absorbed at constant volume. 

By first law of thermodynamics,

ΔU = Q + W 

∴ ΔU = Q_V.

(4) Adiabatic process : In this process, the system does not exchange heat, Q with its surroundings.

∴ Q = 0.

Since by first law of thermodynamics, 

ΔU = Q + W 

∴ ΔU = W_ad.

Hence, 

(i) the increase in internal energy ΔU is due to the work done on the system by surroundings. This results in increase in energy and temperature of the system.

(ii) if the work is done by the system on surroundings, like expansion, then there is a decrease in internal energy (-ΔU) and temperature of the system decreases.

(1) Intensive properties : 

Temperature, Density, Refractive index. Pressure, Viscosity.

(2) Extensive properties :

Enthalpy, Mass, Energy, Volume, Weight.

Extensive property: Mass

Intensive property: Temperature

The properties of a system are classified as,

(A) Extensive property and 

(B) Intesive property.

(A) Extensive property : It is defined as a property of a system whose magnitude depends on the amount of matter present in the system.

Explanation :

• More the quantity (or amount) of the matter of the system, more is the magnitude of extensive property, e.g., mass, volume, heat, energy, enthalpy, etc.
•  The extensive properties are additive.

(B) Intensive property : It is defined as a property of a system whose magnitude is independent of the amount of matter present in the system.

Explanation :

• Intensive property is characteristic of the system, e.g., refractive index, density, viscosity, temperature, pressure, boiling point, melting point, freezing point of a pure liquid, surface tension, etc.
•  The intensive properties are not additive. 

∴ Heat absorbed at constant volume

q_v = ∆U …(i)

Where ∆U Change in internal energy but most of chemical reactions are carried out not at constant volume, but in flask under constant atmospheric pressure. So, at constant pressure, e.g., (i) becomes

∆U = q_p − p∆V …(ii)

Where q_p = heat absorbed by the system

−p∆V = Expansion work done by the system

Or U₂ − U₁ = q_p − p (V₂ − V₁ )

On rearranging, we get

q_p = (U₂ + pV₂ ) − (U₁ − pV₁ ) …(iii)

Now, we know that enthalpy

H = U + pV …..(iv)

So, equation (iii) becomes,

q_p = H₂ − H₁ = ∆H

For finite changes at constant Pressure, equation

(iv) Can be written as:

∆H = ∆U + p∆V .....…(v)

Since P is constant, so equation (v) becomes

∆H = ∆U + p∆V ..............(vi)

The equation (vi) shows the relationship between ∆H and ∆U.

Extensive property : 

Moles, entropy, heat capacity.

Intensive property : 

Molar heat capacity.

Extensive Properties: Heat capacity and molar heat capacity (because these properties depend upon the quantity of matter) 

Intensive Properties: Density and temperature (Because these properties do not depend upon quantity of matter)

(1) The process is carried out at constant temperature.

(2) During the complete process, driving force is infinitesimally greater than opposing force.

(3) Throughout the process, the system exists in equilibrium with its surroundings.

(4) The work obtained is maximum. This is given by,

W_max = -2.303 nRT log₁₀\(\frac{V_2}{V_1}\)

OR

W_max = -2.303 nRT log₁₀\(\frac{P_1}{P_2}\)

Where n, P, V and T represent number of moles, pressure, volume and temperature respectively.

(5) ΔU = 0, ΔH = 0.

(6) The heat absorbed in reversible manner Q_rev is completely converted into work.

Q_rev = W_max.

Hence work obtained is maximum.

Given : 

Cross sectional area = A = 100 cm²

Displacement of a piston = 1 = 10 cm

External pressure = P = 1.0 bar

Work = W = ?

Volume change = A × l 

∴ ΔV = 100 × 10 

= 1000 cm³

= 1 dm³

Work = W = -P × ΔV

= -1 × 1

= -1 dm³ bar

= – 1 × 100 J

= -100 J

∴ Work = W = -100 J

(1) Spontaneous process : It is defined as a process that takes place on its own or without the intervention of the external agency or influence. For example, expansion of a gas or flow of a gas from higher pressure to low pressure or a flow of heat from higher temperature to lower temperature.

(2) Characteristics :

• It occurs on its own and doesn’t require external agency.
• It proceeds in one direction and can’t be completely reversed by external stimulant.
• These processes may be fast or slow. 
• These processes proceed until an equilibrium is reached. 

H₂ (g) + ½ O₂ (g) ---> H₂O (g), Δ_f H⁰ = -286kj mol^-1 

This means that when 1 mol of H₂O (l), is formed , 286 kj of heat is released. This heat is absorbed by the surroundings, i.e., q rev = +286 kj mol^-1 . 

So ΔS = qsurr /T =286KJmol^-1 /298K 

=0.9597kjK^-1 mol^-1 

= 959.7 JK^-1 mol

∆S = q _rev/T

As ΔG = ΔH – TΔS. Thus, ΔG = ΔH only when either the reaction is carried out at 0⁰ K or the reaction is not accompanied by any entropy change, i.e., ΔS = 0

This is a straight-forward application of Equation 2, followed by substitution of the Appropriate values from a table. 

 CaCO₃(s) → CaO (s) + CO₂ (g) 

ΔS° = (1 mol) [ΔS° for CaO (s)] + (1 mol) [ΔS° for CO2(g)] – (1 mol)[ΔS° for CaCO3(s)] 

= (1 mol) (39.8 J/mol K) + (1 mol) (213.7 J/mol K) – (1 mol) (92.9 J/mol K) = 160.6 J/K.

Open system :

1. An open system can exchange matter with the surroundings.

2. It can exchange energy with the surroundings.

3. In this, the total amount of energy does not remain constant.

4. Example : Hot water kept in an open beaker.

Isolated system :

1. An isolated system cannot exchange matter with the surroundings.

2. It cannot exchange energy with the surroundings.

3. In this, the total amount of energy remains constant.

4. Example : Hot water kept in a thermos flask.

(i) Thermodynamic process : It is defined as a transition by which a state of a system changes from initial equilibrium state to final equilibrium state.

The process is carried out by changing the state functions or thermodynamic variables like pressure, volume and temperature. During the process one or more properties of the system change.

(ii) Types of processes :

• Isothermal process 
• Isobaric process 
• Isochoric process 
• Adiabatic process 
• Reversible process 
• Irreversible (spontaneous) process. 

Thermodynamic equilibrium : A system is said to have attained a state of thermodynamic equilibrium if there is no change in any thermodynamic functions or state functions like energy, pressure, volume, etc. with time.

For a system to be in thermodynamic equilibrium, it has to attain following three types of equilibrium :

• Thermal equilibrium 
• Chemical equilibrium 
• Mechanical equilibrium 

Energy can neither be created nor destroyed. The energy of an isolated system is constant. ∆U=q+ w.

Correct answer is

(a) Crystallisation of liquid into solid

(i) Human body: open system

(ii) Milk in thermos flask: Isolated system

(iii) Tea in steel kettle: Closed system

(a) When a gas expands under vacuum i.e. no external pressure work on it (P_ex = 0), its expansion is called free expansion.

\(\because\) P_ex = 0

\(\therefore\) W = 0 in case of free expansion of an ideal gas [\(\because\) W= -P_ex (V₂ - V₁) = 0]

It means no work is done

(b) Given

n = 4 moles

P = 2 atm

T = 25 + 273 = 298 K

P_ex = 1 atm

\(\therefore\) W = -P_ex(V_f - V_i)

Now, initial (V_i) =\(\frac{nRT}{P}\)

= \(\frac{4\times0.082\times298}{2}\)

= 48.87 L

\(\because\) Volume becomes 2 times of its original volume.

\(\therefore\) Final volume (V_f = 48.87 × 2 = 97.74 L)

\(\therefore\) W = −1(97.74 − 48.87)

= −1(48.87)

= −48.87 J

For isothermal reversible expansion of ideal gas

W_rev = -2.303 nRT log\(\frac{V_f}{V_i}\)

= −2.303 × 4 × 8.314 × 298 log\(\big(\frac{97.74}{48.87}\big)\)

= −2.303 × 4 × 8.314 × 298 × 0.3010

= −6869.84 J

(i) Reactant, NaHCO₃ is a solid and it has low entropy. Among products, there are one solid and two gases. So, the products represent a condition of higher entropy.

(ii) H₂ (g) → 2H(g)

Here, one molecule of hydrogen gives two hydrogen atoms i.e. number of particles increases leading to more disordered state. Two moles of hydrogen atoms have higher entropy than one mole of dihydrogen molecule.

(a) Standard enthalpy of formation (∆_fH° ) is enthalpy change accompanying the formation of 1 mole of the substance from its constituent elements in their standard state. ∆_fH° can be > 0 or < 0.

(b) Given: N₂ (g) + 3H₂(g) → 2NH₃ (g);

= ∆_fH° = −92.4 kJ mol⁻¹

This is heat evolved for 2 moles of NH₃ (g)

\(\therefore\) Heat evolved for 1 mole of NH₃ (g)

= \(\frac{-92.4}{2}\) = -46.2 kJ

Hence ∆_fH° of NH₃ gas = - 46.2 kJ mol⁻¹

The properties that depend on the quantity of matter contained in the system are called extensive properties. For example, mass, volume etc. On the other hand, the properties which depend on the nature of the substance and not on the amount of substance are called intensive properties.

For example, viscosity etc.

(d) N₂O₄ (g) + 3CO(g) → N₂O (g) + 3CO₂ (g) ΔrH = ?

∆rH =\(\sum\)∆_fH(product) - \(\sum\)∆_fH(reactants)

= [∆_fH(N₂O) + 3 × ∆_fHCO₂ ] − [∆_fH(N₂O₄ ) + 3 × ∆_fCO]

= [81 + 3 × (−393)] − [9.7 + 3(−110)]

= −777.7 kJ/mol

(a) Reversible Process: It is defined as a system where change takes place infinitesimally slow and the direction of which at any point can be reversed by infinitesimal change in the state of the system.

(b) (i) For q = 0, heat is constant, so the process is adiabatic process.

(ii) For ∆U = 0, internal energy is constant which takes place at constant temperature, so, the process is isothermal process.

(iii) For ∆V = 0, volume is constant, so the process is isochoric process.

(iv) For ∆P = 0, pressure is constant, so the process is isobaric process.

(c) H₂O(g) → H₂ (g) + \(\frac{1}{2}\)O₂(g);

∆_rH^Θ = 286.2 kJ mol⁻¹

H₂ (g) + \(\frac{1}{2}\)O₂ (g) → H₂O(s);

∆_rH^Θ = −286.2 kJ mol⁻¹

It is referred to as first law of thermodynamics. According to this law, “energy can neither be created nor be destroyed but can be converted from one form to another"

(b) zero

The enthalpy of formation for all elements in their standard states is zero.

A function whose value is independent of path. eg. P, V, E, H.

In standard state enthalpies of all elements is zero. 

A process which can take place of its own or upon initiation under some condition. eg. Common salt dissolves in water of its own.  

A real crystal has some disorder due to the presence of defects in their structural arrangement, and Ideal crystal does not have any disorder. 

Real crystal has more entropy because it has more disorderness. 

Enthalpy of atomisation (∆_atomH) :

It is the enthalpy change accompanying the dissociation of one mole of gaseous substance into its atoms at constant temperature and pressure.

For example :

CH_4(g) → C_(g) + 4H_(g) ∆_atomH

= 1660 kJ mol^-1

Option : (a) Work

Option : (b) molar heat capacity

Option : (c) Energy

Path Function: Those properties which depend upon path followed.e.g, work. 

Sate Function: Those properties which depend upon the initial and final stages of the system and not upon the path followed. E.g, enthalpy.

2C(s) + 3H₂ (g) + \(\frac{1}{2}\)O2 (g) → C₂H₅OH(l)∆H_f⁰ =?

C(s) + 2H₂ (g) + 2H₂ (g) → CO₂ (g) + 2H₂O (l)

∆H = [2X(−286) + (−394.5)]

= −572 − 394.5

= −966.5 kJ/mol

2C(s) + 3H₂ (g) + \(\frac{1}{2}\)O₂ (g) → C₂H₂OH(l)

∆H = −966.5 − (−1380)

= −966.5 + 1380

= 413.5 kJ/mole

Given : 

Since the heat is absorbed by the system, the work is of expansion. 

Q = 520 J 

W= -210 J 

ΔV = ? 

ΔU = Q + W 

= 520 + (- 210) 

= 310 J

∴ Internal energy change = ΔU = 310 J

Given, K = 2

R = 8.314 JK⁻¹ mol⁻¹

T = 400 K

∆G° = ?

∵ ∆G° = −2.303 RT log K

= −2.303 × 8.314 × 400 × log 2

= −2.303 × 8.314 × 400 × 0.3010

= −2305.31 J mol⁻¹

= −2.05 kJ mol⁻¹

Internal energy : It is defined as the total energy constituting potential energy and kinetic energy of the molecules present in the system.

Explanation :

• The internal energy of a system is a state function and thermodynamic function. It is denoted by U.
• Its value depends on the state of a system.
• The change in internal energy (ΔU) depends only on the initial state and the final state of the system. Δ U = U – U
•  It is an extensive property of the system.
•  It has same unit as heat and work.
•  Total internal energy U of the system is, Total energy = Potential energy + Kinetic energy