4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 32 InterviewSolution

1551.	What is the aim of the study of chemical thermodynamics?
Answer» The main aim of the study of chemical thermodynamics is to learn Transformation of energy from one form into another form. Utilization of various forms of energies Change in the properties of system produced by chemical or physical effects.

1551.

What is the aim of the study of chemical thermodynamics?

Answer»

The main aim of the study of chemical thermodynamics is to learn

Transformation of energy from one form into another form.
Utilization of various forms of energies
Change in the properties of system produced by chemical or physical effects.

1552.	Force is `"…..........."` property whereas pressure is `"…......................."`property in thermodynamics .
Answer» extensive, intensive

Discussion

1553.	What is the Gibbss Helmhotz equation ?
Answer» `DeltaG = DeltaH-T DeltaS` where `DeltaG , DeltaH` and`DeltaS` are free energy change and entropy change respectively.

Discussion

1554.	Mathematically, the third law of thermodynamics is expressed as …(a) \(\lim\limits_{T\to 0} S=0\)(b) \(\lim\limits_{T\to 0} S=1\)(c) AU = q + w (d) G = H – TS
Answer» \((a)\,\,\lim\limits_{T\to 0} S=0\)

Discussion

1555.	Gibbs’s free energy is defined as ……(a) G = H + TS (b) G = H x TS (c) G = H – TS (d) G = H / TS
Answer» Answer: (c) G = H – TS

Discussion

1556.	When solid melts there is ……(a) an increase of entropy (b) a decrease in entropy (c) an increase in free energy (d) an increase of heat of fusion
Answer» (a) an increase of entropy

Discussion

1557.	Match the List-I and List-II using the correct code given below the list.List-IList - IIA. 2O3(g) → 3O21. Spontaneous of hight temperatureB. 2H2O(g) + O2(g) → 2H2O2(I)2. Spontaneous at all temperatureC. Solid → liquid3. Non-spontaneous of high temperatureD. Adsorption of gasesNon-spontaneous at all temperatureCode: (a) A - 4, B - 2, C - 1, D - 3(b) A - 2, B - 4, C - 1, D - 3(c) A - 1, B - 3, C - 2, D - 4(d) A - 3, B - 1, C - 4, D - 2
Answer» (b) A - 2, B - 4, C - 1, D - 3

Discussion

1558.	What are spontaneous reaction? Give three examples for spontaneous reaction.
Answer» Reaction that does occur under the given set of conditions is called a spontaneous reaction. Example: A waterfall runs downhill, but never up, spontaneously. Heat flows from hotter object to a colder one. Ageing process.

1558.

What are spontaneous reaction? Give three examples for spontaneous reaction.

Answer»

Reaction that does occur under the given set of conditions is called a spontaneous reaction.

Example:

A waterfall runs downhill, but never up, spontaneously.
Heat flows from hotter object to a colder one.
Ageing process.

Discussion

1559.	The unit of entropy is ……(a) J K-1 mol-1 (b) J mol-1 (c) J K mol-1 (d) J-1 K-1
Answer» (a) J K^-1 mol^-1

Discussion

1560.	If G = 0. then the process is ……(a) equilibrium (b) spontaneous(c) non-spontaneous (d) none of these
Answer» (a) equilibrium

Discussion

1561.	1 mole of an ideal gas is maintained at 4.1 atm and at a certain temperature absorbs 3710 J heat and expands to 2 litres. Calculate the entropy change in expansion process.
Answer» Pressure of an ideal gas P_i = 4.1 atm. Expansion in volume = ∆V = 2 litres Heat absorbed = q = 3710 J Entropy change = ∆S = ? For an ideal gas PV = RT for one mole. T = PV/R = 4.1x₂/0.830 = 100°C T = 100 + 273 = 373 K. ∆S = q/T_(K)= 3710/373 Entropy change = 9.946 JK^-1

1561.

1 mole of an ideal gas is maintained at 4.1 atm and at a certain temperature absorbs 3710 J heat and expands to 2 litres. Calculate the entropy change in expansion process.

Answer»

Pressure of an ideal gas P_i = 4.1 atm.

Expansion in volume = ∆V = 2 litres

Heat absorbed = q = 3710 J

Entropy change = ∆S = ?

For an ideal gas PV = RT for one mole.

T = PV/R = 4.1x₂/0.830 = 100°C

T = 100 + 273 = 373 K.

∆S = q/T_(K)= 3710/373

Entropy change = 9.946 JK^-1

Discussion

1562.	The standard conditions for G° are ……(a) 1 mm Hg / 25°C (b) 1 atm /25 K (c) 1 atm / 298 K (d) 1 atm / 0 K
Answer» (c) 1 atm / 298 K

Discussion

1563.	What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure?
Answer» It condenses to solid directly without passing through the liquid phase.

Discussion

1564.	If system moves from ordered state to disordered state, its entropy ……(a) decreases (b) increases (c) become zero (d) increases then decreases
Answer» (b) increases

Discussion

1565.	What do you understand by thermodynamic variables ?
Answer» The variables (Pressure, volume, temperature, internal energy and the number of moles) which are required to specify the state of thermodynamic system are called thermodynamic variables.

Discussion

1566.	What is the effect on pressure on change of state in a system in an isobaric process?
Answer» In an isobaric process, the pressure remains constant hence the change in pressure ∆P = 0.

Discussion

1567.	Explain the various processes of thermodynamics and work done in these processes.
Answer» Every equilibrium state of a thermodynamical system is completely described by specific values of some macroscopic variables. If in a thermodynamic system, variables like temperature (T), pressure (P), volume (V), etc. changes with time, then the process is called thermodynamical process. There are various types of thermodynamical processes : (i) Isothermal process (ii) Isochoric process (iii) Isobaric process (iv) Adiabatic process (v) Cyclic process (i) Isothermal process : An isothermal process is a change of a system, in which the temperature remains constant, ∆T = O.This typically occurs when a system is in contact with an outside thermal reservoir and the change takes place slowly to allow the system to continually adjust to the reservoir’s temperature through heat exchange. Isothermal process takes place in any type of system that regulates the temperature. In the thermodynamic analysis of chemical reactions, it is first analyzed what happens under isothermal conditions. In an isothermal process, the internal energy of an ideal gas is constant. This is due to the fact that there are no intermolecular forces in an ideal gas. In the isothermal compression of a gas, there is work done on the system to decrease the volume and increase the pressure. Doing work on the gas, can increase the internal energy and increase the temperature. For constant temperature, energy must leave the system. For an ideal gas, the amount of energy entering the environment is equal to the amount of work done on the gas, as the internal energy does not vary. (ii) Isochoric process : If in a physical change of thermodynamic system, there is change in pressure (P) and temperature but volume remains constant, then, it is called an isochoric process. So, no work is done by the gas in this process. No work is done on the gas because V is constant. Then, dV = 0 (iii) Isobaric process : If temperature (T) and volume (V) changes in a physical change in a thermodynamic system and pressure (P) is constant, the process is called isobaric i.e, when the state of the substance changes, then the pressure (P) is constant. For example, freezing of water, formation of steam, etc. (iv) Adiabatic process : Adiabatic process is that process in which changed, but there is no transfer of heat between a thermodynamical system and its surroundings. In this process, energy is transferred only as work. This process provides a rigorous conceptual basis to explain the first law of Thermodynamics. A process in which transfer of heat is not involved in the system, so that Q = 0, is called an adiabatic process. For example, the compression of a gas within an engine’s cylinder is assumed to take place so fast that on the time scale of the process of compression, little energy of the system can be transferred out as heat. Essential conditions for an adiabatic process : The process of expansion or compression should be sudden, so that heat does not get time to get exchanged with the surroundings. The walls of the container must be perfectly insulated so that there cannot be any exchange of heat between the gas and surroundings. (v) Cyclic process : The process of thermodynamics in which system undergoes various changes and comes back to the initial state, is called a cyclic process. In this process, dU = 0. Since, internal energy is a function of state. Now, according to the first law of Thermodynamics: dQ = dU + dW dQ = 0 + dW dQ = dW This means that the total absorbed heat in the system is equal to the work done by the system.

1567.

Explain the various processes of thermodynamics and work done in these processes.

Answer»

Every equilibrium state of a thermodynamical system is completely described by specific values of some macroscopic variables. If in a thermodynamic system, variables like temperature (T), pressure (P), volume (V), etc. changes with time, then the process is called thermodynamical process.

There are various types of thermodynamical processes :
(i) Isothermal process
(ii) Isochoric process
(iii) Isobaric process
(iv) Adiabatic process
(v) Cyclic process

(i) Isothermal process :
An isothermal process is a change of a system, in which the temperature remains constant, ∆T = O.This typically occurs when a system is in contact with an outside thermal reservoir and the change takes place slowly to allow the system to continually adjust to the reservoir’s temperature through heat exchange.

Isothermal process takes place in any type of system that regulates the temperature. In the thermodynamic analysis of chemical reactions, it is first analyzed what happens under isothermal conditions. In an isothermal process, the internal energy of an ideal gas is constant. This is due to the fact that there are no intermolecular forces in an ideal gas. In the isothermal compression of a gas, there is work done on the system to decrease the volume and increase the pressure. Doing work on the gas, can increase the internal energy and increase the temperature. For constant temperature, energy must leave the system. For an ideal gas, the amount of energy entering the environment is equal to the amount of work done on the gas, as the internal energy does not vary.

(ii) Isochoric process :
If in a physical change of thermodynamic system, there is change in pressure (P) and temperature but volume remains constant, then, it is called an isochoric process. So, no work is done by the gas in this process. No work is done on the gas because V is constant. Then, dV = 0

(iii) Isobaric process :
If temperature (T) and volume (V) changes in a physical change in a thermodynamic system and pressure (P) is constant, the process is called isobaric i.e, when the state of the substance changes, then the pressure (P) is constant. For example, freezing of water, formation of steam, etc.

(iv) Adiabatic process :
Adiabatic process is that process in which changed, but there is no transfer of heat between a thermodynamical system and its surroundings. In this process, energy is transferred only as work. This process provides a rigorous conceptual basis to explain the first law of Thermodynamics.

A process in which transfer of heat is not involved in the system, so that Q = 0, is called an adiabatic process. For example, the compression of a gas within an engine’s cylinder is assumed to take place so fast that on the time scale of the process of compression, little energy of the system can be transferred out as heat.

Essential conditions for an adiabatic process :

The process of expansion or compression should be sudden, so that heat does not get time to get exchanged with the surroundings.
The walls of the container must be perfectly insulated so that there cannot be any exchange of heat between the gas and surroundings.

(v) Cyclic process :
The process of thermodynamics in which system undergoes various changes and comes back to the initial state, is called a cyclic process. In this process, dU = 0. Since, internal energy is a function of state. Now, according to the first law of Thermodynamics:
dQ = dU + dW
dQ = 0 + dW
dQ = dW
This means that the total absorbed heat in the system is equal to the work done by the system.

Discussion

1568.	Explain heat, work and internal energy of a system.
Answer» Temperature determines the direction of flow of heat when two bodies are in thermal contact. Heat flows from the body at a higher temperature to a lower temperature. The flow stops when the two bodies are in thermal equilibrium. This type of heat transfer is a non-mechanical energy transfer. When thermal energy is transferred from an object into the atmosphere then the thermal energy of the object is taken to be negative while on giving heat to an object from the atmosphere the thermal energy of the object is taken to be positive. If the temperature of the object remains constant then due to the exchange of heat in the object the state of the object changes.

1568.

Explain heat, work and internal energy of a system.

Answer»

Temperature determines the direction of flow of heat when two bodies are in thermal contact. Heat flows from the body at a higher temperature to a lower temperature. The flow stops when the two bodies are in thermal equilibrium. This type of heat transfer is a non-mechanical energy transfer.

When thermal energy is transferred from an object into the atmosphere then the thermal energy of the object is taken to be negative while on giving heat to an object from the atmosphere the thermal energy of the object is taken to be positive. If the temperature of the object remains constant then due to the exchange of heat in the object the state of the object changes.

Discussion

1569.	1 mole of `CO_(2)` gas as at 300 K is expanded under reversible adiabatic condition such that its volume becomes 27 times.
Answer» Correct Answer - `T_(2)= 100 k ; w =- 5.016 kJ` `T_(1)V_(1)^(gamma-1) = T_(2)V_(2)^(gamma-1)` `T_(2) = 100 K` `W= nC_(V) (T_(2)-T_(1)) =- 5016 J`

Discussion

1570.	The ideal gas equation for an adiabatic process isA. `PV^(gamma)`=constantB. `TV^(gamma+1)`=constantC. `P^(gamma-1)`=constantD. `P^(gamma+1)`T=constant
Answer» Correct Answer - A

Discussion

1571.	Which one of the following graphs represents variation of specific heat capacity of water with temperature?A. B. C. D.
Answer» Correct Answer - A

Discussion

1572.	A monoatomic ideal gas of two moles is taken through a cyclic process starting from A as shown `(V_(B))/(V_(A))=2` and `(V_(D))/(V_(A))=4` Temperature `T_(A)` at `A` is `27^(@)C`. Q. Work done during complete cyclic processA. 1200RB. 1500RC. 1400RD. 1000R
Answer» Correct Answer - B,D

Discussion

1573.	Calculate the enthalpy of combustion of ethylene at 300 K at constant pressure, If Its heat of combustion at constant volume (∆U) is – 1406 KJ.
Answer» The complete ethylene combustion reaction can be written as, C₂H_4(g) + 3O_2(g) → 2CO_2(g) + 2H₂O_(I) ∆U = -1406 kJ ∆n = n_p(g) – n_r(g) ∆n = 2 – 4 = -2 ∆H = U + RT ∆n_(g) ∆H = -1406 + (8.314 x 10^-3 x 300 x (-2)) ∆H = – 1410.9 kJ

1573.

Calculate the enthalpy of combustion of ethylene at 300 K at constant pressure, If Its heat of combustion at constant volume (∆U) is – 1406 KJ.

Answer»

The complete ethylene combustion reaction can be written as,

C₂H_4(g) + 3O_2(g) → 2CO_2(g) + 2H₂O_(I)

∆U = -1406 kJ

∆n = n_p(g) – n_r(g)

∆n = 2 – 4 = -2

∆H = U + RT ∆n_(g)

∆H = -1406 + (8.314 x 10^-3 x 300 x (-2))

∆H = – 1410.9 kJ

Discussion

1574.	Chemical reactions are invariably associated with the transfter of energy either in the form of heat or light. In the laboratory, heat changes in physical and chemical processes are measured with an instrument called calorimeter. Heat change in the process is calculated as {:(q = ms DeltaT,,s =Specific heat),(=cDeltaT,,c =Heat capacity):}` Heat of reaction at constant volume is measured using bomb calorimeter. `q_(V) = DeltaU =` Internal energy change Heat of reaction at constant pressure is measured using simple or water calorimeter. `q_(p) = DeltaH` `q_(p) = q_(V) +P DeltaV` `DeltaH = DeltaU +DeltanRT` For which reaction will `DeltaH = DeltaU`? Assume each reaction is carried out in an open container.A. `4CO(g) +2O(g) rarr 4CO_(2)(g)`B. `H_(2)(g) +Br_(2)(g) rarr 2HBr(g)`C. `C(s)+2H_(2)O(g) rarr 2H_(2)(g)+CO_(2)(g)`D. `2PCI_(5)(g) rarr 2PCI_(3)(g)+2CI_(2)(g)`
Answer» `DeltaU = nC_(V) dT` moles `(n) = (0.1g)/(16) = 6.25 xx 10^(-3)` `C_(V) = 500 J K^(-1)` `T = 273 +2 = 275 K` `:. DeltaU =- 160kJ`

Discussion

1575.	`He,N_(2)`, and `O_(3)` are expanded adiabatically and their expansion curves between `P` and `V` are plotted under silimar conditions. About the ratio of the slope, which one is not correct?A. The ratio of slopes of `P-V` curves for `He` and `O_(3) is 1.25`.B. The ratio of slopes of `P-V` curves for `He` and `N_(2) is 1.20`.C. The ratio of slopes of `P-V` curves for `N_(2)` and `O_(3) is 1.05`.D. The slope of `He` is least steeper and for `O_(3)` is most steeper.
Answer» Ratio of slope `=(gamma "for " gas I)/(gamma "for " gas II)`, slope for `He, N_(2)` and `O_(3)` are `7//5, 5//4`, and `4//3`, respectively.

Discussion

1576.	In any cyclic process,A. `sumW=0`B. `sumQ=0`C. `sumQ_(+ve) gt sumW`D. `sumDeltaU=0`
Answer» Correct Answer - C::D

Discussion

1577.	A monatomic gas is expanded adiabatically and due to expansion volume becomes 8 times, thenA. Pressure becomes `(1)/(32)` timesB. Total K.E. of gas molecules becomes `(1)/(4)` timesC. Ratio of translational and roational kinetic energy changes with variation in temperatureD. None of These
Answer» Correct Answer - A::B

Discussion

1578.	A thermally insulated vessel containing diatomic gas of molar mass M is moving with velocity v. the temperature of gas is T. if it is suddenly stopped, thenA. Internal energy of gas is equal to `(5nRT)/(2)` when vessel is movingB. There is no change in temperature if vessel is suddenly stoppedC. There is rise in temperature if vessel is suddenly stoppedD. Temperature increase is due to heat input
Answer» Correct Answer - A::C

Discussion

1579.	`A + B rarr C+D, Delta H = -10,000 J mol^(-1) , Delta S = - 33.3 J mol^(-1) K^(-1)` (i) At what temperature the reaction will occur spontaneously from left to right ? (ii) At what temperature , the reaction will reverse?
Answer» `Delta G = DeltaH -T DeltaS` At equilibrium , `Delta G =0` so that `Delta H = T Delta S` or `T = (DeltaH)/(DeltaS) = (- 10000J mol^(-1))/( - 33.3JK^(-1)mol^(-1))= 300.3K` (i) For spontaneity from left to right, `Delta G` should be -ve for the given reaction. This willbe so if `T lt 300.3K` ( becasue `Delta H` and `Delta S` both are -ve ) (ii) For reverse reaction to occur, `Delta G` should be `+` be for forward reaction. This will be so if `T gt 300.3` K.

Discussion

1580.	At `0^(@)C`, ice and water in equilibrium and `Delta H =6.00 kJ mol^(-1)` for the process`H_(2)O (s) rarr H_(2)O(l)`. What will be `Delta S` and `Delta G` for the conversino of ice to liquid water ?
Answer» Since the given process inequilibrium , `Delta G = 0` Putting this value in the relationship , `Delta G = DeltaH - T DeltaS ` weget `0= Delta H-T Delta S` or `T DeltaS = DeltaH ` or `Delta S = (DeltaH)/(T)` We are given `DeltaH = 6.0 kJ mol^(-1) = 6000 J mol^(-1) `and `T = 0^(@) C = 273 K` `:. Delta S = ( 6000 J mol^(-1))/( 273K ) = 21.98 J K^(-1) mol^(-1)`

Discussion

1581.	`50` students sitting in the room of `5 xx 10 xx 3m^(3)` dimensions. The air inside the room is at `27^(@)C` and `1atm` pressure. If each student loses `100` watt heat per second assuming the walls, ceiling floor, and all the material present inside the room is perfectly insulated as well as neglecting loss of air to the outside as the temperature is raised, how much rise in temperature will be noticed in `10min`? Given `C_(P) = (7)/(2)R` for air.
Answer» Volume of the room `5 xx 10 xx3= 150 m^(3) = 15 xx 10^(4)L` Moles of air in the room at `27^(@)C` and `1atmP`, `n = (PV)/(RT) = (1xx15 xx 10^(4))/(0.0821xx300) = 6.1 xx 10^(3)` Heat produced per second by `50` students `= 100 xx 50` `= 5000 W s^(-1)` Heat produced in `10min = 5000 xx 10 xx 60W` `= 3 xx 10^(6) W or J s^(-1)` `:.` Change in enthalpy of air, `DeltaH = n xx C_(P) xx DeltaT` `3 xx 10^(6) = 6.1 xx 10^(3) xx (7)/(2) xx 8.314 xx DeltaT rArr DeltaT = 16.90` Thus, a rise of `16.90^(@)C` will be noticed in air tempertaure.

Discussion

1582.	The combustion of 1.22 g benzoic acid (M=122) in a bomb calorimeter at 300 K caused a temperature rise of 3 K, While combustion of 0.88 g ethyl ethanote (M=88) caused a temperature rise of 2 K. Calculate the enthalapy change of combustion of `CH_(3)COOC_(2)H_(5)(l)` at 300 K. Given : Internal energy change of combustion for benzoic acid =-300kJ`mol^(-1)` at 300 K and R=8.3` JK^(-1)mol^(-1)`A. `-2000 kJ//mol`B. `-2002.49 kJ//mol`C. `-2006.5 kJ//mol`D. `-3002.5 kJ//mol`
Answer» Correct Answer - B

Discussion

1583.	For a reaction : `3A(g)+B(g)to2C(g)+4D(l)` `Delta U^(@)=50kcal//"mole" and DeltaS^(@)=-400cal//"mol-K".` Calculate `DeltaG^(@) at 200K:`A. 129.2kcalB. 130kcalC. 130.8kcalD. `-30kcal`
Answer» Correct Answer - A

Discussion

1584.	Which of the following expression expressions is true for an ideal gas?A. `((delV)/(delT))_(p)=0`B. `((delP)/(delT))_(V)=0`C. `((delU)/(delP))_(T)=0`D. `((delU)/(delT))_(T)=0`
Answer» Correct Answer - C

Discussion

1585.	The efficiencty of a carnot engine working between `127^(@) C and 77^(@) C` isA. 0.105B. 0.115C. 0.268D. 0.135
Answer» Correct Answer - C Here ice point `T_(2)=0^(@)C=0+273 =273 K` and steam point `T_(1)=100^(@)C=100+273=373 K` Efficiency of the carnot engine `eta=1-(T_(2))/(T_(1))=1-(273)/(373)=(100)/(373)xx100%=26.8%`

Discussion

1586.	The coefficient of performance of refrigerator, whose efficiencty is 25% isA. 1B. 3C. 5D. 7
Answer» Correct Answer - B The efficiency of refrigerator is `eta`=25% =0.25 `therefore` coeffiecient of performance `beta=(1-eta)/(eta)=(1-0.25)/(0.25)=(0.75)/(0.75)=3`

Discussion

1587.	If the coefficient of performance of a refrigerator is 5 and operates at the room temperature (`27^(@)` C), find the temperature inside the refrigerator.A. 240 KB. 250 KC. 230 KD. 260 K
Answer» Correct Answer - B Here coefficient of performance `beta`=5 `t_(1)=27^(@)C=(27+273)K=300K` As `beta =(T_(2))/(T_(1)-T_(2)), rArr 5 =(T_(2))/(300-T_(2))` or `1500-5T_(2)=t_(2)or 6T_(2)=1500` `therefore T_(2)=(1500)/(6)=250 K`

Discussion

1588.	If the co-efficient of performance of a refrigerator is 5 and operates at the room temperature `(27^(@)C)`, find the temperature indide the refrigerator.
Answer» Here, `COP = beta = 5, T_(1) = 270^(@)C = (27+273)K = 300K , T_(2) = ?` As `beta= (T_(2))/(T_(1)-T_(2)) :. 5= (T_(2))/(300-T_(2)) , T_(2)= 250K = (250-273)^(@)C = -23^(@)C`

Discussion

1589.	What is entropy?
Answer» A measure of degree of disorder of a system.

Discussion

1590.	What is the unit of entropy?
Answer» J/K (Joule per Kelvin)

Discussion

1591.	Give an example of a reaction in which AH = AE.
Answer» C(s) + O₂(g) → CO₂(g)

Discussion

1592.	What is Gibbs-Helmholtz equation?
Answer» ΔG° = ΔH° – TΔS°.

Discussion

1593.	The standard enthalpies of formation of Hydrogen fluoride and Hydrogen chloride are -268.6 kJ and -92.3 kJ respectively. Which is between the two is stable?
Answer» Hydrogen fluoride is more stable.

Discussion

1594.	Give an expression for the work done in a reversible isothermal expansion of an ideal gas.
Answer» w = -2.303 nRT log V₂/V₁

Discussion

1595.	Calculate the mean bond enthalpy of C-H bond in methane.
Answer» CH₄ (g) → C (g) + 4H(g), ΔH = ? Mean bond enthalpy of C-H bond = \(\frac{1}{4}\) x Heat energy required to break four C-H bonds in CH₄ (g) = \(\frac{1}{4}\) x ΔH The ΔH for the reaction, CH₄ (g) → C (g) + 4H (g) is the enthalpy of atomisation of methane ΔH = ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ = (425 + 470 + 416 + 336) kJ mol^-1 = 1647 kJ mol^-1 Mean bond enthalpy of C-H bond in CH₄ (g) = \(\frac{1}{4}\) x ΔH = \(\frac{1}{4}\) x 1647 kJ/mol = 411.7 kJ mol^-1

1595.

Calculate the mean bond enthalpy of C-H bond in methane.

Answer»

CH₄ (g) → C (g) + 4H(g), ΔH = ?

Mean bond enthalpy of C-H bond

= \(\frac{1}{4}\) x Heat energy required to break four C-H bonds in CH₄ (g)

= \(\frac{1}{4}\) x ΔH

The ΔH for the reaction, CH₄ (g) → C (g) + 4H (g) is the enthalpy of atomisation of methane

ΔH = ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄

= (425 + 470 + 416 + 336) kJ mol^-1

= 1647 kJ mol^-1

Mean bond enthalpy of C-H bond in CH₄ (g)

= \(\frac{1}{4}\) x ΔH = \(\frac{1}{4}\) x 1647 kJ/mol

= 411.7 kJ mol^-1

Discussion

1596.	What is the value of the standard enthalpy of formation of an element?
Answer» The value of the standard enthalpy of formation of an element is Zero.

Discussion

1597.	Given that `,` `A(s) rarr A(l)DeltaH=x` `A(l) rarr A(g), DeltaH=y` The heat of sublimation of `A` will be `:`A. x - yB. x + yC. x or yD. `-x + y`
Answer» Correct Answer - B

Discussion

1598.	Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A isA. `2gamma ^(-1)`B. `(1)/(2)^(gamma-1)`C. `(1)/(1-gamma)^(2)`D. `(1)/(gamma-1)^(2)`
Answer» Correct Answer - A The gas in container A is comprssed isothermally, `therefore P_(1)V_(1)=P_(2)V_(2)` ltbr. Or `P_(2)=(P_(1)V_(1))/(V_(2))=P_(1)(V_(1))/(V_(1//2))=2P_(1)` Again the gas in container B is compressed adiabatically, `therefore P_(1)V_(1)^(gamma)=P_(2)(V_(2))^(gamma)` `P_(2)=P_(1)(V_(1)^(gamma))/((V_(2))^(gamma))=P_(1)((V_(1)/(V_(1)//(2))=2^(gammaP_(1)` Hence `(P_(2))/(P_(2))=(2^(gammaP_(1))/(2P_(1))=2^(gamma-1)`

Discussion

1599.	Making use of the Maxwell distribution function, find `lt lt 1//v gt gt`, the mean value of the reciprocal of the velocity of molecules in an ideal at a temperature `T`, it the mass of eacd molecule is equal to `m`. Compare the value obtained with the reciprocal of the mean velocity.
Answer» `lt lt (1)/(v) gt gt = int_0^oo ((m)/(2 pi kT))^(3//2) e^((-mv^2)/(2 kT)) 4 pi v^2 dv (1)/(v)` =`((m)/(2 pi kT))^(3//2) 4 pi (1)/(2) (2 kT)/(m) int_0^oo e^-x dx` =`2((m)/(2 pi kT))^(1//2) = ((2 m)/(pi kT))^(1//2) = ((16)/(pi^2) (m pi)/(8 kT))^(1//2) = (4)/(pi lt v gt )`.

Discussion

1600.	Using the Maxwell distribution function, calculate the mean velocity projection `(v_x)` and the mean value of the modules of this projection `lt lt \|v_x\| gt gt` if the mass of each molecule is equal to `m` and the gas temperature is `T`.
Answer» `lt v_x gt = 0` bt symmetry `lt \|v_x\| gt = int_(- oo)^oo \|v_x\| e-(mv_x^2)/(2 kT) dv_x// int_0^oo (-mv_x^2)/(e 2kT) dv_x = int_0^oo v_x (-mv_x^2)/(e 2 kT) dv_x//int_0^oo (-mv^2)/(e 2 kT)dv_x` =`sqrt((2kT)/(m)) int_0^oo ue^(-u^2) du//int_0^oo e^(-u^2) du` =`sqrt((2kT)/(m)) int_0^oo (1)/(2) e^(-x) dx//int_0^oo e^(-x) (dx)/(2 sqrt(x))` =`sqrt((2 kT)/(m)) Gamma (1) //Gamma((1)/(2)) = sqrt((2 kT)/(m pi))`.

Discussion

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Making use of the Maxwell distribution function, find `lt lt 1//v gt gt`, the mean value of the reciprocal of the velocity of molecules in an ideal at a temperature `T`, it the mass of eacd molecule is equal to `m`. Compare the value obtained with the reciprocal of the mean velocity.

Using the Maxwell distribution function, calculate the mean velocity projection `(v_x)` and the mean value of the modules of this projection `lt lt |v_x| gt gt` if the mass of each molecule is equal to `m` and the gas temperature is `T`.