4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 20 InterviewSolution

951.	Write down the expression for standard free energy change in terms of equilibrium constant. State meaning of the symbols used.
Answer» `DeltaG^(@) = - RT` ln K where `DeltaG^(@)` is standard free energy change, K is equilibrium constant, R is gas constant and T is temperature.

Discussion

952.	Molar heat capacity of aluminium will be `"…..............."` times its specific heat.
Answer» 27 (At massof`Al=27 ,i.e.,`1 mole of`Al = 27 g)`

Discussion

953.	If in a reaction at constant temperature, `n_(1)` moles of the gaseous reactants change into`n_(2)` moles of gaseous products , the difference in the enthalpy change and internal energy change of the reaction will be equal to `"…............."`.
Answer» `DeltaH- DeltaV = (n_(2)-n_(1))RT =Deltan_(g)RT`

Discussion

954.	The volume of a gas is reduced adibatically to `(1//4)` of its volume at `27^(@)C`. if `gamma = 1.4` . The new temperature will beA. `340xx4^(0.4)K`B. `300xx4^(0.4)K`C. `150xx4^(0.4)K`D. None of these
Answer» Correct Answer - B

Discussion

955.	State true or false. Wrk done by an ideal gas in aiabatic expansion is less than that in isothermal process (for same temperature range)
Answer» Correct Answer - 1

Discussion

956.	A system is provided 50 J of heat and work can be done on the system is 10 J. The change in internal energy during the process is:A. 40 JB. 60 JC. 80 JD. 50 J
Answer» Correct Answer - B

Discussion

957.	`DeltaH` for the reaction, `OF_(2)+H_(2)OrarrO_(2)+2HF` (B.E. of `O-F, O-H, H-F and O=O` are 44, 111, 135 and `"119 kcal mol"^(-1)` respectively)A. `-222" kcal"`B. `-"88 kcal"`C. `-"111 kcal"`D. `-"79 kcal"`
Answer» Correct Answer - D `DeltaH=SigmaB.E._(R)-SigmaB.E._(P)` `=(2xx44+2xx111)-(119+2xx135)=-"79 kcal"`

Discussion

958.	A gas expands slowely against a variable pressure given by `p=(10)/(V)` bar, where V is the volume of gas at each stage of expansion. During expansion from volume 10 L to 100 L the gas undergoes an increase in internal energy of 400 J. How much heat is absorbed by gas during expansion ?A. 1900 JB. 2300 JC. 2700 JD. 423 J
Answer» Correct Answer - C

Discussion

959.	What is the effect of decreasing of pressure on fusion and B.P. of CO2?
Answer» Both B.P. and fusion decrease with decrease in pressure.

Discussion

960.	How does the melting point of water changes with increase in pressure?
Answer» The B.P. of water increases with increase in pressure. While M.P. of ice decreases with increase in pressure.

Discussion

961.	Standard free energy change`(Delta G^(@))` of a reaction is related to its equilibrium constant as ` : K= 10^(x)` where ` x = "…................."`
Answer» `- (DeltaG^(@))/( 2.303RT) `

Discussion

962.	Write expression for the work done by 1 mole of the gas in each of the following cases `:` (i) For irreversible expansion of the gas from volume `V_(1)` to `V_(2)`. (ii) For reversible isothermal expansion of the gas from volume `V_(1)` to`V_(2)`. (iii)For expansion of the gas into an evaluated vessel. (iv) For reversible isothermal compression of the gas from pressure `P_(1)` to `P_(2)` (v)For adiabatic expansion resulting into change of temperature from `T_(1)` to `T_(2)`.
Answer» (i)Irreversible expansion takes place when external pressure `(P_(ext))` remains constant `w_("irrev")= - P_("ext") (V_(2)-V_(1))= -P_(ext) DeltaV` (ii) Reversible expansion takes place when internal pressure is infinitesimally greater than exterhanl pressure `(P_("int")~=P_(ext))` at every stage. Thsu, externalpressurehas to be adjusted throughout. `w_(rev)= -nRT ln. (V_(2))/(V_(1))` (iii) The expansion is irreversible. Further, as `P_(ext) =0`, therefore, `w= -P_(ext)DeltaV= 0 xx ( DeltaV) = 0` (iv) When gas is compressed, work is done on the gas. For isothermal reversible compression, `w= + nRT ln. (P_(2))/(P_(1))` (v) For adiabatic expansion, temperature falls, i.e.,`T_(2) ltT_(1))` `w=C_(v) (T_(2)-T_(1)),i.e., w` is -ve.

Discussion

963.	Justify the following statements `:` (a)Reactions with `DeltaG^(@) lt 0`always have an equilibrium constant greater than 1. (b)Many thermodynamically enthalpy feasible reactions do not occur under ordinary conditions. (c ) At low temperature , enthalpy change dominates the `DeltaG` expression and at high temperatures, it is entropy whch dominatest the value of `DeltaG`.
Answer» (a)Aready discussed already (b) Under ordinary conditions, the average energy of the reactants may be less than threshold energy. They requiresome activation energy to initiate the reaction. (c )`DeltaG = DeltaH - T DeltaS`. At low temperature, `TDeltaS` is small . Hence,`DeltaH ` dominates. At high temperatures,`T DeltaS` is large, ie., `DeltaS` dominates the value of `DeltaG`.

Discussion

964.	A reaction was nono-spontaneous at high temperature but became spontaneousat low temperature . The reaction is `"…............"`
Answer» Correct Answer - exothermic

Discussion

965.	Find out the value of equilibrium constant for the following reaction at 298K `2NH_(3)(g) + CO_(2)(g) hArr NH_(2)CONH_(2)(aq)+H_(2)O(l)` Standard Gibbs energy change, `Delta_(f)G^(@) ` at the given temperature is `-13.6kJmol^(-1)`
Answer» Correct Answer - `2.4 xx 10^(2)` `Delta_(r)G^(@) = - 2.303 RT log K, i.e., - 13600 = - 2.303 xx 8.314 xx 298 log K` or log K `= 2.38 :. K = ` antilog ` 2.38 xx 2.4 xx 10^(2)`

Discussion

966.	A reaction at 1 bar is non-spontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following:A. Both `DeltaH` and `DeltaS` are positiveB. `DeltaH` is negative while `DeltaS` is positiveC. `DeltaH` is positive while `DeltaS` is positiveD. Both `DeltaH` and `DeltaS` are negative
Answer» Correct Answer - A `DeltaG = DeltaH - T. DeltaS` If `DeltaH` & `Delta S` are both positive, then `DeltaG` may be negative at height temperature hence reaction becomes spontaneous at higher temperature.

Discussion

967.	Gvien the following value for the standard molar Gibbs free energy changes? `G^(@)`, at `25^(@)C` `Ag^(+) (aq) + 2NH_(3) (l) hArr Ag (NH_(3))_(2)^(+) (aq) ? G^(2) = -41.0KJ (mol Ag^(+))^(-1) Ag^(+) (aq) + Cl^(-)(aq) hArr AgCl (s) ? G^(@) = - 55.6 kJ (mol Ag^(+))^(-1)` Calculate the value for the thermodynamic equilibrium constant. K, at `25^(@)C` for the reactions: `AgCl (s) + 2NH_(3) (l) hArr Ag (NH_(3))_(2)^(+) (aq) + Cl (aq)`A. `3.62 xx 10^(2)`B. `1.00`C. `9.94 xx 10^(-1)`D. `2.76 xx 10^(-3)`
Answer» Correct Answer - D Reversing the 2nd equation and adding into 1 st equation gives `Ag Cl (s) + 2NH_(3) (l) hArr Ag (NH_(3))_(2)^(+) (aq) + Cl^(-) (aq)` `Delta G^(@) = 14.6 kJ` Also, `Delta G^(2) = 14.6 xx 100 J` `=-2.303 xx 298 xx 8.314` log `K rArr K = 2.76 xx 10^(-3)`

Discussion

968.	one mole of an ideal gas at 300k in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant presses of 3.0 atm. In this process. The change in entropy of surrroundings `(DeltaS)` in `J^(-1)` is (1 L atm = 101.3 J)A. `5.763`B. `1.013`C. `-1.013`D. `-5.763`
Answer» Correct Answer - C In Isothermal process, `DeltaU=0`, then `q_("irr")= - W_("irr")` `q_("irr")= - (-P_(ect) DeltaV)` `= 3` L atm `=3xx101.3 J=303.9 J` `DeltaS_(surr)=q_(surr)/T=(-q_(sys))/T=(-303.9 J)/(300 K)=-1.013 J K^(-1)` `C(s)+O_(2)(g) rarrCO_(2)(g), DeltaH=-393.5 kJ mol^(-1)` ...(i)

Discussion

969.	Consider the reaction : `2H_(2) (g) + O_(2) (g) hArr2H_(2)O (g)` When hyudrogen is ignited in the presence of oxygen, a lot of heat is released and water is produced. At room temperature, the `DeltaH_(r xx n)^(@)` is about `-484 k J//mol` and `Delta S_(r xx n)^(@)` is about -87 kJ/mol. Given this data how would you interpet this thermodynamic process ? Select the correct answer:A. The reaction is exothermic (heat is released) and it is becoming more orderedB. The reaction is endothermic (heat is absorbed) and it is becoming more disorderdeC. The reaction is exothermic (heat is released) and it is becoming more disorderedD. the reaction is endothermic (heat is absorbed) and it is becoming more ordered.
Answer» Correct Answer - A `Delta H^(@) = -484 kJ`/mol indicates exothermic nature of reaction while `Delta S^(@) = - 87 J//K (lt 0)` indicates that the system is becoming more ordered

Discussion

970.	Match column - I with Colimn - II
Answer» Correct Answer - A-Q, R, B-P,R, C-P, R, S, D-P,R,TU (A) Dissolution of glucose is endothermic ltbr. (B) Dissolution of quick lime is exothermic (C) All combustion reaction are exothermic (D) Graphite is the standard state of carbon

Discussion

971.	`H_(2)(g) + Cl_(2) rarr 2HCl (g), Delta H = -44K.cals 2 Na(s)+2HCl (g)rarr 2NaCl(s)+H_(2)(g), Delta H = -152K.cal` `Na(s)+(1)/(2)Cl_(2)(g) rarr NaCl(s), Delta H = ?`A. `+108K.cal`B. `-196 K.cal`C. `-98 K.cal`D. `50 K.cal`
Answer» Correct Answer - C `Delta H = H_(P-H_(R))`

Discussion

972.	If `S(g)+e^(-) rarr S^(1-)(g), Delta H = -207.6 KJ` `S(g) + 2e^(-) rarr S^(2-)(g), Delta H = +335.2 KJ`. The enthalpy for the reaction `S^(1-)(g)+e^(-) rarr S^(2-)(g)`:A. `+127.6 KJ`B. `-127.6 KJ`C. `+542.8 KJ`D. `-641.8 KJ`
Answer» Correct Answer - C `Delta H = H_(P-H_(R))`

Discussion

973.	The enthalpies of combustion of carbon and carbon monoxide are `-393.5` and `-283` kJ `mol^(-1)` respectively. The enthaly of formation of carbon monoxide per mole is :A. `676.5`B. `-676.5`C. `110.5`D. `110.5`
Answer» Correct Answer - C `CO(g)+1/2 O_(2) (g) rarr CO_(2) (g), DeltaH=-283.5 kJ mol^(-1)` ...(ii) Eq (i) - Eq (ii) gives `C(s)+1/2O_(2) (g) rarr CO(g), Delta_(f)H=-110.5 kJ mol^(-1)`

Discussion

974.	The enthalpies of combustion of carbon and carbon monoxide are `-393.5` and `-283` kJ `mol^(-1)` respectively. The enthaly of formation of carbon monoxide per mole is :A. `-676.5` KJB. `-110.5` KJC. 110.5 KJD. 676.5 KJ
Answer» Correct Answer - B `C_((s))+(1)/(2)O_(2)rarr CO_((g)) , Delta H_(f) = ?` `Delta H_(f)=Sigma (Delta H_("Comb."))_(R )-Sigma (Delta H_("Comb."))_(P)=(-393.5)-(-283)`

Discussion

975.	Calculate the standard heat of formation of carbon disulphide `(l)`. Given that the standard heats of combusion of carbon `(s)`, sulphur `(s)` and acrbon disulphide `(l)` are `-390, 290.0`, and `-1100.0kJ mol^(-1)`. Respectively.
Answer» Required equation is `C(s) +2S(s) rarr CS_(2)(l), Delta_(f)H^(Theta) = ? ` Given a. `C(s) + O_(2) (g) rarr CO_(2)(g), DeltaH^(Theta) =- 390.0kJ` b. `S(s) + O_(2) (g) rarr SO_(2)(g), DeltaH^(Theta) =- 290.00kJ` c. `CS_(2)(l) + 3O_(2)(g) rarr CO_(2)(g) + 2SO_(2)(g), DeltaH^(Theta) =- 1100.00kJ` Multiply the equation (b) by `2`. d. `2S(s) +2O_(2)(g) rarr 2SO_(2)(g), DeltaH =- 580.0kJ` Adding equation (a) and (d) and subtracting (c ), `[C(s) +2S(s) + 3O_(2)(g) -CS_(2)(l) -3O_(2)(g)` `rarr CO_(2)(g) + 2SO_(2)(g) - CO_(2)-2SO_(2)]` `C(s) +2S(s) rarr CS_(2)(l)` This is the required equation. Thus `Delta_(f)H^(Theta) =- 390.0- 580.0 + 1100.0 = 130.00 kJ` Standared heat of formation of `CS_(2)(l) = 130.00 kJ`

Discussion

976.	Calculate the standard heat of formation of carbon disulphide (l). Given that the standard heats of combustion of carbon (s), sulphur (s) and carbon disulphide (l) are -393.3, -293.72 and -1108.76 kJ `mol^(-1)` respectively.A. `-128.02 "KJ mole"^(-1)`B. `+12.802 "KJ mol"^(-1)`C. `+128.02 "KJ mol"^(-1)`D. `-12.802 KJ mol"^(-1)`
Answer» Correct Answer - C `C_((s))+2S_((s))rarr CS_(2(l)) , Delta H_(f) = ?` `Delta H_(f)=Sigma (Delta H_("Comb."))_(R )-Sigma (Delta H_("Comb."))_(P)` `Delta H_(f)=[1xx(-393.3)+2xx(-293.72)]-(-1108.76)` `Delta H_(f)=+128.02 kJ mol^(-1)`

Discussion

977.	For a reversible reaction at constant temperature and at constant pressure the equilibrium composition of reaction mixture corresponds to the lowest point on Gibbls energy Vs progress of reaction diagrams as shown. At equilibrium Gibbs energy of reaction is equal to zero. For a reaction `M_(2)O_((s))^(3//4) rarr 2 M_((S)) + (1)/(2) O_(2 (g)) Delta H = 30 KJ//mol` and `Delta S = 0.07 KJ//mol //K` at 1 atm. The reaction would not be spontaneous at temperatureA. `gt 428 K`B. `gt 428 K`C. `lt 100 K`D. `gt 100 K`
Answer» Correct Answer - B `Delta G^() = Delta H^(0) - T Delta S^(0) gt 0` for a non-spontaneous process

Discussion

978.	The standard heat of formation of carbon disulphide (l) given that standard heat of combustion of carbon (s), sulphur (s) and carbon disulphide (l) are -393.3, -293.72 and `-1108.76 KJ mol^(-1)` respectively isA. `-12.502 KJ. Mol^(-1)`B. `+128.02 KJ.mol^(-1)`C. `-128.02 KJ.mol^(-1)`D. `+12.802 KJ.mol^(-1)`
Answer» Correct Answer - B `C+S_(2) rarr CS_(2)`

Discussion

979.	For a reversible reaction at constant temperature and at constant pressure the equilibrium composition of reaction mixture corresponds to the lowest point on Gibbls energy Vs progress of reaction diagrams as shown. At equilibrium Gibbs energy of reaction is equal to zero. Which diagram represents the large value of equilibrium constant for the reversible reactionA. B. C. D.
Answer» Correct Answer - C `K_(eq)` is high `rArr` Reaction is almost completed towards products side

Discussion

980.	For a reversible reaction at constant temperature and at constant pressure the equilibrium composition of reaction mixture corresponds to the lowest point on Gibbls energy Vs progress of reaction diagrams as shown. At equilibrium Gibbs energy of reaction is equal to zero. The value of `"log"_(10) K_(eq)` is equal to [`k_(eq)` is the equilibrium coustant]A. `- (Delta G^(@))/(RT)`B. `(T Delta S^(@) - Delta H^(@))/(2.303 RT)`C. `(Delta H^(@) - T Delta S^(@))/(RT)`D. `(RT)/(T Delta S^(@) - Delta H^(@))`
Answer» Correct Answer - B `Delta G^(0) = Delta H^(0) - T Delta S^(0) = - 2.303 RT` log Keq

Discussion

981.	In order to decompose 9 grams of water 142.5 KJ heat is required. Hence the enthalpy of formation of water isA. `+142.5 KJ`B. `-142.5 KJ`C. `+285 KJ`D. `-285 KJ`
Answer» Correct Answer - D `9 gm rarr 142.5 KJ` `18 gm rarr` ?

Discussion

982.	Calculate the standard entropy change for the following reaction `: ` `H_(2)(g) + Cl_(2)(g) rarr 2HCl(g)` at 298 K Given `S_(H_(2))^(@) = 131J K^(@) mol^(_1), S_(Cl_(2))^(@) = 223 JK^(-1) mol^(-1)` and `S_(HCl)^(@) =187 JK^(-1) mol^(-1)`
Answer» Correct Answer - `20 JK^(-1) mol^(-1)`

Discussion

983.	Which of the following reaction defines `Delta H _(f)^(@)` ?A. `C_("diamond") + O_(2)(g) rarr CO_(2)(g)`B. `(1)/(2)H_(2)(g)+(1)/(2)F_(2)(g) rarr HF(g)`C. `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`D. `CO(g)+(1)/(2)O_(2)(g)rarr CO_(2)(g)`
Answer» Correct Answer - B

Discussion

984.	Thermodynamically, the most stable form of carbon isA. diamondB. graphiteC. fullerencesD. coal
Answer» Correct Answer - b

Discussion

985.	What is meant by toral entropy change of a process? Assuming the thermodynamic relationship ΔG = ΔH - TΔS, derive the relationship ΔG = TΔStotal for a system.
Answer» Total entropy change of a process if entropy change taking place in universe, i.e. ΔS_system + ΔS_surrounding = ΔS_Total ΔS_Total = ΔS_sys – q/T = ΔS_Sys– ΔH/T TΔS_Total = TAS_Sys – ΔH [∵ ΔG = ΔH – TΔS] TΔS_Total = -ΔG ΔG = -TΔS_Total

985.

What is meant by toral entropy change of a process? Assuming the thermodynamic relationship ΔG = ΔH - TΔS, derive the relationship ΔG = TΔStotal for a system.

Answer»

Total entropy change of a process if entropy change taking place in universe, i.e.

ΔS_system + ΔS_surrounding = ΔS_Total

ΔS_Total = ΔS_sys – q/T = ΔS_Sys– ΔH/T

TΔS_Total = TAS_Sys – ΔH [∵ ΔG = ΔH – TΔS]

TΔS_Total = -ΔG

ΔG = -TΔS_Total

Discussion

986.	Starting with the thermodynamics relationship G = H - TS, derive the following relationship ΔG = -TΔSTotal
Answer» At initial state of the system G_i = H_i - TS_i At final state of the system G_f = H_f - TS_f ∴ Change is free energy ΔG = G_f - G_i = (H_f - H_i) – T (S_t - S_i) ΔG = ΔH – TΔS

986.

Starting with the thermodynamics relationship G = H - TS, derive the following relationship ΔG = -TΔSTotal

Answer»

At initial state of the system G_i = H_i - TS_i

At final state of the system G_f = H_f - TS_f

∴ Change is free energy ΔG = G_f - G_i = (H_f - H_i) – T (S_t - S_i)

ΔG = ΔH – TΔS

Discussion

987.	Explain the help of example, the difference between bond dissociation energy and bond energy.
Answer» Bond disoociation energy is energy required to break 1 mole of bonds e.g. H - H(g) → 2H(g) ΔH = 436 kJ mol^-1 Bond energy is energy released when 1 mole of bonds are formed example 2H →H₂(g) ΔH = -436 kJ mol^-1 In diatomic molecule both bond dissociation energy and bond energy are equal in magnitude but opposite in sign.

987.

Explain the help of example, the difference between bond dissociation energy and bond energy.

Answer»

Bond disoociation energy is energy required to break 1 mole of bonds e.g.

H - H(g) → 2H(g) ΔH = 436 kJ mol^-1

Bond energy is energy released when 1 mole of bonds are formed example

2H →H₂(g) ΔH = -436 kJ mol^-1

In diatomic molecule both bond dissociation energy and bond energy are equal in magnitude but opposite in sign.

Discussion

988.	Why most of the exothermic process (reactions) spontaneous?
Answer» ΔG = ΔH - ΔH - TΔS; For exothermic reactions ΔH is -ve. For a spontaneous process AG is to be -ve. Thus decrease in enthalpy (-ΔH) contributes significantly to the driving force (to make AG negative).

988.

Why most of the exothermic process (reactions) spontaneous?

Answer»

ΔG = ΔH - ΔH - TΔS; For exothermic reactions ΔH is -ve. For a spontaneous process AG is to be -ve.

Thus decrease in enthalpy (-ΔH) contributes significantly to the driving force (to make AG negative).

Discussion

989.	Proceeding from the heat of formation of gaseous carbon dioxide. (ΔH°f = -393.5 kJ mol-1) and the thermochemical equation:C(graphite) + 2N2O(g) → CO2(g) + 2N2(g)ΔH° = -557.5 kJCalculate the heat of formation of N2O(g)
Answer» C_(graphite) + 2N₂O(g) → CO₂(g) + 2N₂(g) ΔH° = -557.5 kJ ΔH° = ΣΔH°_f(products) - ΣΔH°_freactants ΔH = ΣCO₂(g) + 2ΔH_fN₂ - 2ΔH_fN₂O(g) - ΔH°_(graphite) -557.5 kJ = -393.5 kJ + 2 x 0 - 2ΔH_fN₂O(g) = 0 2ΔH_fN₂O(g) = -393.5 kJ + 557.5 kJ ΔH_fN₂O(g) = \(\frac{164.0}{2}\)kJ = 82.0 kJ mol^-1

989.

Proceeding from the heat of formation of gaseous carbon dioxide. (ΔH°f = -393.5 kJ mol-1) and the thermochemical equation:C(graphite) + 2N2O(g) → CO2(g) + 2N2(g)ΔH° = -557.5 kJCalculate the heat of formation of N2O(g)

Answer»

C_(graphite) + 2N₂O(g) → CO₂(g) + 2N₂(g)

ΔH° = -557.5 kJ

ΔH° = ΣΔH°_f(products) - ΣΔH°_freactants

ΔH = ΣCO₂(g) + 2ΔH_fN₂ - 2ΔH_fN₂O(g) - ΔH°_(graphite)

-557.5 kJ = -393.5 kJ + 2 x 0 - 2ΔH_fN₂O(g) = 0

2ΔH_fN₂O(g) = -393.5 kJ + 557.5 kJ

ΔH_fN₂O(g) = \(\frac{164.0}{2}\)kJ = 82.0 kJ mol^-1

Discussion

990.	A certain volume of dry air at NTP is allowed to expand 4 times of its original volume under (a) isothermal conditions, (b) adiabatic conditions. Calculate the final pressure and temperature in each case if γ = 1.4.
Answer» Given : Suppose V₁ = V V₂ = 4V P₁ = 76 cm of Hg P₂ = ? γ = 1.4 T₁ = 273K T₂ = ? (a) For isothermal expansion : P₁V₁ = P₂V₂ or, P₂ = \(P_1\frac{V_1}{V_2}\) = 19 cm of Hg. As the process is isothermal, therefore the final temperature will be the same as the initial temperature, i.e., T₂ = 273 K (b) Adiabatic expansion : From the relation P₁V₁^γ = P₂V₂^γ ∴ P₂ = P₁\((\frac{V_1}{V_2})^γ\) \(= 76\times(\frac{1}{4})^{0.04}\) \(= 76\times(0.25)^{1.4}\) = 10.91 cm of Hg From the relation, T₁V₁^γ−1 = T₂V₂^γ−1 ⇒ T₂= T₁\((\frac{V_1}{V_2})^{γ-1}\) ⇒ T₂= 273\((\frac{1}{4})^{0.04}\) = \(\frac{273}{(4)^{0.04}}\) = 156.8 K

990.

A certain volume of dry air at NTP is allowed to expand 4 times of its original volume under (a) isothermal conditions, (b) adiabatic conditions. Calculate the final pressure and temperature in each case if γ = 1.4.

Answer»

Given :

Suppose V₁ = V

V₂ = 4V

P₁ = 76 cm of Hg

P₂ = ?

γ = 1.4

T₁ = 273K

T₂ = ?

(a) For isothermal expansion :

P₁V₁ = P₂V₂

or, P₂ = \(P_1\frac{V_1}{V_2}\)

= 19 cm of Hg.

As the process is isothermal, therefore the final temperature will be the same as the initial temperature,

i.e., T₂ = 273 K

(b) Adiabatic expansion : From the relation

P₁V₁^γ = P₂V₂^γ

∴ P₂ = P₁\((\frac{V_1}{V_2})^γ\)

\(= 76\times(\frac{1}{4})^{0.04}\)

\(= 76\times(0.25)^{1.4}\)

= 10.91 cm of Hg

From the relation,

T₁V₁^γ−1 = T₂V₂^γ−1

⇒ T₂= T₁\((\frac{V_1}{V_2})^{γ-1}\)

⇒ T₂= 273\((\frac{1}{4})^{0.04}\)

= \(\frac{273}{(4)^{0.04}}\)

= 156.8 K

Discussion

991.	A cylinder containing one gram molecule of the gas was compressed adiabaticaly untill its tempertaure rose from `27^(@)C` to `97^(@)C`. Calculate the work done and heat produced in the gas `(gamma=1.5)`.
Answer» Here, initial temperature, `T_(1)= 27^(@)C= 273+27= 300K` final temperature, `T_(2)= 97^(@)C= 273+97= 370K`. When a gas is compressed adiabatically, work done on gas is given by `W=R/((1-gamma))(T_(2)-T_(1))` `=(8.3xx(370-300))/(1-1.5)` `= -11.62xx10^(2)J` `:.` Heat produced, `H= (-W)/J= (11.62xx10^(2))/(4.2)= 276.7cal`.

Discussion

992.	A gram molecule of gas at `27^(@)C` expands isothermally untill its volume is doubled. Find the amount of work done and heat absorbed. Take `R=8.31J mol e^(-1) K^(-1)`.
Answer» Here, `T= 27^(@)C= 27+273= 300 K, V_(2)= 2V_(1)` `W=2.3026RT "log"_(10)(V_(2))/(V_(1)) 2.3026xx8.31xx300 log_(10) 2= 2.3026xx8.31xx300xx0.3010` `W=1.727xx10^(3) "joule"` Amount of heat absorbed `=(1.727xx10^(3))/(4.2)cal. 411.4 cals`

Discussion

993.	1 mole of gas expands isothermally at `37^(@)C`. The amount of heat is absorbed by it until its volume doubled is `(R= 8.31 J mol^(-1) K^(-1))`A. 411.25 calB. 418.50 calC. 420.25 calD. 425.40 cal
Answer» Correct Answer - D Here `T=37^(@)c=37+273=310 K V_(2)=2V` work done in siothermal process is `W=nRTln(V_(2))/(V_(1))=1xx8.31xx310xxln(2V_(1))/(V_(1))` `=8.31xx310xxln2=1.786xx10^(3)` Amount of heat absorbedc =`(1.786xx10^(3))/(4.2)` cal =425.4 cal

Discussion

994.	Three moles of an ideal gas at `127^(@)C` expands isothermally untill the volume is doubled. Calculate the amount of work done and heat absorbed.
Answer» Correct Answer - `6912 J, 6912 J` Here, `n=3, V_(2)= 2V_(1), dW= ?, dQ=?` As the process is isothermal, therefore, `dW= nRT "log"_(e)((V_(2))/(V_(1)))` `=3xx8.31(127+273)log_(e)2` `=3xx8.31xx400xx0.6931= 6912J` `dQ= dU+dW` `= 0+6912= 6912J`

Discussion

995.	Which of the following is true for ideal solution :-A. `Delta H_(("mix"))=0`B. `Delta S_(("mix"))=0`C. `Delta G_(("mix"))=0`D. `Delta A_(("mix"))=0`
Answer» Correct Answer - A

Discussion

996.	Based on first law of thermodynamics , which one of the following is correct ?A. For an isochoric proces`,DeltaU =-q`B. For an adiabatic process, `DeltaU = -w`C. For an isothermal process`,q= +w`D. For a cyclic process,`q= -w`
Answer» Correct Answer - D (a) `DeltaU = q+w`. For an isochoric process, `w = - P DeltaV =0`. Hence,`DeltaU = q_(v)` (b) For an adiabatic process , `q=0`. Hence, `DeltaU =w` (c ) For an isothermal process, `DeltaU =0` Hence, `q=-w` (d) For a cyclic process , `DeltaU =0`. Hence, `q= - w`

Discussion

997.	Enthalpies of formation of `CO(g), N_(2)O(g)` and `N_(2)O_(4)(g)` are -110, -393, 81 and `9.7 kJ mol^(-1)` respectively . Find the value of `Delta_(r)H` for the reaction`:` `N_(2)O_(4)(g)+3CO(g) rarr N_(2)O(g) +3CO_(2)(g)`
Answer» `Delta_(r)H = Sigma Delta_(f)H(`Products) `-Sigma Delta_(f)H(` Reactants) `= [ Delta _(f) H( N_(2)O) +3Delta_(f)H(CO_(2))]- [ Delta_(f)H(N_(2)O_(4)) +3Delta_(f) H(CO)]` `= [8] +3 ( -393)] - [9.7 +3( -110) ] kJ = -777.7 kJ`

Discussion

998.	The enthalpy of formation of `CO(g), CO_(2)(g),N_(2)O(g)` and `N_(2)O_(4)(g)` is `-110,-393,+81` and 10 kJ / mol respectively. For the reaction `N_(2)O_(4)(g)+3CO(g)rarr N_(2)O(g)+3CO_(2)(g). Delta H_(r )` isA. `-212`B. `+212`C. `+778`D. `-778`
Answer» Correct Answer - D

Discussion

999.	For adiabatic process which is correct -A. `Delta T = 0`B. `Delta S = 0`C. q = 0D. `q_(p) = 0`
Answer» Correct Answer - C

Discussion

1000.	Which of the following is not thermodynamic function -A. Internal energyB. work doneC. EnthalpyD. Entropy
Answer» Correct Answer - B

Discussion

Explore topic-wise InterviewSolutions in .

Write down the expression for standard free energy change in terms of equilibrium constant. State meaning of the symbols used.

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The volume of a gas is reduced adibatically to `(1//4)` of its volume at `27^(@)C`. if `gamma = 1.4` . The new temperature will beA. `340xx4^(0.4)K`B. `300xx4^(0.4)K`C. `150xx4^(0.4)K`D. None of these

State true or false. Wrk done by an ideal gas in aiabatic expansion is less than that in isothermal process (for same temperature range)

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Match column - I with Colimn - II

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The enthalpies of combustion of carbon and carbon monoxide are `-393.5` and `-283` kJ `mol^(-1)` respectively. The enthaly of formation of carbon monoxide per mole is :A. `-676.5` KJB. `-110.5` KJC. 110.5 KJD. 676.5 KJ

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Calculate the standard entropy change for the following reaction `: ` `H_(2)(g) + Cl_(2)(g) rarr 2HCl(g)` at 298 K Given `S_(H_(2))^(@) = 131J K^(@) mol^(_1), S_(Cl_(2))^(@) = 223 JK^(-1) mol^(-1)` and `S_(HCl)^(@) =187 JK^(-1) mol^(-1)`

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Starting with the thermodynamics relationship G = H - TS, derive the following relationship ΔG = -TΔSTotal

Explain the help of example, the difference between bond dissociation energy and bond energy.

Why most of the exothermic process (reactions) spontaneous?

Proceeding from the heat of formation of gaseous carbon dioxide. (ΔH°f = -393.5 kJ mol-1) and the thermochemical equation:C(graphite) + 2N2O(g) → CO2(g) + 2N2(g)ΔH° = -557.5 kJCalculate the heat of formation of N2O(g)

A certain volume of dry air at NTP is allowed to expand 4 times of its original volume under (a) isothermal conditions, (b) adiabatic conditions. Calculate the final pressure and temperature in each case if γ = 1.4.

A cylinder containing one gram molecule of the gas was compressed adiabaticaly untill its tempertaure rose from `27^(@)C` to `97^(@)C`. Calculate the work done and heat produced in the gas `(gamma=1.5)`.

A gram molecule of gas at `27^(@)C` expands isothermally untill its volume is doubled. Find the amount of work done and heat absorbed. Take `R=8.31J mol e^(-1) K^(-1)`.

1 mole of gas expands isothermally at `37^(@)C`. The amount of heat is absorbed by it until its volume doubled is `(R= 8.31 J mol^(-1) K^(-1))`A. 411.25 calB. 418.50 calC. 420.25 calD. 425.40 cal

Three moles of an ideal gas at `127^(@)C` expands isothermally untill the volume is doubled. Calculate the amount of work done and heat absorbed.

Which of the following is true for ideal solution :-A. `Delta H_(("mix"))=0`B. `Delta S_(("mix"))=0`C. `Delta G_(("mix"))=0`D. `Delta A_(("mix"))=0`

Based on first law of thermodynamics , which one of the following is correct ?A. For an isochoric proces`,DeltaU =-q`B. For an adiabatic process, `DeltaU = -w`C. For an isothermal process`,q= +w`D. For a cyclic process,`q= -w`

Enthalpies of formation of `CO(g), N_(2)O(g)` and `N_(2)O_(4)(g)` are -110, -393, 81 and `9.7 kJ mol^(-1)` respectively . Find the value of `Delta_(r)H` for the reaction`:` `N_(2)O_(4)(g)+3CO(g) rarr N_(2)O(g) +3CO_(2)(g)`

The enthalpy of formation of `CO(g), CO_(2)(g),N_(2)O(g)` and `N_(2)O_(4)(g)` is `-110,-393,+81` and 10 kJ / mol respectively. For the reaction `N_(2)O_(4)(g)+3CO(g)rarr N_(2)O(g)+3CO_(2)(g). Delta H_(r )` isA. `-212`B. `+212`C. `+778`D. `-778`

For adiabatic process which is correct -A. `Delta T = 0`B. `Delta S = 0`C. q = 0D. `q_(p) = 0`

Which of the following is not thermodynamic function -A. Internal energyB. work doneC. EnthalpyD. Entropy