4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 12 InterviewSolution

551.	Pick the extensive property from the given optionsA. ViscosityB. Refractive indexC. DensityD. Heat capacity
Answer» Correct Answer - D

Discussion

552.	If a process is both endothermic and spontaneous thenA. `Delta S gt 0`B. `Delta S lt 0`C. `Delta H lt 0`D. `Delta G gt 0`
Answer» Correct Answer - A

Discussion

553.	The standard entropies of `N_(2)(g),H_(2)(g)` and `NH_(3)(g)` are `191.5, 130.5,192.6JK^(-1)mol^(-1)` . The value of `DeltaS^(0)` during the formation of 1 mole of ammonia isA. `-98.9 JK^(-1) mol^(-1)`B. ZeroC. `+129.4 JK^(-1) mol^(-1)`D. `-29.4 JK^(-1) mol^(-1)`
Answer» Correct Answer - A

Discussion

554.	The specific heat of gas is found to be 0.075 calories at constant volume and its formula weight is 40. The atomicity of the gas would beA. OneB. TwoC. ThreeD. Four
Answer» Correct Answer - A

Discussion

555.	Specific heat of an elementary gas is found to be 0.313 J at constant volume. If the molar mass of the gas is 40 g `mol^(-1)` , what is the atomicity of the gas ?
Answer» Correct Answer - monoatomic `C_(v) = 0.313 xx 40 = 12.52J mol^(-1) , C_(p) = C_(v)+R = 12.52 + 8.31= 20.83J mol^(-1)` `:. C_(p) //C_(v) = 20.83 // 12.52 = 1.66`

Discussion

556.	A monotomic ideal gas of two metal is taken through a cyclic process straining from A as shown `V_(B)//V_(A)=2 and V_(D)//V_(A)=4` Temperature `T_(A) is 27^(@)C` The work done during process `B to C` (approx) is A. 1000RB. 800RC. 832RD. 945R
Answer» Correct Answer - C

Discussion

557.	Calculate the heat of reaction of the following reaction :CO2(g) + H2(g) → CO(g) + H2O(g) Given that the ΔfH°CO(g) = -110.5kJ, ΔfH°CO2(g) = -393.8kJ, ΔfH°H2O(g) = -241.8kJ respectively.
Answer» The required equation is CO₂(g) + H₂(g) → CO(g) + H₂O(g) ΔH – Σ ΔH_f(products) — Σ ΔH_f(reactants) = ΔH_f²CO(g) + ΔH_f²H₂O(g) – ΔH_f²CO₂(g) - ΔH_f²H₂(g) = -110.5 kJ - 241.8 kJ - (-398.3 kJ) - 0 = -352.3 kJ + 393.8 kJ = 41.5 kJ.

557.

Calculate the heat of reaction of the following reaction :CO2(g) + H2(g) → CO(g) + H2O(g) Given that the ΔfH°CO(g) = -110.5kJ, ΔfH°CO2(g) = -393.8kJ, ΔfH°H2O(g) = -241.8kJ respectively.

Answer»

The required equation is CO₂(g) + H₂(g) → CO(g) + H₂O(g)

ΔH – Σ ΔH_f(products) — Σ ΔH_f(reactants)

= ΔH_f²CO(g) + ΔH_f²H₂O(g) – ΔH_f²CO₂(g) - ΔH_f²H₂(g)

= -110.5 kJ - 241.8 kJ - (-398.3 kJ) - 0 = -352.3 kJ + 393.8 kJ = 41.5 kJ.

Discussion

558.	When a bomb calorimeter is used to determine the heat of recation is most likely of the system under investigation is most likely to remain constant ?A. number of moleculesB. pressureC. temperatureD. volume
Answer» Correct Answer - d

Discussion

559.	One mole of an ideal monoatomic gas at 1 atm, 300 K undergo a process in which pressure of gas is 8 atm then calculate W (in cal).
Answer» Correct Answer - 1200

Discussion

560.	Which of the following is /are correct ?A. `DeltaH=DeltaU+Delta(PV)` when P and V both changesB. `DeltaH=DeltaU+PDeltaV` when pressure is constantC. `DeltaH=DeltaU+VDeltaP` when volume is constantD. `DeltaH=DeltaV+PDeltaV+VDeltaP` when nP and V both changes
Answer» Correct Answer - a,b,c

Discussion

561.	One mole of an ideal monoatomic gas undergo process from the state A to state C as : `underset(300 k)underset(1 atm)"State"Aunderset(process)overset("isochoric")tounderset(600K)"state"Bunderset("process")overset("Adiabatic reversible")tounderset(64 atm)"State" C` Choose the correct option (s).A. Pressure of gas at state B in 2 atmB. `DeltaH_(AB)=900 cal`C. `DeltaU_(AC)=6300 cal`D. `W_(BC)=5400 cal`
Answer» Correct Answer - a,c,d

Discussion

562.	Use the information in the table to calculate the enthalpy of this reaction . `C_(2)H_(6)(g)+(7)/(2)O_(2)(g)to2CO_(2)(g)+3H_(2)O(l)` `{:(Reaction ,DeltaH_(f)^(@)KJ.mol^(-1)),(2C(s)+3H_(2)(g)toC_(2)H_(6)(g),-84.7),(2C(s)+O_(2)(g)toCO_(2)(g),-393.5),(H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),-285.8):}`A. `-764KJ`B. `-1560KJ`C. `-1664KJ`D. `-3120KJ`
Answer» Correct Answer - b

Discussion

563.	An ideal gas expanded adiabaticlly such that , T `prop V^(-1//2)` then :A. `gamma` of gas will be `1.5`B. `VpropP^(2//3)`C. `Pprop T^(3)`D. `P prop T^(2)`
Answer» Correct Answer - a,b,c

Discussion

564.	1 mole of an ideal monatomic gas is subjected toi the following reversible charge of state `Aunderset("Expansion")overset("Isothermal")toBunderset("Cooling") overset("Isochoric")to` (5 atm , 500K) `Cunderset("compression")overset("Adiabatic")toA` (300K) Then , which of the following are correct? [ Given `(0.6)^(2.5) =0.3 , "in" 2=0.7`] `(R=0.08 L-atm//"mol"-K=2cal//"mol"-k)`A. Pressure of point B is 2.5 atmB. Volume at point C is 16 litreC. Magnitude of work involved in complete process is 100 calD. Change in enthalpy of process of process `C-A` is `+1` kcal .
Answer» Correct Answer - a,b,c,d

Discussion

565.	Consider this reaction . `2N_(2)H_(4)(l)to3N_(2)(g)+4H_(2)O(g)` `DeltaH=-1078Kj` how much energy is released by this reaction during the formation of 140 g of `N_(2)(g)`?A. 1078KJB. 1797KJC. 3234KJD. 5390KJ
Answer» Correct Answer - b

Discussion

566.	Enthalpy of neutralisation of NH4OH with HC is 51.45 kJ. Calculate the enthalpy of ionization of ammonium hydroxide.
Answer» Enthalpy of ionzators of NH₄OH = -51 - 41 - (57.3) = 5.8500 J

Discussion

567.	One mole of an ideal gas is heated at constant pressure from `0^(@)C` to `100^(@)C` (A) Calculate work done. (B) If the gas were expanded isothermally and reversibly at `0^(@)C` from 1 atm to some other pressure `P_(1)`, water must be the final pressure if the maximum work is equal to the work involned in (a)?
Answer» Correct Answer - molar internal energy change =37.5 kj/mol, molar enthaply of vaporization =40.6kj/mol Solution (a) work involed in heating of gas `W_(a) = - p.Delta V = - P (V_(2) - V_(1)) - 198.7` `= -2.303 xx1xx9.87xx 273 log_(10). (1)/(P_(1)) :. P_(1) = 0.694 atm`

Discussion

568.	Which of the following will involve evolution of heat ?A. Dissolution of conc. `H_(2)SO_(4)` in waterB. Formation of NO in atmosphereC. Conversion of molecular hydrogen to atomic hydrogenD. Formation of water gas from coal and steam
Answer» Correct Answer - A Dissolution of sulphuric acid in water is exothermic

Discussion

569.	The bond dissociation energies for`Cl_(2),I_(2)` and ICIare 242.3,151.0 and211.3 `kJ // ` mole respectively . The enthalpy of sublimation of iodine is62.8 kJ `//` mole.What is the standard enthalpy of formation ofICI(g) nearly equal to ?A. `- 211.3 kJ //`moleB. `- 14.6kJ //`moleC. 16.8 kJ`//` moleD. `33.5 kJ//` mole
Answer» Correct Answer - C We aim `: (1)/(2) I_(2)(s) + (1)/(2) Cl_(2) (g) rarr ICI (g) ` We are given `:` (i) `Cl_(2)(g)rarr 2Cl(g), DeltaH=242.3 kJ` (ii) `I_(2)(g) rarr 2I(g),DeltaH = 151 kJ` (iii) `ICI (g) rarr (i) + Cl(g), DeltaH =211.3 kJ` (iv) `I_(2)(s) rarrI_(2)(g), DeltaH =62.8 kJ` `(1)/(2) (ii) (1)/(2) (i)- (iii) +(1)/(2) ` gives the required equation with `DeltaH =(1)/(2) (151) +(1)/(2) (292.3) -211.3 + (1)/(2) (62.8) = 16.75kJ`

Discussion

570.	Water is boiled under a pressure of `1.0 atm`. When an electric current of `0.50A` from a `12V` supply is passed for `300`s through a resistance in thermal contact with it, it is found that `0.789 g` of water is vaporied. Calculate the molar internal enegry and enthalpy chnages at boiling point `(373.15K)`.
Answer» Solution : The vaporization occurs at constant pressure therefore the enthalpy change is equal to the work done by the heater, `Delta H = 0.50 xx 12 xx 300 = 1800 J = + 1.8 kJ` `:.` Molar enthalpy of vaporization `Delta H = (Delta H)/("mole of " H_(2)O) = (Delta H)/(n_(H_(2))O)` `= (1.8)/(((0.798)/(18))) = 40.6 kJ mol^(-1)` `= Delta U + PDelta V = Delta U + Delta n_(g) RT = Delta U + RT` also. `( :. H_(2)o(l) hArr H_(2)O (g) , Delta n_(g) = 1)` `Delta U =` molar int ernal enrgy change `= Delta H - RT = 40.6 -8.314 xx 10^(-3) xx 373.15 = 37.5 kJ mol^(-1)`

Discussion

571.	The sublimation energy of `I_(2)` (solid) is 57.3 KJ/mole and enthalpy of fusion is 15.5 KJ/mole. The enthalpy of vapourisation of `I_(2)` isA. 41.8 KJ/moleB. `-41.8 KJ//"mole"`C. `72.8 KJ//"mole"`D. `-72.8 KJ//"mole"`
Answer» Correct Answer - A `I_(2)(S) rarr I_(2)(l) rarr I_(2)(g)` `Delta H_(SE) = Delta H_("Fusion") + Delta H_("Eva")`

Discussion

572.	The sublimation energy of `I_(2)(s)` is 57.3 kJ `//` mol and the enthalpy of fusion is 15.5 kJ `//` mol. The enthalpy of vaporisation of `I_(2)` isA. 41.8 kJ `//` molB. `-41.8 kJ //` molC. 72.8 kJ`//` molD. `-72.8 kJ // `mol
Answer» Correct Answer - a `DeltaH_("sub")= DeltaH_("fusion") + DeltaH _(vap)`

Discussion

573.	Will entropy increase or decrease in the following changes and give reasons for your answer ? (i) Sugar dissolves in water (ii) Water freezes to form ice.
Answer» Correct Answer - (i) increase(ii) decrease

Discussion

574.	For vaporization of water at 1 atmospheric pressure the values of `DeltaH` and `DeltaS` are `40.63KJmol^(-1)` and `108JK^(-1)mol^(-1)` , respectively. The temperature when Gibbs energy change `(DeltaG)` for this transformation will be zero isA. 393.4 KB. 373.4 KC. 293.4 KD. 273.4 K
Answer» Correct Answer - B

Discussion

575.	Thermodynamics mainly deals with:A. interrelation of various from of energy and their transformation from one form to another .B. energy change in the processes which depend only on initinal and final states of the microscophi system containing a few moleculas .C. how and what rate these energy transformations are carried out.D. the system in equilibrium state or moving from one equlildrium state to another equilibrium state.
Answer» Correct Answer - a,d

Discussion

576.	The enthalpy change change for which reaction represents the standard enthalpy of formation for hydrogen cyanide, HCN?A. `H(g)+C_("graphite")+N(g)to HCN(g)`B. `(1)/(2)H_(2)(g)+C_("graphite")(1)/(2)N_(2)(g) to HCN(g)`C. `HCN(g)to (1)/(2)H_(2)(g)+C_("graphite")+(1)/(2)N_(2)(g)`D. `H_(2)(g)+2C_("graphite")+N_(2)(g)to2HCn(g)`
Answer» Correct Answer - b

Discussion

577.	The value of `K_(p)` for the water gas reaction, `CO +H_(2)O hArr CO_(2) +H_(2)`is `1.06 xx 10^(5)` at `25^(@)C` . Calculate the standard free energy change for the reaction at `25^(@)C`.
Answer» Correct Answer - `-28.38 kJ mol^(-1)` `Delta_(r)G^(@) =- 2.303 rt LOG k = - 2.303 xx 8.314 xx 298 xx log ( 1.06 xx 10^(5))`

Discussion

578.	Calculate the equilibrium constant for the following reaction at 298 K and 1 atm pressure `:` `NO(g) + (1)/(2) O_(2)(g) hArr NO_(2)(g)` Given `: Delta_(f) H^(@) `at 298K For `NO(g) = 90.4 kJ mol^(-1), ` For `NO_(2)(g) = 33.8 kJ mol^(-1)` `Delta S^(@) ` at 298 K for the reaction ` = - 70.8 J K^(-1) mol^(-1)` Gas constant, `R= 8.31 JK^(-1) mol^(-1)`
Answer» Correct Answer - `1.679xx 10^(6)` `Delta_(r)H^(@) = Delta_(f) H^(@) ( NO_(2)) -[Delta_(f) H^(@) (NO)+(1)/(2) Delta_(f) H^(@) (O_(2))]= 33.8 -[90.4 + 0] = - 56.6kJ mol^(-1)` `= - 56600 J mol^(-1) - 298 K ( - 70.8 JK^(-1) mol^(-1)) = - 35501.6 J mol^(-1)` `:. - 35501.6 = - 2.303 xx 8.31 xx298 log K ` or `log K = 6.2250` or `K = 1.679 xx 10^(6)`

Discussion

579.	Define heat of enthalpy of a solution.
Answer» Heat of a solution is the change in enthalpy produced when one mole of the solute is dissolved in excess of a solvent so that further dilution does not produce any heat exchange.

Discussion

580.	Write the thermochemical equation for the combustion of glucose.
Answer» C₆H₁₂O_6(s) + 6O_2(g)→ 6CO_2(g) + 6H₂O_(l) ΔH = -xkJ

Discussion

581.	A gas fired power plant burns methane, `CH_(4)(g)` for which standard enthalpy of combustion is -890 kJ `mol^(-1)`. How many moles of `CO_(2)` gas are produced for every `1.0 ` MJ ( mega joule) of energy produced by it ?
Answer» Correct Answer - `1.12mol` `1 MJ = 10^(6) J`

Discussion

582.	The heat evolved in the combustion of methane is given by the equation `:` `CH_(4)(s) + 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l), DeltaH = - 890 .3 kJ mol^(-1)` (a ) How manygrams of methane would be required to produce 445.15 kJ of heat of combustion ? (b ) How many grams of carbon dioxide would beformed when 445.15 kJ of heat is evolved ? (c ) What volume of oxygen at STP would be used in the combustion process (a ) or (b) ?
Answer» (a) From the given equation , 890.3 kJ of heat is produced from 1 mole of`CH_(4), i.e., 12+4 = 16 g ` of `CH_(4)` (b ) `:. 445.15kJ` of heat is produced from 8g of `CH_(4)` (b) From the given equation, when 890.3 kJ of heat is evolved , `CO_(2)` formed `=1` mole `=44 g ` `:. `When 445.15 kJ of heat is evolved , `CO_(2)` formed `= 22g` (c ) From the equation , `O_(2)` used in the production of 890.3 kJ of heat `=2` moles `= 2 xx 22.4 ` litres at STP `= 44.8` litres at STP Hene, `O_(2)` used in the production of 445.1 5 kJ of heat = 2 22.4 litres at STP.

Discussion

583.	Explain the Heat of combustion of methane is -890 kJ.
Answer» Heat of combustion of methane is -890 kJ means when one mole of methane (CH₄) i.e., 16 grams of methane is completely burnt in oxygen, 890 kJ of heat is liberated.

Discussion

584.	Write the thermochemical equation for the reaction involving the burning of hydrogen gas in excess of air forming water and liberating heat.
Answer» 2H_2(g) + O_2(g)→ 2H₂O_(l); ΔH = -xkJ

Discussion

585.	Chloroform is passed from methane according to the reaction `:` `CH_(4)(g) + 3Cl_(2)(g) rarr CHCl_(3)(l) + 3HCl(g)` Calculate `Delta H ` for the reaction given that enthalpies of formation of HCl(g), `CH_(4)(g) ` and `CHCl_(3)(g)` are - 92.0 ,-74.9 and -134.3kJ per mole respectively.
Answer» Correct Answer - `-335.4kJ mol^(-1)` `Delta_(r) H^(@) = [ Delta_(r) H^(@)(CHCl_(3)) + 3Delta_(f)H^(@) (HCl) ]- [ Delta_(f)H^(@)(CH_(4)) + 3Delta_(f) H^(@) (Cl_(2))]` `[ ( - 134.3) + 3 (-92.0 ) ] -[( - 74.9 ) + 0]` ` = - 335.4kJ mol^(-1)`

Discussion

586.	The standard heat of formation of Fe2O2 (s) is 824.2 kJ mol-1. Calculate heat change for the reaction4Fe (s) + 3O2 (g) → 2Fe2O2 (s)
Answer» ΔH° = ΣΔH°_f(products) - ΣΔH°_f(reactants) = [2 x ΔH°_fFe₂O₃] - [4ΔH°_fFe(s) + 3ΔH°_fO₂(g)] = 2 (-824.2 kJ) - [4 x zero + 3 x zero] = -1648.4 kJ

586.

The standard heat of formation of Fe2O2 (s) is 824.2 kJ mol-1. Calculate heat change for the reaction4Fe (s) + 3O2 (g) → 2Fe2O2 (s)

Answer»

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

= [2 x ΔH°_fFe₂O₃] - [4ΔH°_fFe(s) + 3ΔH°_fO₂(g)]

= 2 (-824.2 kJ) - [4 x zero + 3 x zero]

= -1648.4 kJ

Discussion

587.	The molar heat of formation of `NH_(4)NO_(3)(s)` is `-360.0kJ` and those of `N_(2)O(g)`and `H_(2)O(l)` are `+80.00kJ` and `-285.00kJ`, respectively, at `25^(@)C` and `1atm`. Calculate `DeltaH` and `DeltaU` for the reaction. `NH_(4)NO_(3)(s) rarr N_(2)O(g) +2H_(2)O(l)`
Answer» `DeltaH^(Theta) = DeltaH = Delta_(f)H^(Theta) ("products") - Delta_(f)H^(Theta) ("reactants")` `=[Delta_(f)H_((N_(2)O))^(Theta)+2xxDelta_(f)H_((H_(2)O))^(Theta)-[Delta_(f)H_((NH_(4)NO_(3)))^(Theta)]` `= 80.00 + 2 xx (-280.00) - (-360.0)` `= 80.0 - 560.0 + 360.0 =- 120.0kJ` We know that `DeltaH = DeltaU + DeltanRT` or `DeltaU = DeltaH - DeltanRT` `Deltan = 1,R = 8.314 xx 10^(-3) kJ mol^(-1) K^(-1), T = 298K` `DeltaU =- 120.0 -(1) (8.314xx10^(-3)) (298)` `=- 120.0 - 2.477 =- 122.477 kJ`

Discussion

588.	The standard enthalpy of formation of `Fe_(2)O_(3)(s)` is - 824.2 kJ `mol^(-1)` . Calculate the enthalpy change for the reaction `: 4 Fe (s) + 3O_(2) (g) rarr 2Fe_(2) O_(3) (s)`
Answer» Correct Answer - `-1648.4kJ` `Delta_(r)H^(@) = [ 2 xx Delta_(f) H^(@) ( Fe_(2)O_(3))] - [ Delta_(f)H^(@) ( Fe)+ 3Delta_(f) H^(@) (O_(2))]` `= [2(-824.2) ] -[0+0]` `= - 1648.4kJ`

Discussion

589.	What is open system? Give example.
Answer» An open system is one in which there is an exchange of both matter and energy with its surroundings. Example : Water in an open beaker.

589.

What is open system? Give example.

Answer»

An open system is one in which there is an exchange of both matter and energy with its surroundings.

Example : Water in an open beaker.

Discussion

590.	The enthalpy of combustion of ethyl alcohol is -1360 kJ/ mol. Write the thermochemical equation of the reaction.
Answer» C₂H₅OH_(l) + 3O_2(g) → 2CO_2(g) + 3H₂O_(l) ΔH = -1360 kJ/mole.

Discussion

591.	For the equilibriumPCl5 (g) == PCl3 (g) + Cl2 (g)at 298 K, K = 1.8 x 10-7. What is ΔG° for the reaction?
Answer» ΔG° = -2.303 RT log K = -2.303 x 8.31 JK^-1 mol^-1 x 298 K x log (1.8 x 10^-7) = -2.303 x 8.31 JK^-1 mol^-1 x 298 K x (0.0334 - 7) = -2.303 x 8.31 x 298 K x (-6.9664) J mol^-1 = + 39730 J mol^-1 = + 39.73 kJ mol^-1

591.

For the equilibriumPCl5 (g) == PCl3 (g) + Cl2 (g)at 298 K, K = 1.8 x 10-7. What is ΔG° for the reaction?

Answer»

ΔG° = -2.303 RT log K

= -2.303 x 8.31 JK^-1 mol^-1 x 298 K x log (1.8 x 10^-7)

= -2.303 x 8.31 JK^-1 mol^-1 x 298 K x (0.0334 - 7)

= -2.303 x 8.31 x 298 K x (-6.9664) J mol^-1

= + 39730 J mol^-1

= + 39.73 kJ mol^-1

Discussion

592.	The enthalpy of combustion of sulphur is 297 kJ. Write the thermochemical equation for combustion of sulphur. What is the value of ΔfH of SO2?
Answer» S(s) + O₂(g) → SO₂(g); ΔH = -297kJ and Δ_fH of SO₂ = -297 kJ mol^-1.

Discussion

593.	One mole of an ideal monoatomaic gas is taken from `A` to `C` along the path `ABC` . The temperature of the gas at `A` is `T_(0)` . For the process `ABC` : A. work done by the gas is `RT_(0)`B. change in internal energy of the gas is `(11)/(2) RT_(0)`C. heat absorbed by the gas is `(11)/(2) RT_(0)` D. heat sbsorbed by the gas is `(13)/(2) R T_(0)` . (`R`=uniyersal gas constant)
Answer» Correct Answer - A::C Process `AB:P` =constant `:. Vprop P V_(B)=2V_(A)` `T_(B)=2T_(A)=2T_(0) DeltaW_(AB)=P_(0)V_(0)=RT_(0) DeltaQ_(AB)=C_(p)DeltaT=((5)/(3)R)(2T_(0)-T_(0))=(5)/(2)RT_(0)` `:. DeltaU_(AB)=DeltaQ_(AB)-DeltaW_(AB)=(3)/(2)RT_(0)` . Process `BC:V`=constant `:. P propT , P_(C)=2P_(B)` `:. T_(C)=2T_(B)=4T_(0) , DeltaW_(BC)=0` `:. DeltaQ_(BC)=DeltaU_(BC)=C_(v)DeltaT` `=((3)/(2)R)(rT_(0)-2T_(0))=3RT_(0)` `:. DeltaW_("net") =RT_(0),DeltaQ_("net")=(11)/(2)RT_(0)` and `DeltaU_("net")=(9)/(2)RT_(0)`.

Discussion

594.	What is a refrigerator?
Answer» A refrigerator is a device that uses work to transfer energy in the form of heat from a cold reservoir to a hot reservoir as it continuously repeats a thermodynamic cycle. Thus, it is a heat engine that runs in the backward direction.

Discussion

595.	Write the Mayer’s relation.
Answer» Mayer’s relation ⇒ C_p – C_v = R where R = universal gas constant C_p = molar specific heat at constant pressure C_v = molar specific heat at constant volume

Discussion

596.	What is Mayer’s relation?
Answer» Mayer’s relation is C_p − C_v = R.

Discussion

597.	What are the causes of irreversibility?
Answer» 1. Some processes such as a free expansion of a gas or an explosive chemical reaction or burning of a fuel take the system to non-equilibrium states. 2. Most processes involve dissipative forces such as friction and viscosity (internal friction in fluids). These forces can be minimized, but cannot be eliminated.

597.

What are the causes of irreversibility?

Answer»

1. Some processes such as a free expansion of a gas or an explosive chemical reaction or burning of a fuel take the system to non-equilibrium states.

2. Most processes involve dissipative forces such as friction and viscosity (internal friction in fluids). These forces can be minimized, but cannot be eliminated.

Discussion

598.	Draw a diagram to illustrate that the work done by a system depends on the process even when the initial and final states are the same.Different ways to change a system
Answer» In the above diagram, the initial state of a gas is characterized by (P_i, V_i) [corresponding to point A] and the final state of the gas is characterized by (P_f, V_f) [corresponding to point B]. Path 1 corresponds to constant temperature. Path 2 corresponds to the combination AC [P constant] + CB [V constant]. Path 3 corresponds to the combination AD [V constant] + DB [P constant]. The work done by the gas (W) is the area under the curve and is different in each case.

Discussion

599.	What is a quasistatic process ?
Answer» A quasistatic process is an idealised process which occurs infinitely slowly such that at all times the system is infinitesimally close to a state of thermodynamic equilibrium. Although the conditions for such a process can never be rigorously satisfied in practice, any real process which does not involve large accelerations or large temperature gradients is a reasonable approximation to a quasistatic process.

Discussion

600.	Draw P-V diagram for positive work at constant pressure.Positive work of constant pressure
Answer» In this case, during the expansion, the work done by the gas, W = \(\int^{V_2}_{V_1}\)pdV = P(V₂ - V₁) is positive as V₂ > V₁.

Discussion

Explore topic-wise InterviewSolutions in .

Pick the extensive property from the given optionsA. ViscosityB. Refractive indexC. DensityD. Heat capacity

If a process is both endothermic and spontaneous thenA. `Delta S gt 0`B. `Delta S lt 0`C. `Delta H lt 0`D. `Delta G gt 0`

The standard entropies of `N_(2)(g),H_(2)(g)` and `NH_(3)(g)` are `191.5, 130.5,192.6JK^(-1)mol^(-1)` . The value of `DeltaS^(0)` during the formation of 1 mole of ammonia isA. `-98.9 JK^(-1) mol^(-1)`B. ZeroC. `+129.4 JK^(-1) mol^(-1)`D. `-29.4 JK^(-1) mol^(-1)`

The specific heat of gas is found to be 0.075 calories at constant volume and its formula weight is 40. The atomicity of the gas would beA. OneB. TwoC. ThreeD. Four

Specific heat of an elementary gas is found to be 0.313 J at constant volume. If the molar mass of the gas is 40 g `mol^(-1)` , what is the atomicity of the gas ?

A monotomic ideal gas of two metal is taken through a cyclic process straining from A as shown `V_(B)//V_(A)=2 and V_(D)//V_(A)=4` Temperature `T_(A) is 27^(@)C` The work done during process `B to C` (approx) is A. 1000RB. 800RC. 832RD. 945R

Calculate the heat of reaction of the following reaction :CO2(g) + H2(g) → CO(g) + H2O(g) Given that the ΔfH°CO(g) = -110.5kJ, ΔfH°CO2(g) = -393.8kJ, ΔfH°H2O(g) = -241.8kJ respectively.

When a bomb calorimeter is used to determine the heat of recation is most likely of the system under investigation is most likely to remain constant ?A. number of moleculesB. pressureC. temperatureD. volume

One mole of an ideal monoatomic gas at 1 atm, 300 K undergo a process in which pressure of gas is 8 atm then calculate W (in cal).

Which of the following is /are correct ?A. `DeltaH=DeltaU+Delta(PV)` when P and V both changesB. `DeltaH=DeltaU+PDeltaV` when pressure is constantC. `DeltaH=DeltaU+VDeltaP` when volume is constantD. `DeltaH=DeltaV+PDeltaV+VDeltaP` when nP and V both changes

An ideal gas expanded adiabaticlly such that , T `prop V^(-1//2)` then :A. `gamma` of gas will be `1.5`B. `VpropP^(2//3)`C. `Pprop T^(3)`D. `P prop T^(2)`

Consider this reaction . `2N_(2)H_(4)(l)to3N_(2)(g)+4H_(2)O(g)` `DeltaH=-1078Kj` how much energy is released by this reaction during the formation of 140 g of `N_(2)(g)`?A. 1078KJB. 1797KJC. 3234KJD. 5390KJ

Enthalpy of neutralisation of NH4OH with HC is 51.45 kJ. Calculate the enthalpy of ionization of ammonium hydroxide.

Which of the following will involve evolution of heat ?A. Dissolution of conc. `H_(2)SO_(4)` in waterB. Formation of NO in atmosphereC. Conversion of molecular hydrogen to atomic hydrogenD. Formation of water gas from coal and steam

The sublimation energy of `I_(2)` (solid) is 57.3 KJ/mole and enthalpy of fusion is 15.5 KJ/mole. The enthalpy of vapourisation of `I_(2)` isA. 41.8 KJ/moleB. `-41.8 KJ//"mole"`C. `72.8 KJ//"mole"`D. `-72.8 KJ//"mole"`

The sublimation energy of `I_(2)(s)` is 57.3 kJ `//` mol and the enthalpy of fusion is 15.5 kJ `//` mol. The enthalpy of vaporisation of `I_(2)` isA. 41.8 kJ `//` molB. `-41.8 kJ //` molC. 72.8 kJ`//` molD. `-72.8 kJ // `mol

Will entropy increase or decrease in the following changes and give reasons for your answer ? (i) Sugar dissolves in water (ii) Water freezes to form ice.

For vaporization of water at 1 atmospheric pressure the values of `DeltaH` and `DeltaS` are `40.63KJmol^(-1)` and `108JK^(-1)mol^(-1)` , respectively. The temperature when Gibbs energy change `(DeltaG)` for this transformation will be zero isA. 393.4 KB. 373.4 KC. 293.4 KD. 273.4 K

The value of `K_(p)` for the water gas reaction, `CO +H_(2)O hArr CO_(2) +H_(2)`is `1.06 xx 10^(5)` at `25^(@)C` . Calculate the standard free energy change for the reaction at `25^(@)C`.

Define heat of enthalpy of a solution.

Write the thermochemical equation for the combustion of glucose.

A gas fired power plant burns methane, `CH_(4)(g)` for which standard enthalpy of combustion is -890 kJ `mol^(-1)`. How many moles of `CO_(2)` gas are produced for every `1.0 ` MJ ( mega joule) of energy produced by it ?

Explain the Heat of combustion of methane is -890 kJ.

Write the thermochemical equation for the reaction involving the burning of hydrogen gas in excess of air forming water and liberating heat.

Chloroform is passed from methane according to the reaction `:` `CH_(4)(g) + 3Cl_(2)(g) rarr CHCl_(3)(l) + 3HCl(g)` Calculate `Delta H ` for the reaction given that enthalpies of formation of HCl(g), `CH_(4)(g) ` and `CHCl_(3)(g)` are - 92.0 ,-74.9 and -134.3kJ per mole respectively.

The standard heat of formation of Fe2O2 (s) is 824.2 kJ mol-1. Calculate heat change for the reaction4Fe (s) + 3O2 (g) → 2Fe2O2 (s)

The molar heat of formation of `NH_(4)NO_(3)(s)` is `-360.0kJ` and those of `N_(2)O(g)`and `H_(2)O(l)` are `+80.00kJ` and `-285.00kJ`, respectively, at `25^(@)C` and `1atm`. Calculate `DeltaH` and `DeltaU` for the reaction. `NH_(4)NO_(3)(s) rarr N_(2)O(g) +2H_(2)O(l)`

The standard enthalpy of formation of `Fe_(2)O_(3)(s)` is - 824.2 kJ `mol^(-1)` . Calculate the enthalpy change for the reaction `: 4 Fe (s) + 3O_(2) (g) rarr 2Fe_(2) O_(3) (s)`

What is open system? Give example.

The enthalpy of combustion of ethyl alcohol is -1360 kJ/ mol. Write the thermochemical equation of the reaction.

For the equilibriumPCl5 (g) == PCl3 (g) + Cl2 (g)at 298 K, K = 1.8 x 10-7. What is ΔG° for the reaction?

The enthalpy of combustion of sulphur is 297 kJ. Write the thermochemical equation for combustion of sulphur. What is the value of ΔfH of SO2?

What is a refrigerator?

Write the Mayer’s relation.

What is Mayer’s relation?

What are the causes of irreversibility?

Draw a diagram to illustrate that the work done by a system depends on the process even when the initial and final states are the same.Different ways to change a system

What is a quasistatic process ?

Draw P-V diagram for positive work at constant pressure.Positive work of constant pressure