540 + Interview Questions in CHEMICAL-THERMODYNAMICS IN GENERAL AWARENESS Page 11 InterviewSolution

501.	Heat of neutralisation of a strong acid by a strong base is equal to `DeltaH` ofA. `H^(+)+OH^(-)=H_(2)O`B. `H_(2)O+H^(+)=H_(3)O^(+)`C. `2H_(2)O_(2)=2H_(2)O`D. `CH_(3)COOH+NaOH=CH_(3)COONa+H_(2)O`
Answer» Correct Answer - C `NaOH + HCl rarr NaCl + H_(2)O` `Na^(+) + OH^(-) + H^(+) + Cl^(-) rarr + Na^(+) + Cl^(-) + H_(2)O` `OH^(-) + H^(+) rarr H_(2)O` `Delta H_("neut") = 57.1 kl//mol`

Discussion

502.	A process is taking place at constant temperature and pressure. ThenA. `DeltaH=DeltaE`B. `DeltaH=TDeltaS`C. `DeltaH=0`D. `DeltaS=0`
Answer» Correct Answer - C For an isothermal, isobaric expansion. `DeltaT=DeltaP=0` `:. DeltaH=DeltaE=0`

Discussion

503.	The standard heat of combustion of solid boron is equal toA. `DeltaH_(f)^(@)(B_(2)O_(3))`B. `1//2DeltaH_(f)^(@)(B_(2)O_(3))`C. `2DeltaH_(f)^(@)(B_(2)O_(3))`D. `-1//2DeltaH_(f)^(@)(B_(2)O_(3))`
Answer» Correct Answer - B `4B+3O_(2)rarr2B_(2)O_(3)` or `B+(3)/(4)O_(2)rarr(1)/(2)B_(2)O_(3)` `:. DeltaH_("comb")^(@)(B)=(1)/(2)DeltaH_(f)^(@)(B_(2)O_(3))`

Discussion

504.	One mole of a perfect gas expands isothermally to ten times its original volume. The change in entropy isA. `0.1 R`B. `2.303 R`C. `10.0 R`D. `100.0 R`
Answer» Correct Answer - B `q_(rev)=-W_(rev)=2.303nRTlog"(V_(2))/(V_(1))` `DeltaS=(q_(rev))/(T)=2.303 nR log"(V_(2))/(V_(1))` Here `n=1`, `(V_(2))/(V_(1))=10` `:. DeltaS=2.303xx1xxRxxlog10` `=2.303xxRxx1=2.303R`

Discussion

505.	If `H^(+)+OH^(-)rarrH_(2)O+13.7Kcal` , the heat of neutralisation for complete neutralisation of 1 mole of `H_(2)SO_(4)` by base will beA. `13.7Kcal`B. `27.4Kcal`C. `6.85Kcal`D. `3.425Kcal`
Answer» Correct Answer - B 1 mole or 2 eq. of `H_(2)SO_(4)` is neutralised.

Discussion

506.	The value of `Delta H` and `Delta S` for five different reaction are given below. `\|{:("Reaction",Delta H(kJ mol^(-)),Delta S(JK^(-)mol^(-))),(I,+98.0,+14.8),(II,+55.5,+14.8),(III,+28.3,-84.8),(IV,-40.5,+24.6),(V,+34.7," 0.0"):}\|` On the basis of these values, predict which one of these will be spontaneous at all temperature?A. Reaction `I`B. Reaction `II`C. Reaction `III`D. Reaction `IV`
Answer» Correct Answer - D For a reaction to be spontaneous at all temperatures `DeltaS` should be `+ve` and `DeltaH` should be `-ve` (both the factors must favour the process). Thus, only reaction `(IV)` will be spontaneous at all temperatures.

Discussion

507.	A cylinder of gas supplied by Bharat Petroleum is assumed to contain `14 kg` of butane. If a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder lasts `(Delta_(C ) H^(@)` of `C_(4) H_(10) = - 2658 kJ mol^(-1))`A. `15` daysB. `20` daysC. `50` daysD. `40` days
Answer» Correct Answer - D Molar mass of butane, `C_(4)H_(10)=12xx4+10` `=48+10=58 g mol^(-1)` `58g` of butane gives `2658 kJ` of heat energy `:. 14 kg` of butane will give heat energy `=(2658kJxx(14xx10^(3)g))/(58g)=641.5862xx10^(3)kJ` Daily energy requirement for cooking `=20,000 kJ` `=2xx10^(4) kJ "day"^(-1)` `:. ` No. of days cylinder will last `=(641.5862xx10^(3)kJ)/(2xx10^(4)kJ"day^(-1))=32.08"days"`

Discussion

508.	`Delta G^(@)` for a reaction is `46.06 kcal mol^(-)`. `K_(P)` for the reaction at `300 K` isA. `10^(-8)`B. `10^(-22.22)`C. `10^(-33.55)`D. none of these
Answer» Correct Answer - C `DeltaG^(@)=46-.06 kcal mol^(-1)` `=46.06xx1000xx4.814 J mol^(-1)` `DeltaG^(@)=-RT ln K_(P)=-2.303 RT log K_(P)` `46.04xx1000xx4.184` `=-2.303xx8.31xx300 log K_(P)` or `log K_(P)=-33.55` or `K_(P)=10^(-33.55)`

Discussion

509.	For the reaction `CO (g) + (1)/(2) O_(2) (g) rarr CO_(2) (g)` `Delta H` and `Delta S` are `-283 kJ` and `-87 J K^(-1)`, respectively. It was intended to carry out this reaction at 1000,1500,3000, and 3500 K. At which of these temperatures would this reaction be thermodynamically spontaneous?A. 1500 and 3500 KB. 3000 and 3500 KC. 1000, 1500 and 3000KD. 1500, 3000 and 3500 K
Answer» Correct Answer - C `DeltaG=DeltaH-TDeltaS` At 1000 K, `DeltaG=-283-(-1000xx0.087)`(Spontaneous) `=-196kJ` AT 1500K `DeltaG=-283-(-1500xx0.087)` (spontaneous) `=-152.5kJ` At 3000 K, `DeltaG=-283-(-3000xx0.087)` (spontaneous) `=-22kJ` At 3500 K, `DeltaG=-283-(-3500xx0.087)` (Non-spontaneous) =`+21.5kJ`

Discussion

510.	The equilibrium constant for a reaction is `10`. What will be the value of `DeltaG^(Θ)`? `R=8.314 J K^(-1) mol^(-1), T=300 K`.A. `-5527 kJ mol^(-1)`B. `-5.527 kJ mol^(-1)`C. `-55.27 kJ mol^(-1)`D. `+5.527 kJ mol^(-1)`
Answer» Correct Answer - B `DeltaG^(@)=-RT ln K = -2.303 RT log K` `=-2.303xx8xx300=-5527 J mol^(-1)` `=-5.527 kJ mol^(-1)`

Discussion

511.	One mole of an ideal gas expanded freely and isothermally at `300 K` from `5` litres to `10` litres. If `DeltaE` is `0`, then `DeltaH` isA. `5 cal`B. `5xx300cal`C. zeroD. `-2xx5xx300cal`
Answer» Correct Answer - C As the gas expands freely at constant temperature `:. DeltaH=0`

Discussion

512.	For the reaction `C_(6)H_(12)(l)+9O_(2)(g)to6H_(2)O(l)+6CO_(2)(g)` , `DeltaH=-936.9kcal` Which of the following is true ?A. `-936.9=DeltaE-(2xx10^(-3)xx298xx3)kcal`B. `+936.9=DeltaE+(2xx10^(-3)xx298xx3)kcal`C. `-936.9=DeltaE+(2xx10^(-3)xx298xx2)kcal`D. `-936.9=DeltaE-(20.0821xx298xx3)kcal`
Answer» Correct Answer - A `DeltaH=DeltaE+Deltan_(g)RT` `Deltan_(g)=6-9=-3`, `R=0.002 kcal`, `T=298 K`

Discussion

513.	If `50` calories are added to a system and system does work of `30` calories on surroundings, the change in internal energy of system isA. `20cal`B. `50cal`C. `40cal`D. `30cal`
Answer» Correct Answer - A `q=DeltaE+w` `50=DeltaE+30` `:. DeltaE=20cal`

Discussion

514.	How many calories are required to heat `40 g` of argon from `40^(@)C` to `100^(@)C` at constant volume? `(R = 2 cal mol^(-1) K^(-1))`A. `120`B. `2400`C. `1200`D. `180`
Answer» Correct Answer - D For a monoatomic gas like argon, `C_(V)=3 kcal mol^(-1)`, `C_(P)=5cal mol^(-1)`. Molar heat capacity is the heat required to raise the temperature of `1` mole of the gas by `1^(@)`. `40 g` of argon `=1` mole.Hence , heat required to raise the temperature from `40^(@)` to `100^(@)C` at constant volume `=3xx(100-40)=180cal`

Discussion

515.	Which one of the following sets of units represents the smallest and the largest amout of energy, respectively?A. `J` and `erg`B. `erg` and `cal`C. `cal` and `eV`D. lit atm and `J`
Answer» Correct Answer - D Smallest `=eV` and largest `=It`. Atm `1 eV=1.6xx10^(-19)J`, `1 cal=4.186 J` `1 erg = 10^(-7) J`, `1 It atm=101.3 J`

Discussion

516.	The enthalpy of formation of methane at constant pressure and `300 K` is `-75.83 kJ`. What will be the heat of formation at constant volume? `[R = 8.3 J K^(-1) mol^(-1)]`
Answer» Correct Answer - `-73.34kJ`

Discussion

517.	The correct thermodnamic conditions for the spontaneous reaction at all temperature is:A. `DeltaHlt0` and `DeltaSlt0`B. `DeltaHlt0` and `DeltaS=0`C. `DeltaHgt0` and `DeltaSlt0`D. `DeltaHlt0` and `DeltaSgt0`
Answer» Correct Answer - C `DeltaG=DeltaH-TDeltaS` For spontaneous process `(DeltaG=-ve)` at all temperature, `DeltaHlt0` and `DeltaSgt0` are favourable factors `DeltaG=DeltaH-TDeltaS=(-ve)-T(+ve)`

Discussion

518.	A `1.0g ` sample of substance `A` at `100^(@)C` is added to `100 mL` of `H_(2)O` at `25^(@)C`. Using separate `100 mL` portions of `H_(2)O`, the procedure is repeated with substance `B` and then with substance `C`. How will the final temperatures of the water compare ? `{:("Substance","Specific heat"),(A,0.60 J g^(-1) "^(@)C^(-1)),(B,0.40 J g^(-1) "^(@)C^(-1)),(C,0.20 J g^(-1) "^(@)C^(-1)):}`A. `T_(C ) gt T_(B) gt T_(A)`B. `T_(B ) gt T_(A) gt T_(C )`C. `T_(A ) gt T_(B) gt T_(C )`D. `T_(A ) = T_(B) = T_(C )`
Answer» Correct Answer - C Heat evolved or absorbed, `Q=m s Deltat` where `m=` mass of the substance, `s=` specific heat, `Deltat=` temperative difference. Therefore more the specific heat more is the heat content and hence more the rise in temperature. Thus correct order of final temperature `T_(A) gt T_(B) gt T_(C )`.

Discussion

519.	During winters, moisture condenses in the from of dew and can be seen on plant leaves and grass. The entropy of the system in such cases decreases as liquids process lesser disorder as compared to gases. With reference to the second law, which statement is correct. for the above process?A. The randomness of the universe decreasesB. The randomness of the surroundings decreasesC. Increases in randomness of surrroundings equals ito the decreases in randomness of systemD. The increases in randomness of the surroundings is greater as compared to the decreases in randomness of the system.
Answer» Correct Answer - A As dew formation is spontaneous process, therefore, entropy or randomness of the universe will increase. As randomness of the system has decreased but randomness of the surrounding will increase significantly so that change is positive.

Discussion

520.	At `27^(@)C` latent heat of fusion of a compound is `2930mol` . Entropy change during fusion isA. `9.77J//molK`B. `10.77J//molK`C. `9.07J//molK`D. `0.977J//molK`
Answer» Correct Answer - B `DeltaS=(DeltaH_(f))/(T)=(2930)/(300)=9.77J mol^(-1)K^(-1)`

Discussion

521.	For the gas phase reaction `PCl_(5)rarrPCl_(3)(g)+Cl_(2)(g)` which of the following conditions are correct?A. `DeltaH=0` and `DeltaSlt 0`B. `DeltaHgt0` and `DeltaSgt 0`C. `DeltaHlt 0` and `DeltaSlt 0`D. `DeltaHgt 0` and `DeltaSlt 0`
Answer» Correct Answer - D `DeltaH=DeltaE+ngRT` For the reaction, `Deltan=2-1=1` Thus, the value of `DeltaH` is positive or `gt 0` . `DeltaG=DeltaH-TDeltaS` For a spontaneous reaction, `DeltaG` must be negative. Since in this reaction `DeltaH` is positive, so for the negative value o0f `DeltaG , DeltaS` must be positive or `gt 0` . Hence `DeltaHgt 0 , DeltaSgt 0`

Discussion

522.	For the gas phase reaction, `PCl_(5)(g)rarrPCl_(3)(g)+CL_(2)(g)` Which of the following conditions are correct?A. `DeltaHgt0`` and `DeltaSgt0`B. `DeltaHlt0`` and `DeltaSlt0`C. `DeltaHgt0`` and `DeltaSgt0`D. `Delta=0` and `DeltaSlt0`
Answer» Correct Answer - C `DeltaH=+ve` and `DeltaS=+ve` . The disored increases with increase in moles. `PCl_(5) hArr PCl_(3) +Cl_(2), DeltaH=+ve` (disordered)

Discussion

523.	For the gas phase reaction `PCl_(5)rarrPCl_(3)(g)+Cl_(2)(g)` which of the following conditions are correct?A. `DeltaH=0` and `DeltaS lt 0`B. `DeltaHgt0` and `DeltaS gt 0`C. `DeltaHlt0` and `DeltaS lt 0`D. `DeltaHgt0` and `DeltaS lt 0`
Answer» Correct Answer - B Since dissociation of `PCl_(5)` is an endothermic reaction, `DeltaH=+ve` i.e., `DeltaH gt 0`. The reaction is accompanied by randomness, `DeltaS gt 0`

Discussion

524.	If `DeltaH_(f)^(@)` of `Icl(g)`, `Cl(g)` and `I(g)` is `17.57`, `121.34` and `106.96 J mol^(-1)` respectively. Then bond dissociation energy of `I-Cl` bond isA. 17.57B. 210.73C. 35.15D. 106.96
Answer» Correct Answer - B The process will be: `I(g)+Cl(g)toI-Cl(g)," "DeltaH^(@)=17.57" J "mol^(-1)` `DeltaH=`Heats of atomisation of I(g) and Cl(g)`-`Bond energy of I-Cl bond `17.57=121.34+106.96-x` `x=210.73" J "mol^(-1)`

Discussion

525.	In the reaction `A^(+)+BtoA+B^(+)` there is no entropy change. If enthalpy change is 22 kJ of `A^(+)`, calculate `DeltaG` for the reaction.
Answer» For the given reaction, `DeltaH=22kJ,DeltaS=0` `therefore` From `DeltaG=DeltaH-TDeltaS` `DeltaG=22-Txx0=22kJ" "mol^(-1)`

Discussion

526.	For the reaction `2NO(g)+O_(2)(g)to2NO_(2)g` Calculate `DeltaG` at 700K when enthalpy and entropy changes are `-113.0kJ" "mol^(-1) and `-145J" "K^(-1)mol^(-1)` respectively.
Answer» We know that `DeltaG=DeltaH-TDeltaS` Given, `DeltaH=-113kJ" "mol^(-1)` `=-113000J" "mol^(-1)` `DeltaS=-145J" "K^(-1)" "mol^(-1)` `T=700K` Substituting these values in the above equation, we get `DeltaG=-113000-700xx(-145)` `=-11500J" "mol^(-1)` `=-11.5kJ" "mol^(-1)`

Discussion

527.	`DeltaH` and `DeltaS` for the reaction: `Ag_(2)O(s) rarr 2Ag(s) +(1//2)O_(2)(g)` are `30.56 kJ mol^(-1)` and `66.0 J JK^(-1) mol^(-1)` respectively. Calculate the temperature at which free energy change for the reaction will be zero. Predict whether the forward reaction will be favoured above or below this temperature.
Answer» We know that, `DeltaG=DeltaH-TDeltaS` At equilibrium, `DeltaG=0` so that `0=DeltaH-TDeltaS` ltbr or `T=(DeltaH)/(DeltaS)` Given that, `DeltaH=30.56kJ" "mol^(-1)` `=30560J" "mol^(-1)` `DeltaS=66.0J" "K^(-1)" "mol^(-1)` `T=(30560)/(66)=463K` Above this temperature `,DeltaG` will be negative and the process will be spontaneous in forward direction.

Discussion

528.	For a reaction, `Ag_(2)O(s)hArr2Ag(s)+(1)/(2)O_(2)(g)` `DeltaH,DeltaS and T ` are 40.63 kJ `mol^(-1)`, 108.8 J `K^(-1)mol^(-1) and 373.3K` respectively. Predict the feasibility of the reaction:A. feasibleB. non-feasibleC. remains at equilibriumD. not predicted
Answer» Correct Answer - C `DeltaG=DeltaH-TDeltaS` `=40.63xx1000-373.4xx108.8=4.08`

Discussion

529.	Given two processes: `(1)/(2)P_(4)(s) +3CI_(2)(g) rarr 2PCI_(3)(l), DeltaH =- 635 kJ` `PCI_(3) (l) +CI_(2)(g) rarrPCI_(5)(s), DeltaH =- 137 kJ` The value of `DeltaH^(Theta)` of `PCI_(5)` isA. `454.5kJ" "mol^(-1)`B. `-454.5kJ`C. `-772kJ" "mol^(-1)`D. `-498kJ" "mol^(-1)`
Answer» Correct Answer - B `(1)/(4)P_(4)(s)+(3)/(2)Cl_(2)(g)toPCl_(5)(l),DeltaH=-(635)/(2)kJ` `PCl_(3)(l)+Cl_(g)toPCl_(5)(s),DeltaH=-137kJ` On adding, `overline((1)/(4)P_(4)(s)+(5)/(2)Cl_(2)(g)toPCl_(5)(s),DeltaH=-137kJ)`

Discussion

530.	Assertion: The zeroth law of thermodynamics was know before the first law of thermodynamics. Reason: The zeroth law concerning thermal equilibrium appeared after three laws `(I, II and III)` of thermodynamics and thus was named zeroth law.A. If both the assertion and reason are true and the reason is the correct explanation of the assertion.B. If both the assertion and reason are true but reason is not the correct explantion of the assertion.C. If the assertion is true but reason is false.D. If assertion is false but reason is true.
Answer» Correct Answer - B The reason in an explanation for the name zeroth law.

Discussion

531.	The heat of combustion of solid benzoic acid at constant volume is `-321.30 kJ` at `27^(@)C`. The heat of combustion at constant pressure isA. `-321.30 -300R`B. `-321.30+300R`C. `-321.30-150R`D. `-321.30+900R`
Answer» Correct Answer - C `C_(6)H_(5)COOH(s)+(15)/(2)O_(2)(g)to7CO_(2)(g)+3H_(2)O(l)` `Deltan+7-(15)/(2)=-(1)/(2)` `DeltaH=DeltaU+DeltanRT` `=-321.30-(1)/(2)xxRxx300=-321.30-150R`

Discussion

532.	i. `H_(2)(g) +CI_(2)(g)rarr 2HCI(g),DeltaH =- x kJ` ii. `NaCI +H_(2)SO_(4) rarr NaHSO_(4) +HCI, DeltaH =- y kJ` iii. `2H_(2)O +2CI_(2) rarr 4HCI +O_(2),DeltaH =- z kJ` From the above equations, the value of `DeltaH` of `HCI` isA. `-x kJ`B. `-y kJ`C. `-z kJ`D. `-x//2 kJ`
Answer» Correct Answer - D As per definition, the equation `(i)` represents the formation of `2` moles of `HCl`. Hence `DeltaH_(f)` of `HCl(g)=-X//2kJ mol^(-1)`.

Discussion

533.	0.5 g of benzoic acid was subjected to combustion in a bomb calorimeter at `15^(@)C` when the temperature of the calorimeter system (including water) was found to rise by `0.55^(@)C`. Calculate the heat of combustion of benzoic acid (i) at constant volume and (ii) at constant pressure. the thermal capacity of the calorimeter including water was found to be 23.85 kJ.
Answer» (i). Heat of combustion at constant volume, `DeltaU` =Heat capacity of calorimeter and its constent `xx`rise in temperature`xx("Mol. Mass of compound")/("Mass of compound")` `=23.85xx0.55xx(122)/(0.5)=3200.67kJ` i.e., `DeltaU=-3200.67kJ" "mol^(-1)` (ii) we know that, `DeltaH=DeltaU+DeltanRT` `C_(6)H_(5)COOH(s)+(15)/(2)O_(2)(g)to7CO_(2)(g)+3H_(2)O(l)` `Deltan=7-7.5=-0.5,R=8.314xx10^(-3)kJ" "K^(-1)mol^(-1)," "T=288K` Substituting the values in the above equation, `DeltaH=-3200.67-8.314xx10^(-3)xx(-0.5)xx288` `=-3200.67-1.197=-3201.867kJ" "mol^(-1)`

Discussion

534.	A fixed mass `m` of a gas is subjected to transformation of state: `K` to L to `M` and back to `K` as shown in the figure. The succeeding operations that enabel this transformation of state are A. hating,cooling,heating,coolongB. cooling,heating,cooling,heatingC. heating,cooling,cooling,heatingD. cooling,heating,heating,cooling
Answer» Correct Answer - C `KtoL` (volume increases at constant pressure, hence the gas will be heated.) `LtoM` (Pressure decreases at constant volume, it will take place by cooling). `MtoN` (Volume decreases at constant pressure, it will take place by cooling the gas.) `NtoK` (Pressure of gas increases at constant volume, it will take place by heating the gas).

Discussion

535.	One kilogram water at `0^(@)C` is brought into contact with a heat reservoir at `100^(@)C`. Find
Answer» Correct Answer - (a). `312cal" "K^(-1),` (b). `-268cal" "K^(-1)` (c) ` 43.9cal" "K^(-1)` (d) spontaneous (a) `DeltaS=2.303nC_(P)log" "((T_(2))/(T_(1)))` `=2.303xx(1000)/(18)xx18" log "((373)/(273))=312cal" "K^(-1)` (b) `DeltaS_("reservoir")=-(DeltaQ)/(T)` `DeltaQ=msDeltat=1000xx1xx1000=10^(5)cal` `DeltaS=(-10^(5))/(373)=-268.1cal" "K^(-1)` (c) `DeltaS_("universe")=312-268.1=43.9cal" "K^(-1)` (d) `DeltaS gt 0`, the process will be spontaneous.

Discussion

536.	For the given reactions `SiO_(2) +4HF rarr SiF_(4) +2H_(2)O, DeltaH =- 10.17 kcal` `SiO_(2) +4HCI rarr SiCI_(4) +2H_(2)O, DeltaH = 36.7 kcal` It may be concluded thatA. HF and HCl both will react with silicaB. Only Hf will react with silicaC. Only HCl will react with silicaD. Neither HF nor HCl will react with silica
Answer» Correct Answer - B Exothermic reactions are spontaneous.

Discussion

537.	A fixed mass m of a gas is subjected to transfromation of states from K to L to M and back to K as shown in the figure. The pair of isochoric processes among the transfromation of states isA. K to L and L to MB. L to M and N to KC. L to M and M to ND. M to N and N to K
Answer» Correct Answer - B Isochoric process means the process at constant volume. `thereforeLtoM and Nto K` are the pair of isochoric processes.

Discussion

538.	Identify different steps in the following cyclic process:
Answer» (i) A `to`B (Temperature and pressure are constant) (ii) `BtoC` it is adiabatic expansion in which temperatrue falls from `T_(1)` to `T_(2)` (iii) `CtoD` (Temperature and volume are constant). `therefore` This process is isothermal and isochoric. ltbr. (iv) `DtoE` (temperature and pressure are constant) `therefore` It is isothermal and isobaric contraction (v) `EtoF` (it is adiabatic compression in whcih temperature increases from `T_(2)` to `T_(1)`) (vi). `FtoA` (Temperature and volume are constant). `therefore` It is isothermal and isochoric process.

Discussion

539.	A mono-atomic ideal gas of two moles is taken through a cyclic process starting from `A` as shwon in the figure below. The volume ratios are `V_(B)//V_(A) = 2` and `V_(D)//V_(A) = 4`. If the temperature `T_(A)` at `A` is `27^(@)C`. Calculate a. The temperature of gas at `B`. b. Heat absorbed or evolved in each process. c. Total work done in cyclic process.
Answer» `AtoB` (It is isobaric process) `(V_(A))/(T_(A))=(V_(B))/(T_(B))` `thereforeT_(B)=(V_(B))/(V_(A))xxT_(A)=2xx300=600K` `q_(AB)=nC_(P)DeltaT=2xx(5)/(2)RDeltaT` `=2xx(5)/(2)xx2xx300=3000cal` `BtoC`, (Isothermal process) `DeltaU=0` `thereforeq_(BC)=W=2.303nRT" log "((V_(C))/(V_(B)))` `=2.303xx2xx2xx600log((4)/(2))` `=1.663xx10^(3)cal` C`to`D (Isobaric process) `q_(CD)=nC_(V)DeltaT=2xx(3)/(2)xx2(-300)=-1800cal` `DtoA` (Isothermal process) ltbr `q_(DA)=2.303nRT_(A)" log "(V_(A))/(V_D))` `=2.303xx2xx2xx300" log "(1)/(4)` `=-1 .663xx10^(3)cal` ,brgt Total heat change `=3000+1.663xx10^(3)-1800-1.663xx10^(3)` `=1200cal` work done `=-1200cal`

Discussion

540.	`1mol` of a mono-atomic gas is subjected to following cyclic process: a. Calculate `T_(1)` and `T_(2)`. b. Calculate `DeltaU,q,` and `W` in calories in each step of cyclic process.
Answer» (a) At A `PV=nRT` `20xx1=1xx0.0821xxT_(1)` `T_(1)=243.6K` At B: `PV=nRT` `20xx10=1xx0.0821xxT_(2)` `T_(2)=2436.05K` (b) Path AB: isobaric process `(DeltaU=0,q=w)` `w=PDeltaV=20xx9=180` litre-atm `=(180xx101.3)/(4.185)cal` `=4356.9cal` (work in compression is positive) Path BC: Isochoric process w=0 `q_(V)=DeltaU=nC_(V)DeltaT=1xx(3)/(2)Rxx(2436-243.6)` `=(3)/(2)xx2xx2192.4=6577.2cal` It is cooling process, `q_(V)=-6577.2cal` Path CA: It is isothermal compression `DeltaV_(0)` `q=w=2.303nRtlog(V_(2))/(V_(1))` `q=w=2.303xx1xx2xxlog((10)/(1))=1122.02cal`

Discussion

Explore topic-wise InterviewSolutions in .

Heat of neutralisation of a strong acid by a strong base is equal to `DeltaH` ofA. `H^(+)+OH^(-)=H_(2)O`B. `H_(2)O+H^(+)=H_(3)O^(+)`C. `2H_(2)O_(2)=2H_(2)O`D. `CH_(3)COOH+NaOH=CH_(3)COONa+H_(2)O`

A process is taking place at constant temperature and pressure. ThenA. `DeltaH=DeltaE`B. `DeltaH=TDeltaS`C. `DeltaH=0`D. `DeltaS=0`

The standard heat of combustion of solid boron is equal toA. `DeltaH_(f)^(@)(B_(2)O_(3))`B. `1//2DeltaH_(f)^(@)(B_(2)O_(3))`C. `2DeltaH_(f)^(@)(B_(2)O_(3))`D. `-1//2DeltaH_(f)^(@)(B_(2)O_(3))`

One mole of a perfect gas expands isothermally to ten times its original volume. The change in entropy isA. `0.1 R`B. `2.303 R`C. `10.0 R`D. `100.0 R`

If `H^(+)+OH^(-)rarrH_(2)O+13.7Kcal` , the heat of neutralisation for complete neutralisation of 1 mole of `H_(2)SO_(4)` by base will beA. `13.7Kcal`B. `27.4Kcal`C. `6.85Kcal`D. `3.425Kcal`

A cylinder of gas supplied by Bharat Petroleum is assumed to contain `14 kg` of butane. If a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder lasts `(Delta_(C ) H^(@)` of `C_(4) H_(10) = - 2658 kJ mol^(-1))`A. `15` daysB. `20` daysC. `50` daysD. `40` days

`Delta G^(@)` for a reaction is `46.06 kcal mol^(-)`. `K_(P)` for the reaction at `300 K` isA. `10^(-8)`B. `10^(-22.22)`C. `10^(-33.55)`D. none of these

The equilibrium constant for a reaction is `10`. What will be the value of `DeltaG^(Θ)`? `R=8.314 J K^(-1) mol^(-1), T=300 K`.A. `-5527 kJ mol^(-1)`B. `-5.527 kJ mol^(-1)`C. `-55.27 kJ mol^(-1)`D. `+5.527 kJ mol^(-1)`

One mole of an ideal gas expanded freely and isothermally at `300 K` from `5` litres to `10` litres. If `DeltaE` is `0`, then `DeltaH` isA. `5 cal`B. `5xx300cal`C. zeroD. `-2xx5xx300cal`

For the reaction `C_(6)H_(12)(l)+9O_(2)(g)to6H_(2)O(l)+6CO_(2)(g)` , `DeltaH=-936.9kcal` Which of the following is true ?A. `-936.9=DeltaE-(2xx10^(-3)xx298xx3)kcal`B. `+936.9=DeltaE+(2xx10^(-3)xx298xx3)kcal`C. `-936.9=DeltaE+(2xx10^(-3)xx298xx2)kcal`D. `-936.9=DeltaE-(20.0821xx298xx3)kcal`

If `50` calories are added to a system and system does work of `30` calories on surroundings, the change in internal energy of system isA. `20cal`B. `50cal`C. `40cal`D. `30cal`

How many calories are required to heat `40 g` of argon from `40^(@)C` to `100^(@)C` at constant volume? `(R = 2 cal mol^(-1) K^(-1))`A. `120`B. `2400`C. `1200`D. `180`

Which one of the following sets of units represents the smallest and the largest amout of energy, respectively?A. `J` and `erg`B. `erg` and `cal`C. `cal` and `eV`D. lit atm and `J`

The enthalpy of formation of methane at constant pressure and `300 K` is `-75.83 kJ`. What will be the heat of formation at constant volume? `[R = 8.3 J K^(-1) mol^(-1)]`

The correct thermodnamic conditions for the spontaneous reaction at all temperature is:A. `DeltaHlt0` and `DeltaSlt0`B. `DeltaHlt0` and `DeltaS=0`C. `DeltaHgt0` and `DeltaSlt0`D. `DeltaHlt0` and `DeltaSgt0`

At `27^(@)C` latent heat of fusion of a compound is `2930mol` . Entropy change during fusion isA. `9.77J//molK`B. `10.77J//molK`C. `9.07J//molK`D. `0.977J//molK`

For the gas phase reaction `PCl_(5)rarrPCl_(3)(g)+Cl_(2)(g)` which of the following conditions are correct?A. `DeltaH=0` and `DeltaSlt 0`B. `DeltaHgt0` and `DeltaSgt 0`C. `DeltaHlt 0` and `DeltaSlt 0`D. `DeltaHgt 0` and `DeltaSlt 0`

For the gas phase reaction, `PCl_(5)(g)rarrPCl_(3)(g)+CL_(2)(g)` Which of the following conditions are correct?A. `DeltaHgt0`` and `DeltaSgt0`B. `DeltaHlt0`` and `DeltaSlt0`C. `DeltaHgt0`` and `DeltaSgt0`D. `Delta=0` and `DeltaSlt0`

For the gas phase reaction `PCl_(5)rarrPCl_(3)(g)+Cl_(2)(g)` which of the following conditions are correct?A. `DeltaH=0` and `DeltaS lt 0`B. `DeltaHgt0` and `DeltaS gt 0`C. `DeltaHlt0` and `DeltaS lt 0`D. `DeltaHgt0` and `DeltaS lt 0`

If `DeltaH_(f)^(@)` of `Icl(g)`, `Cl(g)` and `I(g)` is `17.57`, `121.34` and `106.96 J mol^(-1)` respectively. Then bond dissociation energy of `I-Cl` bond isA. 17.57B. 210.73C. 35.15D. 106.96

In the reaction `A^(+)+BtoA+B^(+)` there is no entropy change. If enthalpy change is 22 kJ of `A^(+)`, calculate `DeltaG` for the reaction.

For the reaction `2NO(g)+O_(2)(g)to2NO_(2)g` Calculate `DeltaG` at 700K when enthalpy and entropy changes are `-113.0kJ" "mol^(-1) and `-145J" "K^(-1)mol^(-1)` respectively.

For a reaction, `Ag_(2)O(s)hArr2Ag(s)+(1)/(2)O_(2)(g)` `DeltaH,DeltaS and T ` are 40.63 kJ `mol^(-1)`, 108.8 J `K^(-1)mol^(-1) and 373.3K` respectively. Predict the feasibility of the reaction:A. feasibleB. non-feasibleC. remains at equilibriumD. not predicted

Given two processes: `(1)/(2)P_(4)(s) +3CI_(2)(g) rarr 2PCI_(3)(l), DeltaH =- 635 kJ` `PCI_(3) (l) +CI_(2)(g) rarrPCI_(5)(s), DeltaH =- 137 kJ` The value of `DeltaH^(Theta)` of `PCI_(5)` isA. `454.5kJ" "mol^(-1)`B. `-454.5kJ`C. `-772kJ" "mol^(-1)`D. `-498kJ" "mol^(-1)`

The heat of combustion of solid benzoic acid at constant volume is `-321.30 kJ` at `27^(@)C`. The heat of combustion at constant pressure isA. `-321.30 -300R`B. `-321.30+300R`C. `-321.30-150R`D. `-321.30+900R`

i. `H_(2)(g) +CI_(2)(g)rarr 2HCI(g),DeltaH =- x kJ` ii. `NaCI +H_(2)SO_(4) rarr NaHSO_(4) +HCI, DeltaH =- y kJ` iii. `2H_(2)O +2CI_(2) rarr 4HCI +O_(2),DeltaH =- z kJ` From the above equations, the value of `DeltaH` of `HCI` isA. `-x kJ`B. `-y kJ`C. `-z kJ`D. `-x//2 kJ`

One kilogram water at `0^(@)C` is brought into contact with a heat reservoir at `100^(@)C`. Find

A fixed mass m of a gas is subjected to transfromation of states from K to L to M and back to K as shown in the figure. The pair of isochoric processes among the transfromation of states isA. K to L and L to MB. L to M and N to KC. L to M and M to ND. M to N and N to K

Identify different steps in the following cyclic process:

`1mol` of a mono-atomic gas is subjected to following cyclic process: a. Calculate `T_(1)` and `T_(2)`. b. Calculate `DeltaU,q,` and `W` in calories in each step of cyclic process.