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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 3301. |
If `Delta E` is the heat of reaction for `C_(2)H_(5)OH_((1))+3O_(2(g)) rarr 2CO_(2(g))+3H_(2)O_((1))` at constant volume, the `Delta H` (Heat of reaction at constant pressure) at constant temperature isA. `Delta H = Delta E + 2RT`B. `Delta H = Delta E - 2RT`C. `Delta H = Delta E + RT`D. `Delta H = Delta E - RT` |
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Answer» Correct Answer - D `Delta n = 2-3=-1` |
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| 3302. |
Consider the following reactions : (a) `H_((aq))^(+)+OH_((aq))^(-)=H_(2)O_((l)), Delta H= - X_(1)"kJ mol"^(-1)` (b) `H_(2(g))+(1)/(2)O_(2(g))=H_(2)O_((l)), Delta H= - X_(2)"kJ mol"^(-1)` (c ) `CO_(2(g))+H_(2(g))=CO_((g))+H_(2)O_((l))- X_(3)"kJ mol"^(-1)` (d) `C_(2)H_(2(g))+(5)/(2)O_(2(g))=2CO_(2(g))+H_(2)O_((l))+X_(4)"kJ mol"^(-1)` Enthalpy of formation of `H_(2)O_((l))` is :A. `+X_(1)"kJ mol"^(-1)`B. `-X_(2)"kJ mol"^(-1)`C. `+X_(3)"kJ mol"^(-1)`D. `-X_(4)"kJ mol"^(-1)` |
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Answer» Correct Answer - B Heat of formation of `H_(2)O` equation is `H_(2(g))+(1)/(2)rarr H_(2)O_((l)) , Delta H = -X_(2) kJ mol^(-1)` |
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| 3303. |
If `Delta E` is the heat of reaction for `C_(2)H_(5)OH_((1))+3O_(2(g)) rarr 2CO_(2(g))+3H_(2)O_((1))` at constant volume, the `Delta H` (Heat of reaction at constant pressure) at constant temperature isA. `DeltaH=DeltaE+RT`B. `DeltaH=DeltaE-RT`C. `DeltaH=DeltaE-2RT`D. `DeltaH=DeltaE+2RT` |
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Answer» Correct Answer - B We know that , `DeltaH=DeltaE+ Deltan_(g)RT` where, `Deltan_(g)` = total number of moles of gaseous product-total number of moles of gaseous reactant. =2-3=-1 So, `DeltaH=DeltaE-RT` |
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| 3304. |
The heat of combusion of glycogen is about `476 kJ mol^(-1)` of carbon. Assume that average heat loss by an adult male is `150W`. If we were to assume that all the heat comes from oxidation of glycogen, how many units of glycogen `(1mol` carbon per unit) must be oxidised per day to provide for this heat loss? |
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Answer» Correct Answer - 27.22 units Total energy required in the day `= (150 xx 24 xx 60 xx 60)/(1000) kJ` (1 watt = J/sec) = 12960 kJ Unit of glycogen required `= (12960)/(476) = 27.22` units. |
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| 3305. |
`H_(2)O_((g)) rarr H_((g))+OH_((g))Delta H = x_(1)` `OH_((g)) rarr H_((g))+O_((g))Delta H = x_(2)` Based on the these value, BE of O-H bond isA. `(x_(1)+x_(2))/(2)`B. `x_(1)+x_(2)`C. `(x_(1)-x_(2))/(2)`D. `2(x_(1)+x_(2))` |
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Answer» Correct Answer - A `Delta H = H_(r) - H_(P)` |
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| 3306. |
At `300K` , the standard enthalpies of formation of `C_(6)H_(5)COOH_((s),CO_(2(g))` and `H_(2)O_((l))` are `-408,-393` and `-286kJ mol^(-1)` respectively. Calculate the heat of combustion of benzoic acid at `(i)` constant pressure, `(ii)` constant volume. `(R=8.31J mol^(-1)K^(-1))` |
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Answer» Correct Answer - `-3199.75 kj` Given, `7C (s) + 3H_(2) (g) + O_(2) (g) rarr C_(6) H_(5) COOH (s), Delta H^(0) = -408 kJ` (i) (i) `C(s) + O_(2) (g) rarr CO_(2) (g)` `Delta H^(0) = -393 kJ`...(ii) `H_(2) (g) + (1)/(2) O_(2) (g) rarr H_(2)O (l)` `Delta H^(0) = -286 kJ`...(iii) `C_(6)H_(5)COOH (s) + (15)/(2) O_(2) (g) rarr 7 CO_(2) (g) + 3H_(2) O (l) Delta H = ?`.....(iv) By (ii) `xx 7 + (iii) xx 3 - (i)` `C_(6)H_(5)COOH (s) + (15)/(2) O_(2) (g) rarr 7CO_(2) (g) + 3H_(2) O (l)` `Delta H^(0) = [-393 xx 7 - 286 xx 3 + 408]` `:. Delta H^(0) = - 3201 kJ` Also `Delta H = Delta U + Delta n_(g) RT` `:gt = 3201 = Delta U + (-0.5) xx 8.314 xx 10^(-3) xx 300` `:. Delta U = - 3201 + 1.2471 = -3199.75 kJ` |
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| 3307. |
For polytropic process `PV^(n)` = constant, molar heat capacity `(C_(m))` of an ideal gas is given by:A. `C_(v,m)+(R)/((n-1))`B. `C_(v,m)+(R)/((1-n))`C. `C_(v,m)+R`D. `C_(p,m)+(R)/((n-1))` |
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Answer» Correct Answer - B `dU=dq+dw` `nC_(v,m).dT=nC_(m).dT-P.dV` `C_(m)=C_(v,m)+(P.dV)/(n.dT)` ….(1) `PV^(n)=K and PV=nRT` `therefore" "KV^(1-n)=nRT` `K(1-n)V^(-n).dV=nRdT` `(dV)/(dT)=(nR)/(K(1-n)V^(-n))` …(2) from Eqs. (1) and (2) `C_(m)=C_(v,m)+(R)/(1-n)` |
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| 3308. |
The limitation of 1 st law of thermodynamics isA. Energy can neight be created nor destroyedB. Impossible to constuct the `1^(st)` kind of perpetual motion machineC. Law of conservation of energyD. Spontaneous nature of a process |
| Answer» Correct Answer - D | |
| 3309. |
Temperature of1 mole of a gas is increased by `1^(@)` at constant pressure. Work done isA. 2 calB. 3 calC. 4 calD. 5 cal |
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Answer» Correct Answer - a At temp. T, PV`=`RT When temp. is increased by `1^(@)`, at constant pressure, suppose volume increased by `DeltaV` `:. P ( V+DeltaV) = R (T+1) `or`PV + P DeltaV =RT + R ` But`PV =RT :. P Delta V = R`, i.e., work done `=R= 2 cal`. |
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| 3310. |
The standard heats of formation of `CH_(4)(g), CO_(2)(g)` and `H_(2)O(g)` are `-76.0, -390.0`, and `-240.0 kJ mol^(-1)`, respectively. Calculate the amount of heat evolved by burning `1m^(3)` of methane measured under normal conditions. |
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Answer» The required for the conustion of mathane is: `CH_(4) +2O_(2)rarr CO_(2)+2H_(2)O,DeltaH=?` `DeltaH^(Theta) = Delta_(f)H^(Theta)("product")-Delta_(f)H^(Theta)("reactants")` `=Delta_(f)H_((CO_(2)))^(Theta)+2xxDelta_(f)H^(Theta)H_((H_(2)O))^(Theta)-Delta_(f)H_((CH_(4)))^(Theta)-2Delta_(f)H_((O_(2)))^(Theta)` `=- 390.0 - 2 xx 240.0 -(-76.0) - 2xx0` `=- 718.0 kJ mol^(-1)`. Heat evolved by buring `22.4L` (1mol) methane `= 718.0 kJ`. So, heat evolved by buring `1000L (1m^(3))` methane `=-(718.0)/(22.4) xx 1000 =- 32053.5 kJ` |
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| 3311. |
Enthalpy of sublimation of a substance is equal to :A. Enthalpy of fusion + enthalpy of vaporisationB. enthalpy of fusionC. enthalpy of vaporisationD. twice the enthalpy of vaporisation |
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Answer» Correct Answer - a |
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| 3312. |
An ideal gas is allowed to expand both reversibly and irreversibly in anisolated system. If`T_(i)` is the initial temperature and`T_(f)` is the final temperature, which of the following statements is correct?A. `(T_(f))_(rev)= (T_(f))_(irrev)`B. `T_(f) = T_(i)` for both reversible and irreversible processesC. `(T_(f))_(irrev) gt (T_(f))_(rev)`D. `T_(f) gt T_(i)` for revrsible process but `T_(f) =T_(i)` for irreversible processes. |
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Answer» Correct Answer - b In an ideal gas, there are no intermolecular forces of attraction. Hence, `T_(f) = T_(i)` for both reversible and irreversible processes. |
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| 3313. |
Enthalpy of sublimation of a substance is equal to :A. enthalpy of fusion `+` enthalpy of vaporisationB. enthalpy of fusionC. enthalpy of vaporisationD. twice the enthalpy of vaporisation |
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Answer» Correct Answer - B Enthaply of sublimation of a substance is equal to enthalpy of fusion `+` enthalpy vaporisation. Subimation is, direct conversion of solid to vapour. Solid `rarr` vapour Writing in two steps, we have solid `rarr` liquid `rarr` vapour, solid `rarr` liquid requires enthaply of fusion liquid `rarr` vapour requires enthalpy of vaporisation |
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| 3314. |
Which of the following equations represents a reaction that provides the enthalpy of formation of `CH_(3)Cl`?A. `C(s)+HCl(g)+H_(2)(g)rarrCH_(3)Cl(g)`B. `C(s)+3H(g)+Cl(g)rarrCH_(3)Cl(g)`C. `C(s)+3//2H_(2)(g)+1//2Cl_(2)(g)rarrCH_(3)Cl(g)`D. `CH_(4)(g)+Cl_(2)(g)rarrCH_(3)Cl(g)+HCl(g)` |
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Answer» Correct Answer - C `C(s)+(3)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)rarrCH_(3)Cl(g)` |
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| 3315. |
Which of the following is not correct ?A. `Delta G` is zero for a reversible reactionB. `Delta G` is positive for a spontaneous reactionC. `Delta G` is negative for a spontaneous reactionD. `Delta G` is positive for a non-spontaneous reaction. |
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Answer» Correct Answer - A::D `Delta G` gives a criteria for spontaneity at constant pressure and temperature. (i) If `Delta G` is negative `(lt 0)`, the process is spontaneous (ii) If `Delta G` is positive `(gt 0)`, the process is non-spontaneous. (iii) If `Delta G` is zero then reaction is equilibrium |
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| 3316. |
Use the given standard enthalpies of formation (in kJ/mol) to determine the enthalpy of reaction of the following reaction : `NH_(3)(g)+3F_(2)(g)rarrNF_(3)+3HF(g)` `DeltaH_(f)^(@)(NH_(3),g)=-46.2, " "DeltaH_(f)^(@)(NF_(3),g)=-113.0 ," "DeltaH_(f)^(@)(HF, g)=-269.0`A. `-335.8 kJ//mol`B. `-873.8 kJ//mol`C. `-697.2 kJ//mol`D. `-890.4 kJ//mol` |
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Answer» Correct Answer - B `Delta_(r )H=(Delta_(f)H)_(P)-(Delta_(f)H)_(R)` `Delta_(r )H=-113 + 3xx(-269)+46.2 =-873.8 " kJ"` |
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| 3317. |
`DeltaH` for `CaCO_(3)(s)rarrCaO(s)+CO_(2)(g)` is `176 kJ mol^(-1)` at `1240 K`. The `DeltaE` for the change is equal toA. `165.6 kJ`B. `160.0 kJ`C. `186.4 kJ`D. `180.0 kJ` |
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Answer» Correct Answer - A `DeltaH- DeltaU= Delta nRT` |
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| 3318. |
Calculate change in enthalpy for the reaction at `27^(@)C` `H_(2)(g)+Cl_(2)(g)rarr2H-Cl(g)` by using the bond energy and energy data Bond energies of `H-H,Cl-Cl` and `H-Cl` bonds are `435 kJ mol^(_1), 240 kJ mol^(-1)` and `430 kJmol^(-1)` respectively. |
| Answer» Correct Answer - `-185 kJ` | |
| 3319. |
Oxygen is of vital importance for all of us . Oxygen enters the body via the lungs and is transported to the tissues in our body by blood . There it can deliver energy by the oxidation of sugars. `C_(6)H_(12)O_(6) + 6O_(2) rarr 6CO_(2) + 6H_(2)O` This reaction releases 400 KJ of energy per mole of oxygen `O_(2)` uptake by blood is at four heme (Hm) group in this protein hemoglobin (Hb). Free Hm consists of an `Fe^(2+)` giving `HmO_(2)` complex. Carbon monoxides can be complexed similarily giving a Hm CO complex . CO is poison as it bonds more strongly to Hm than `O_(2)` does. The equilibrium constant `K_(f)` for the reaction: `Hm+ CO hArr HCO " "........(i)` is 1000 times larger than the equilibrium constant `K_(2)` for the reaction: `Hm + CO_(2) hArr HmO_(2)" " ........(ii)` Each Hb molecules can take up four molecules of `O_(2)` absorbs a fraction of this amount, depending on the oxygen pressure , as shown in figure1 (curve 1) . Also shown are the curve (2) and (3) for blood with two kinds of dificient Hb . These occur in patients with certain hereditary diseases. Relevant data , `O_(2)` pressure in lungs is 15 KPa , in the muscles it is 2KPa . The maximum flow of blood through heart and lungs is `4 xx 10^(-4)m^(-3)s^(-1)` . The red cells in blood occupy 40% of the volume, inside the cells the concentration of Hb has a molar mass of 64 kg `"mol"^(-1)` R=8.314 ` J "mol"^(-1) K^(-1)` , T=298k . Using the relation between K and the standard Gibbs energy `DeltaG^(@)` for a reaction, calculated the difference between the `DeltaG^(@)` values for the home reactions (i) and (ii). |
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Answer» Correct Answer - `DeltaG_(1)^(@) =- RT "In" K_(1)" "DeltaG_(2)^(@) =-RT "In"K_(2)` `DeltaG_(2)^(@) - DeltaG_(1)^(@) = DeltaG_(1)^(@) = RT "In" (K_(1))/(K_(2))` `DeltaG_(2)^(@) - DeltaG_(1)^(@) = (8.314 J "mol"^(-1) K^(-1) xx 298k xx "In" 10000) J =23 KJ "mol^(-1)` |
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| 3320. |
Since 1891 lighting lamps have been manufactured in the Netherlands. The improvement today in comparison to the first lamp is enormous , especially with the intorduction of the gas discharge lamps. The life- time has increased by orders of magnitude . the colour is also an important aspect. Rare earth metal compunds like `CeBr_(3)` are now included to reach a colour temperature of 6000K in the lamp the compounds are ionic solids at room temperature , and upon heating they sublime partially to give a vapour of neutral metal halide molecules . To achieve a high vapour pressure, the sublimation enthalpy should be as low as possible. Calculate the enthalpy of sublimation of `CeBr_(3)` (in intergers, be aware of signs). Attemps to make a better lamp have been undertaken by adding a stoichiometric amount of `CsBr` to the `CeBr_(3)` in the lamp leading at room temperature to solid `CsCeBr_(4)` lattice has a NaCl structure with Cs as cations and tetrahedral `CeBr_(4)^(-)` as complex anions. Sublimation of `CsCeBr_(4)` leads to a vapour of CsBr and `CeBr_(3)` molecules. |
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Answer» Correct Answer - `H_(1) =- (139 xx 3xx1xx2.958)/(0.297) xx (10)/(11) KJ mol^(-1)` `H_(e) =(-3 xx(139xx3xx1)/(0.294)xx(10)/(11)) + (3 xx (139 xx1xx1)/(0.297sqrt(3))xx(10)/(11)) KJ "mol"^(-1) =-3092 KJ "mol"^(-1)` `H_(3) =(-3 xx(139 xx 3xx1)/(0.297)xx(10/(11)) + (3 xx (139 xx 1xx1)/(0.297 sqrt(3))xx (10)/(11)) KJ "mol"^(-1)` |
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| 3321. |
Since 1891 lighting lamps have been manufactured in the Netherlands. The improvement today in comparison to the first lamp is enormous , especially with the intorduction of the gas discharge lamps. The life- time has increased by orders of magnitude . the colour is also an important aspect. Rare earth metal compunds like `CeBr_(3)` are now included to reach a colour temperature of 6000K in the lamp the compounds are ionic solids at room temperature , and upon heating they sublime partially to give a vapour of neutral metal halide molecules . To achieve a high vapour pressure, the sublimation enthalpy should be as low as possible. Give a thermochemical cycle (law of Hess) for sublimation of `CeBr_(3)` , via vapour of mononucler ions (`H_(1) =H_("lattice"), H_(e) = H_("electrostatic"), H_(s) =H_("sublimation") , `H is not absolute , H means `DeltaH`) `H_(s)=-H_(1) + H_(e)` The lattice energy of the solid can be calculated using the Baron -Lande formula, `H_(1)=d(Z+Z-Ae^(2))/(r_(+) +r_(-))(1-(1)/(n))` The factor `Fe^(2)` (necessary in order to calculate the lattice is 2.985. The Borm exponent n is 11. The charge of the ions `Z_(+)` and `Z_(_)` are integer number (Z is negative ). For the calculation of the energy of gaseous `CeBr_(3)` (when formed from ions) the same Born- Lande formula can be used without A. The structure of `CeBr_(3)` in the gas phase is planer triangular . The radius of `Ce^(3+)` is 0.115 nm and of Br is 0.182nm. |
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Answer» Correct Answer - `(CeBr_(3))_("glucose") overset(-H_(1))rarr Ce^(3-) + 3Br^(-)` `Ce^(3+) + 3 Br overset(-He)rarr (CeBr_(3))_("molecules") ___________ `(CeBr_(3))_("tattice") overset(+H_(3)) rarr (CeBr_(3))_("molecules") " "H_(3) =- H_91) + H_(e)` The lattice energy of the solid cn be calculated using the Boron - Lande formula: `H_(1) = d(Z_(+)Z_(-)Ae^(2))/(r_(+)+R_(-))(1-(1)/(n))` The factor `Fe_(2)` (necessary in order to calculate the lattice energy in KJ `"mol"^(-1)` ) amounts of 139 when the ionic radii are substituted in nm . The Madelung constant A for the lattice is 2.985. The Born exponent n is 11 . The charge of the ions `Z_(+)` and `Z_(-)` are integer number (Z is negative) . For the calculation of the energy of gaseous `CeBr_(3)` (when formed from ions ) the same Born-Londe formula can be used without A. The structure of `CeBr_(3)` in the gas phase is planar triangular . The radius of `Ce^(3+)` is 0.115 nm and of Br is 0.182 nm. |
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| 3322. |
Since 1891 lighting lamps have been manufactured in the Netherlands. The improvement today in comparison to the first lamp is enormous , especially with the intorduction of the gas discharge lamps. The life- time has increased by orders of magnitude . the colour is also an important aspect. Rare earth metal compunds like `CeBr_(3)` are now included to reach a colour temperature of 6000K in the lamp the compounds are ionic solids at room temperature , and upon heating they sublime partially to give a vapour of neutral metal halide molecules . To achieve a high vapour pressure, the sublimation enthalpy should be as low as possible. Calculate the enthalpy of sublimation fo `CsCeBr_(4)` (in integers) Use the Born-Lande formula for all steps in the process and report the separate energies also (be aware of the signs). The `CeBr_(4)^(-)` anion is a teteahedron and in which the ratio between the edge and the destance between a corner of the tetrahedron and the centre of gravity (body - radius) amounts to `(2sqrt(6))//3 = 1.33` . The Born exponent of the CsBr is 11. The radius of Cs is 0.181 nm. |
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Answer» Correct Answer - Steps 1 : The lattice energy of `CsCeBr_(4)` with opposite sign is: `H_(1) =(139xx1xx1xx1.75)/(0.617) xx(10)/(11) KJ "mol"^(-1)` step 2: `H_(2) = 4 xx(139.3.1)/(0.297)(10)/(11) -6 xx(139 xx1xx1)/(0.297 xx (2)/(3)xx sqrt(6) xx(10)/(11) KJ "mol"^(-1) = 3543 KJ "mol"^(-1)` Step 3 : The electronic energy in the gas phase of `CeBr_(3)` is (see answer 8.2): `H_(3) =-3xx(139xx 1xx1)/(0.297 xx sqrt(3))xx(10)/(11) KJ "mol"^(-1) =- 3092 KJ "mol"^(-1)` Step 4 : The electrostatic energy in the gas phase of CsBr is `H_(4) =-(139xx1xx1)/(0.363) xx(10)/(11) KJ mol^(-1) =-348 KJ "mol"^(-1)` Total sum . `H_("total") = H_(1) + H_(2) + H_(3) +H_(4) =461 KJ "mol"^(-1)` |
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| 3323. |
The feasibility of a chemical reaction can be explained based on DH, DS and DG, so an swer the following : |
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Answer» Correct Answer - (A-q,t); (B-p,r); (C-r); (D-t) (a) `O_(2)` is more thermodynamically stable than `O_(3)` (c) 0 mol `rarr 3` mol (d) `0^(@)C` is the melting point of ice so ice so its melting is spontaneous at room temperature. |
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| 3324. |
Can a system be heated and its temperature ramains constant ? |
| Answer» Yes, this is possible when the entire heat supplied to the system is utilised in expansion. i.e., its working against the surrondings. | |
| 3325. |
One mole of an ideal gas with heat capacity at constant pressure `C_p` undergoes the process `T = T_0 + alpha V`, where `T_0` and `alpha` are constants. Find : (a) heat capacity of the gas as a function of its volume , (b) the amount of heat transferred to the gas, if its volume increased from `V_1` to `V_2`. |
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Answer» (a) Heat capacity is given by `C = C_V + (RT)/(V)(dV)/(dT) ` (see solution of `2.52`) We have `T = T_0 + alpha V` or, `V = (T)/(alpha) - (T_0)/(alpha)` After differentiating, we get, `(dV)/(dT) = (1)/(alpha)` Hence `C = C_V + (RT)/(V).(1)/(alpha) = (R)/(gamma - 1) +(R(T_0 + alpha V))/(V) .(1)/(alpha)` =`(R)/(gamma - 1)+ R((T_0)/(alpha V) + 1) = (gamma R)/(gamma - 1)+(RT_0)/(alpha V) = C_V + (RT)/(alpha V) = V_p + (RT_0)/(alpha V)` (b) Given `T = T_0 + alpha V` As `T = (pV)/( R)` for one mole of gas `p = (R)/(V) (T_0 + alpha V) = (RT)/(V) = alpha R` Now `A = int_(V_1)^(V_2) pdV = int_(V_1)^(V_2) ((RT_0)/(v) + alpha R) dV` (for one mole) =`RT_0 1n(V_2)/(V_1) + alpha (V_2 - V_1)` `Delta U = C_v (T_2 - T_1)` =`C_V[T_0 + alpha V_2 - T_0 alpha V_1] = alpha C_V (V_2 - V_1)` By the first law of thermodynamics `Q = Delta U + A` =`(alpha R)/(gamma - 1) (V_2 - V_1) + RT_0 1n (V_2)/(V_1) + alpha R (V_2 - V_1)` =`alpha R(V_2 - V_1)[1 + (1)/(gamma - 1)] + RT_0 1n (V_2)/(V_1)` . =`alpha C_p(V_2 - V_1) + RT_0 1n (V_2)/(V_1)` =`alpha C_p (V_2 - V_1) + RT_0 1n (V_2)/(V_1)`. |
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| 3326. |
An ideal gas has a molar heat capacity `C_v` at constant volume. Find the molar heat capacity of this gas as a function of its volume `V`, if the gas undergoes the following process : (a) `T = T_0 e^(alpha v)` , (b) `p = p_0 e^(alpha v)`, where `T_0, p_0`, and `alpha` are constants. |
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Answer» (a) By the first law of thermodynamics `dQ = dU + dA = vC_VdT + pd V` Molar specfic heat according to definition `C = (dQ)/(v dT) = (C_V dT + pdV)/(vdT)` =`(v C_V dT + (vRT)/(V) dV)/(v dT) = C_V + (RT)/(V) (dV)/(dT)`, We have `T = T_0 e^(alpha V)` After differentiating, we get `dT = alpha T_0 e^(alpha V). dV` So, `(dV)/(DT) = (1)/(alpha T_0 e^(alpha V))`, Hence `C= = C_V + (RT)/(V).(1)/(alpha T_0 e^(alpha V)) = C_V + (RT_0 e^(alpha V))/(alpha VT_0 e^(alpha v)) = C_v + (R)/(alpha V)` (b) Process is `p = p_0 e^(alpha V)` `P = (RT)/(V) = p_0 e^(alpha V)` or, `T = (p_0)/(R) e^(alpha v). V` So, `C = C_V + (RT)/(V) (dV)/(dT) = C_V + p_0 e^(alpha V).(R)/(p_0 e^(alpha V)(1 + alpha V)) = C_V + (R)/(1 + alpha V)`. |
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| 3327. |
For the case of an ideal gas find the equation of the process (in the variables `T, V`) in which the molar heat capacity varies as : (a) `C = C_V + alpha T` , (b) `C = C_V + beta V `, ( c) `C = C_v + ap` , where `alpha, beta` and `a` are constants. |
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Answer» Heat capacity is given by `C = C_V + (RT)/(V) (dV)/(dT)` (a) Given `C = C_V + alpha T` So, `C_V + alpha T = C_V + (RT)/(V)(dV)/(dT)`or, `(alpha)/(R) dT = (dV)/(V)` Integrating both sides, we get `(alpha)/(R)T = 1n C_0 = 1n VC_0 , C_0` is a constant. Or, `V. C_0 = e^(alpha T//R)` or `V . e^(alpha T//R) = (1)/(C_0) = constant` (b) `C = C_V + beta V` and `C = C_V + (RT)/(V) (dV)/(dT)` so, `C_V (RT)/(V) (dV)/(dT) = C_V + beta V` or, `(RT)/(V)(dV)/(dT) = beta V` or, `(dV)/(V^2) = (beta)/(R) (dT)/(T)` or, `V^-2 = (dT)/(T)` Integrating both sides, we get `(R)/(beta) (V^-1)/(beta - 1) = 1n T + 1n C_0 = 1n T.C_0` So, `1n T.C_0 = -(R)/(betaV) T. C_0 = e^(-R//beta V) ` or, `T e^(-R//beta V) = (1)/(C_0) = constant` ( c) `C = C_V + ap` and `C = C_V + (RT)/(V) (dV)/(dT)` So, `C_V + ap = C_V + (RT)/(V) (dV)/(dT)` so, `ap = (RT)/(V) (dV)/(dT)` or, `a (RT)/(V) = (RT)/(V) (dV)/(dT)` (as `p = (RT)/(V)` for one mole of gas) or, `(dV)/(dT) =a` ro, `dV = adT`or `dT = (dV)/(a)` So, `T = (V)/(a) + constant` or `V - aT = constant`. |
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| 3328. |
The specific heat of a gas in an isothermal process isA. infiniteB. ZeroC. NegativeD. Remains constant |
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Answer» Correct Answer - A |
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| 3329. |
A thermally insulated container is divided into two parts by a screen. In one part the pressure and temperature are `P` and `T` for an ideal gas filled. In the second part it is vacuum. If now a small hole is created in the screen, then the temperature of the gas will |
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Answer» Correct Answer - Remain same |
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| 3330. |
Two mole of an ideal monatomic gas are confined within a cylinder by a mass less spring loaded with a frictionless piston of negligible mass and crossectional area `4xx10^(-3)m^(2)`. The gas is heated by a heater for some time. During this time the gas expands and does `50J` of work in moving the piston through a distance of `0.01`m. The temperature of gas increases by `50`k. Change in internal energy of the gas isA. `1246.5 J`B. `124.65 J`C. `200 J`D. `12.46 J` |
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Answer» Correct Answer - A When the gas is heated it expands `&` pushes the piston by x. If k is force constanat of spring and `A` is area of crossection of the piston. It `P_(0)` is atmospheric pressure then at equilibrium of piston the pressure of the gas on the piston `P=P_(0)+(kx)/(A)` The increas in the volume of the gas by samll movement of x of piston is `dv=Adx` `W=int_(0)^(X)Pdv=P_(0)Ax+(1)/(2)kx^(2)` Putting `A=4xx10^(-3)m^(2),x=0.1m, W=50 J`. `P_(0)=1.013xx10^(5) Nm^(-2)` in above equation `:. k=1896 Nm^(-1)` |
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| 3331. |
In the cyclic process shown in the figure, the work done by the gas in one cycle is A. `28P_(1)V_(1)`B. `14P_(1)V_(1)`C. `18P_(1)V_(1)`D. `9P_(1)V_(1)` |
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Answer» Correct Answer - D |
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| 3332. |
An ideal gas is taken around the cycle `ABCA` shown in `P - V` diagram. The net work done by the gas during the cycle is equal to A. `12P_(1)V_(1)`B. `6P_(1)V_(1)`C. `3P_(1)V_(1)`D. `2P_(1)V_(1)` |
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Answer» Correct Answer - D |
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| 3333. |
The given figrue shows the variation of force exerted on the piston by an ideal gas (enclosed in a piston-cylinder arrangement) that undergoes a process during which piston position changes from 0.1 to 0.4 m. if the initial internal energy of the system is 3.25 J then find the net heat (in joule) absorbed during the process. |
| Answer» Correct Answer - 5 | |
| 3334. |
Statement-1 : In an adiabatic process, change in internal energy of a gas is equal to work done on/by the gas in the process. Statement-2 : This is because temp.of gas remains constant in an adiabatic process.A. If both, Assertion and Reason are true and Reason is the correct explanation of the Asserrion.B. If both,Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - C Internal energy of an ideal gas does not depends upon volume of gas, because there are no forces of attraction/repulsion amongst the molecules of an ideal gas. Statement-1 is true, but statement-2 is false. |
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| 3335. |
Statement-1: Temperature of a system can be increased without supplying heat. Statement-2: If the initial volume of system is equal to final volume in a process then work done is necessarily zero. Statement-3: If an adiabatic curve for a gas intersect with an isothermal curve for the same gas then at the point of intersection, the ratio of slope of the adiabatic and the isothermal curve is `gamma`.A. FTFB. TFTC. TFFD. TTT |
| Answer» Correct Answer - B | |
| 3336. |
Two glass discs of radius `R = 5.0 cm` were wetted with water and put together so that the thickness of the water layer between them was `h = 1.9 mu m`. Assuming the wetting to be complete. Find the force that has to be applied at right angles to the plates in order to pull them apart. |
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Answer» This is analogous to the previous prolem except that : `A = pi R^2` So `F = (2 pi R^2 alpha)/(h) = 0.6 kN`. |
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| 3337. |
Can we increase the temperature of a gas without supplying heat to it? |
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Answer» Yes, This happens when the gas undergoes, adiabatic compression. From `dQ=dU+dW` as `dQ=0` in an adiabatic process, `:. dU+dW=0` or `dU= -dW` In compression, work is done on the gas, therefore, `dW` is negative. Hence, `dU is +, i.e., internal energy of the gas increases. Hence, temperature of gas increases. |
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| 3338. |
Find the attractive force is newton between two parallel glass plates, separated by a distance `h=0.1mm` after a water drop of mass `m=70mg` was introduced between them. Assume wetting to be complete and surface tension of water, `T=70` dyne/cm |
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Answer» The pressure inside the film is less than that outside by an amount `alpha((1)/(r_1) +(1)/(r_2))` where `r_1` and `r_2` are the principle radii of curvature of the meniscus. One of these is small being given by `h = 2 r_1 cos theta` while the other is large and will be ignored. Then `F ~~ (2 A cos theta)/(h) alpha` where `A =` area of the water film between the plates. Now `A = (m)/(rho h)` so `F = (2 m alpha)/(rho h^2)` when `theta` (the angle of contact) `= 0`. |
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| 3339. |
The adiabatic relation between P and T isA. `P^(1-gamma).T^(gamma)=` constantB. `PT^(gamma)=` constantC. `P^(gamma)T^(1-gamma)=` constantD. `P^(gamma)T^(gamma-1)=` constant |
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Answer» Correct Answer - A Some of the examples of isothermal changes |
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| 3340. |
Standard entropy of `X_(2)` , `Y_(2)` and `XY_(3)` are `60, 40 ` and `50JK^(-1)mol^(-1)` , respectively. For the reaction, `(1)/(2)X_(2)+(3)/(2)Y_(2)rarrXY_(3),DeltaH=-30KJ` , to be at equilibrium, the temperature will be:A. 1250 KB. 500 KC. 1000 KD. 750 K |
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Answer» Correct Answer - D `DeltaS=S_(P)-S_(R), DeltaS =(DeltaH)/(T)` |
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| 3341. |
In conversion of lime-stone to lime, `CaCO_(3(s)) to CaO_((s)) + CO_(2(g))` the values of `DeltaH^@` and `DeltaS^@` are `+179.1 kJmol^(-1)` and `160.2 J//K` respectively at `298K` and 1 bar. Assuming that `DeltaH^@` and `DeltaS^@` do not change with temperature, temperature above which conversion of lime-stone to lime will be spontaneous is :A. 845 KB. 1118 KC. 1008 KD. 1200 K |
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Answer» Correct Answer - B `Delta G = Delta H - T Delta S lt 0` for spontaneity |
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| 3342. |
The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of `10 dm^(3)` to a volume of `100dm^(3)` at `27^(@)C` isA. `35.8 J mol^(-1) K^(-1)`B. `32.3 J mol^(-1) K^(-1)`C. `42.3 J mol^(-1) K^(-1)`D. `38.3 J mol^(-1) K^(-1)` |
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Answer» Correct Answer - D `DeltaS=2.303 nR "log" V_(2)/V_(1)` |
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| 3343. |
Work done by the system in isothermal reversible process is `w_(rev.)= -2.303 nRT "log"(V_(2))/(V_(1))`. Also in case of adiabatic reversible process work done by the system is given by: `w_(rev.) = (nR)/(gamma -1) [T_2 - T_1]`. During expansion disorder increases and the increase in disorder is expressed in terms of change in entropy `DeltaS = q_(rev.)/T`. The entropy changes also occurs during transformation of one state to other end expressed as `DeltaS = DeltaH/T`. Both entropy and enthalpy changes obtained for a process were taken as a measure of spontaniety of process but finally it was recommended that decrease in free energy is responsible for spontaniety and `DeltaG=DeltaH - T DeltaS`. `Ag_2O_((s)) to 2Ag_((s)) + 1/2 O_(2(g))` attains equilibrium at temperature…`K` is : (The `DeltaH` and `DeltaS` for the reaction are `30.5kJ mol^(-1)` and `66J mol^(-1) K^(-1)` )A. `462.12`B. 237C. 373D. 273 |
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Answer» Correct Answer - A At eq, `Delta G = 0 rArr Delta H - T Delta S = 0` `T = (Delta H)/(Delta S)` |
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| 3344. |
In conversion of lime-stone to lime, `CaCO_(3(s)) to CaO_((s)) + CO_(2(g))` the values of `DeltaH^@` and `DeltaS^@` are `+179.1 kJmol^(-1)` and `160.2 J//K` respectively at `298K` and 1 bar. Assuming that `DeltaH^@` and `DeltaS^@` do not change with temperature, temperature above which conversion of lime-stone to lime will be spontaneous is :A. 1118 KB. 1008 KC. 1200 KD. 845 K |
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Answer» Correct Answer - A `DeltaG^(@)=DeltaH^(@)-T DeltaS^(@)` |
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| 3345. |
Work done by the system in isothermal reversible process is `w_(rev.)= -2.303 nRT "log"(V_(2))/(V_(1))`. Also in case of adiabatic reversible process work done by the system is given by: `w_(rev.) = (nR)/(gamma -1) [T_2 - T_1]`. During expansion disorder increases and the increase in disorder is expressed in terms of change in entropy `DeltaS = q_(rev.)/T`. The entropy changes also occurs during transformation of one state to other end expressed as `DeltaS = DeltaH/T`. Both entropy and enthalpy changes obtained for a process were taken as a measure of spontaniety of process but finally it was recommended that decrease in free energy is responsible for spontaniety and `DeltaG=DeltaH - T DeltaS`. A chemical change will definitely be spontaneous if:A. `Delta H = -ve, Delta S = -ve` and low temperatureB. `Delta H = +ve, Delta S = -ve` and high temperatureC. `Delta H = -ve, Delta S = +ve` and any temperatureD. `Delta H = +ve, Delta S = +ve` and `T Delta S lt Delta H` |
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Answer» Correct Answer - C Conditions for spontaniety |
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| 3346. |
Work done by the system in isothermal reversible process is `w_(rev.)= -2.303 nRT "log"(V_(2))/(V_(1))`. Also in case of adiabatic reversible process work done by the system is given by: `w_(rev.) = (nR)/(gamma -1) [T_2 - T_1]`. During expansion disorder increases and the increase in disorder is expressed in terms of change in entropy `DeltaS = q_(rev.)/T`. The entropy changes also occurs during transformation of one state to other end expressed as `DeltaS = DeltaH/T`. Both entropy and enthalpy changes obtained for a process were taken as a measure of spontaniety of process but finally it was recommended that decrease in free energy is responsible for spontaniety and `DeltaG=DeltaH - T DeltaS`. Which statements are correct? (1) The expansion work for a gas into a vacuum is equal to zero. (2) `1` mole of a gas occupying `3` litre volume on expanding to `15` litre at constant pressure of `1atm` does expansion work `1.215 kJ`. (3) The maximum work done during expansion of `16gO_2` at `300K` from `5dm^3` to `25 dm^3` is `2.01 kJ`. (4) The `DeltaS` for `S to L` is almost negligible in comparision to `DeltaS` for `L to G`. (5) `DeltaS = 2.303 nR "log"(V_(2))/(V_(1)).` (at constant `T`)A. 2,3,4,5B. 1,2,3,4,5C. 1,2D. 4,5 |
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Answer» Correct Answer - B (1) In free expansion `P_(ext) = 0 rarr w = 0` (2) `W = -1 [15-3] = -12lit-atm` `= -12 xx 101.3 J` (3) `W_(rev) = -2.303 xx (16)/(32) xx 8.314 xx 300 xx log.(25)/(5)` (4) `Delta S` for `L rarr G` is more (5) `(Delta S)_(T) = 2.303nR log.(v_(2))/(v_(1))` |
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| 3347. |
The standard enthalpy of formation of `NH_(3)` is `-46.0KJ mol^(-1)` . If the enthalpy of formation of `H_(2)` from its atoms is `-436KJ mol^(-1)` and that of `N_(2)` is `-712KJ mol^(-1)` , the average bond enthalpy of `N-H` bond in `NH_(3)` isA. `-96 kJ mol^(-1)`B. `+ 352 kJ mol^(-1)`C. `+1056 kJ mol^(-1)`D. `-1102 kJ mol^(-1)` |
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Answer» Correct Answer - B `N_(2)+3H_(2) rarr 2NH_(3)` |
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| 3348. |
Explain state function. |
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Answer» State function:- State functions are the fundamental properties which determine the state of system. Examples of state functions are pressure (p), Volume (V), Enthalpy (H), Free energy (G), Temperature (T), Internal energy (E), Entropy (S) and number of moles (n). |
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| 3349. |
Show that in the isothermal expansion of an ideal gas , `DeltaU = 0` and `DeltaH = 0` |
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Answer» (a) For one moole of an ideal gas, `C_(v) = ( dU)/( dT) ` or ` dU =C_(v) dT` For a finite change, `DeltaU = C_(v) DeltaT` For isothermal process, `T=` constant so that`DeltaT =0`. Hence, `DeltaU = 0` (b) `DeltaH = DletaU + Delta(PV) =DeltaU+ Delta(RT) =DeltaU+ R DeltaT` But `DeltaU =0` ( proved above) and `DeltaT = 0` ( for isothermal process) , `:. DeltaH =0` |
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| 3350. |
The expression for the entropy change when a gas expands from volume `V_(1)` to volume `V_(2)` at constant temperature T is `DeltaS_(T)= " …..................." ` |
| Answer» `DeltaS_(T) =Rln . (V_(2))/(V_(1))` | |