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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.
| 3651. |
On the basis of the following thermochemical data `:` `(Delta_(f)G^(@)H_((aq.))^(+)=0)` `H_(2)O_((l))rarr H_((aq.))^(+)+OH_((aq.))^(-),DeltaH=57.32kJ` `H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l)),DeltaH=-286.20kJ` The value of enthalpy of formation of `OH^(-)` ion at `25^(@)C` is `:`A. `-22.88 kJ`B. `-228.88 kJ`C. `+228.88 kJ`D. `-343.52 kJ` |
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Answer» Correct Answer - B `H_(2)+1/2 O_(2) rarr H^(+)+OH^(-) Delta H=-228.88 kJ` |
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| 3652. |
On the basis of the following thermochemical data `:` `(Delta_(f)G^(@)H_((aq.))^(+)=0)` `H_(2)O_((l))rarr H_((aq.))^(+)+OH_((aq.))^(-),DeltaH=57.32kJ` `H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l)),DeltaH=-286.20kJ` The value of enthalpy of formation of `OH^(-)` ion at `25^(@)C` is `:`A. `- 228.88 kJ`B. `+ 228.88 kJ`C. `- 22.88 kJ`D. `- 343.52 kJ` |
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Answer» Correct Answer - A According to thermodynamics, we have `Delta_(r) H^(@) = sum a_(i) Delta_(f) H^(@)` )product) `- sum b_(i) Delta_(g) H^(@)` (reactants Applying this relation ship of first equation, we have `Delta_(r) H^(@) = [Delta_(f) H^(@) (H^(+), aq.) + Delta_(f) H^(@) (OH^(-), aq.)] - Delta_(f) H^(@) (H_(2) O, 1)` `57.32 = [0 + Delta_(f) H^(@) (OH^(-) , aq.)]- (286.20)` `:. Delta_(f) H^(@) (OH^(-), aq.) = 57.32 - 286.20 = - 228.88 kJ` |
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| 3653. |
On the basis of the following thermochemical data `:` `(Delta_(f)G^(@)H_((aq.))^(+)=0)` `H_(2)O_((l))rarr H_((aq.))^(+)+OH_((aq.))^(-),DeltaH=57.32kJ` `H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l)),DeltaH=-286.20kJ` The value of enthalpy of formation of `OH^(-)` ion at `25^(@)C` is `:`A. `+228.88 kJ`B. `-343.52 kJ`C. `-22.88 kJ`D. `-228.88 kJ` |
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Answer» Correct Answer - D `H_(2)O_((l))rarr H^(+)(aq)+OH^(-)(aq) , Delta H=57.32` `Delta H=[Delta H_(f)(H^(+))+Delta H_(f)(OH^(-)]-Delta H_(f)(H_(2)O)` From eq. (ii) `Delta H_(f)(H_(2)O)=-286.20 kJ` Given `Delta H_(f)(H^(+)) = 0` `therefore 57.32 =0+Delta H_(f)(OH^(-))-(-286.20)` `Delta H_(f)=-228.88 kJ` |
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| 3654. |
On the basis of the following thermochemical data `:` `(Delta_(f)G^(@)H_((aq.))^(+)=0)` `H_(2)O_((l))rarr H_((aq.))^(+)+OH_((aq.))^(-),DeltaH=57.32kJ` `H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l)),DeltaH=-286.20kJ` The value of enthalpy of formation of `OH^(-)` ion at `25^(@)C` is `:`A. `-228.88 KJ`B. `+228.88 KJ`C. `-343.52 KJ`D. `-22.88 KJ` |
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Answer» Correct Answer - A `H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)O(l)" " =-286.20 KJ` `DeltaH_(r)=DeltaH_(f)(H_(2)O,l) = DeltaH_(f)(H_(2),g) -(1)/(2)DeltaH_(r) (O_(2),g)` `-286.20 = DeltaH_(f)(H_(2)O(l))` So `DeltaH-=_(f) (H_(2)O,l) =-286.20 KJ//"mole"` `H_(2)O(l) rarr H^(+) (aq) + OH^(-)(aq)" "DeltaH = 57.32 KJ` `DeltaH_(r)= DeltaH_(f)^(@) (H^(+),aq) + DeltaH_(f)^(@)(OH^(-),aq) -DeltaH_(f)^(@)(H_(2)O,l)` `57.32 = 0+ DeltaH_(f)^(@)(OH^(-),aq) -(-286.20)` `DeltaH_(f)^(@)(OH^(-), aq) = 57.32 - 286.20 =-228.88 KJ`. |
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| 3655. |
Which molecules, ice at `10^(@)C` or water at `0^(@)C` have greater potential energy and why? |
| Answer» The potential energy of water molecule at `0^(@)C ` is more, because heat spends in melting is used up in increasing the `P-E`. | |
| 3656. |
A system goes from A and B via two processes. I and II as shown in figure. If `DeltaU_1 and DeltaU_2` are the changes in internal energies in the processes I and II respectively, then A. `DeltaU_(2) lt DeltaU_(1)`B. `DeltaU_(2) gt DeltaU_(1)`C. relation between `DeltaU_(1)` and `DeltaU_(2)` cannot be determinedD. `DeltaU_(1)=DeltaU_(2)` |
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Answer» Correct Answer - D As change in internal energy does not depend upon the actual path followed between the two given points on `P-V` diagram therfore, `DeltaU_(1)=DeltaU_(2)` |
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| 3657. |
A system goes from A and B via two processes. I and II as shown in figure. If `DeltaU_1 and DeltaU_2` are the changes in internal energies in the processes I and II respectively, then |
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Answer» As initial and final states A and B for the two processes are the same for two processes and change in internal energy is independent of the path, `:. dU_(1)=dU_(2)` |
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| 3658. |
A system goes from A and B via two processes. I and II as shown in figure. If `DeltaU_1 and DeltaU_2` are the changes in internal energies in the processes I and II respectively, then A. `Deltau_(II)gtDeltaU_(I)`B. `Deltau_(II)ltDeltaU_(I)`C. `Deltau_(II)=DeltaU_(I)`D. Relation between `DeltaU_(I)` and `DeltaU_(II)` can not be determined |
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Answer» Correct Answer - C |
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| 3659. |
What would be the heat released when:A. `0.5` mol of `HCI` is neutralised with `0.5` mol of `NaOH`B. `0.5` mol of `HNO_(3)` is neutralised with `0.3` mol of `NaOH`C. `100 ml` of `0.2M HCI` + `200 ml` of `0.2 M KOH`D. `200 ml` of `0.1 M H_(2)SO_(4)` + `150 ml` of `0.2 M KOH` |
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Answer» a. `0.5 mol` of `HCI` is neutralised by `0.5mol` of `NaOH` `:. DeltaH =- 57.1 xx 0.5 =- 28.55 kJ` b. `0.3 mol` of `HNO_(3)` will neutralise `0.3` mol of `NaOH`. `:. DeltaH =- 57.1 xx 0.3 =- 17.1 kJ` c. `HCI = 100 xx 0.2 = 20 mmol or 20 mEq` `KOH = 200 xx 0.2 = 40 mmol or 40 mEq` `:. 20 mEq` will neutralise or `20 xx 10^(-3) Eq` will neutralise `DeltaH = (-57.1 xx 20 xx10^(-3)) =- 1.14 kJ` d. `H_(2)SO_(4) = 200 xx 0.1 xx2` (since `H_(2)SO_(4)` is diabasic) `= 40 mEq` `KOH = 150 xx 0.2 = 30 mEq` `:. 30 mEq or 30 xx 10^(-3) Eq` will neutralise `:. DeltaH =- 57.1 xx 30 xx 10^(-3) =- 1.713 kJ` |
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| 3660. |
Given: `NaCI(s) +aq rarr Na^(o+) (aq) +CI^(Theta) DeltaH = 3.9 kJ` `Na^(o+)(g) +CI^(Theta)(g) rarr NaCI(s) DeltaH =- 788 kJ` `CI^(Theta)(g)+aq rarr CI^(Theta)(aq) DeltaH =- 394.1 kJ` Calculate the enthalpy of hydration of `Na^(o+)` ions. |
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Answer» `Na^(o+) (g) +aq rarr Na^(o+) (aq) DeltaH = ?` `DeltaH = DeltaH_(1) +DeltaH_(2) - DeltaH_(3)` `= 3.9 - 788 -(-394.1)` `= 3.9 - 788 +394.1 =- 390.0 kJ` |
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| 3661. |
`71g` of chlorine gas is allowed to expand freely into vacuum Calculate w.q `DeltaU` and `DeltaH` . |
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Answer» `P_(ex) =0, w =- P_(ex) (V_(2)-V_(1)) =- 0 (V_(2)-V_(1)) = 0` As tempertaure is not given assume it as constant. `:. DeltaU = 0, DeltaU =q +w, q =0` `0 =q+0 DeltaH = 0` |
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| 3662. |
Given `S^@` for `C_("Graphite"), H_(2(g))` and `CH_(4(g))` are `5.70, 130.7` and `186.3 JK^(-1) mol^(-1)`. Also standard heat of formation of `CH_4` is `-74.81kJmol^(-1).` Calculate the standard Gibbs energy change for the formation of methane at `298K.` `C_("Graphite")+2H_(2(g))toCH_(4(g))` |
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Answer» `Delta_(f)H^(@)CH_(4)=H_(CH_(4))^(@)=-74.81kJmol^(-1)` Also `H^(@)` for `C_("graphite")` and `H_(2(g))=0` (standard state of elements) `C_("graphite")+2H_(2(g))toCH_(4(g)),` `:. DeltaH^(@)=DeltaH_(CH_(4))^(@)-H_(C_("graphite"))^(@)-2xxH_(H_(2))^(@)` `=74.81` Also `DeltaS^(@)` (for the reaction) `=S_(CH_(4))^(@)-S_(C_("graphite"))^(@)-2xxS_(H_(2))^(@)` `=186.3-5.70-130.7xx2=-80.8JK^(-1)mol^(-1)` Using `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `=-74.81-298xx[-80.8xx10^(-3)]` `=-50.73 kJ` |
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| 3663. |
In a spontaneous irreversible process the total entropy of the system and surroundingsA. Remains constantB. IncreasesC. DecreasesD. Zero |
| Answer» Correct Answer - B | |
| 3664. |
Which of the following expression for entropy change of an irreversible process ?A. `dS gt (dq)/(T)`B. `dS = (dq)/(T)`C. `dS lt (dq)/(T)`D. `dS = (dU)/(T)` |
| Answer» Correct Answer - A | |
| 3665. |
The temperature of an ideal gas increases in an:A. adiabatic expansionB. isothermal expansionC. adiabatic compressionD. isothermal compression |
| Answer» Correct Answer - C | |
| 3666. |
If the door of a refrigerator is kept open, then which of the following is trueA. gets cooledB. gets heatedC. neigther gets cooled nor gets heatedD. gets cooled or heated depending on the initial temperature of the room |
| Answer» Correct Answer - B | |
| 3667. |
In the following P-V diagram two adiabatics cut two isothermals at temperature `T_(1) and T_(2)` (fig). The value of `V_(a)/V_(d)` will be A. `(V_(b))/(V_(c))`B. `(V_(c))/(V_(b))`C. `(V_(d))/(V_(a))`D. `V_(b)V_(c)` |
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Answer» Correct Answer - A |
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| 3668. |
An ideal gas can be expanded form an initial state to a certain volume through two different processes `PV^(2) =` constant and (ii) `P = KV^(2)` where `K` is a positive constant. ThenA. Final temperature in (i) will be greater than in (ii)B. Final temperature in (ii) will be greater then in (i)C. Total heat given to the gas in (i) case is greater than in (ii)D. Total heat is given to the gas in (ii) case is greater than in (i) |
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Answer» Correct Answer - B,D |
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| 3669. |
Suppose that the voulme of a certain ideal gas is to be doubled by one the following processes : (1) isothermal expansion (2) adiabatic expansion (3) free expansion in isulated condition (4) expansion at constant pressure . If `E_(1),E_(2) ,E_(3) and E_(4)` respectively are the changes in average kinetic energy of the molecules for the four processes, them :A. `E_(2)=E_(3)`B. `E_(1)=E_(3)`C. `E_(1)gtE_(4)`D. `E_(4)gtE_(3)` |
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Answer» Correct Answer - b,d |
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| 3670. |
An adiabatic process occurs at constantA. TemperatureB. PressureC. HeatD. Temperatuer and Pressure |
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Answer» Correct Answer - C |
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| 3671. |
A gas at NTP is suddenly compressed to one-fourth of its original volume. If. `lambda` is supposed to be `(3)/(2)`, then the final perssure isA. 4 atmosphereB. `(3)/(2)` atomsphereC. 8 atmosphereD. `(1)/(4)` atmosphere |
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Answer» Correct Answer - C |
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| 3672. |
A gas at NTP is suddenly compressed to one-fourth of its original volume. If. `lambda` is supposed to be `(3)/(2)`, then the final perssure is |
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Answer» Correct Answer - 8atm |
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| 3673. |
A polyatomic gas `(lambda=(4)/(3))` is compressed to `(1)/(8)` of its voulme Adiabatically. If initial pressure is `P_(0)`, Its mew perssure will beA. `8P_(0)`B. `16P_(0)`C. `6P_(0)`D. `2P_(0)` |
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Answer» Correct Answer - B |
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| 3674. |
A sample of certain mas s of an ideal polyatomic gas is expanded against constant pressure of 1 atm adiabatically from volume 2L , pressure 6 atm and temperature 300K to state where its final volume is 8L . Then calculate entropy change (in J/k) in the process.(Neglect vibrational degress of freedom)[1 L atm = 100 J ,log 2 =0.3 , log 3 = 0.48 , log e = 2.3] (approximate integer) |
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Answer» Correct Answer - 6 `W=-P_("ext")(V_(f)-V_(i))` =- (1 atm) (8-2)L =- 6L atm as q=0 so `DeltaE=w " " implies " " 3(8P_(f) -12)=-6` `8P_(f) =12 -(6)/(3)=10 implies P_(f)=(5)/(4) atm` so `(T_(f))/(T_(i))=((5)/(4)xx3)/(6xx2)=(10)/(12)` so `DeltaS =3(12)/(300) "In" (10)/(12) + (12)/(300) "In" 8` `=(3xx12)/(300) "In" ((5)/(6)xx2)` `=(12)/(100) "In" ((5)/(3)) xx 100J` `=12 ("In" 5-"In"3)= 12 xx 2.3 xx(0.7 - 0.48)` `=12xx 2.3 xx 0.22 =6.072 J//K` |
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| 3675. |
`10` litre of a monoatomic ideal gas at `0^(@)C` and `10 atm` pressure is suddenly released to `1 atm` pressure and the gas expands adiabatically against this constant pressure. The final volume `(L)` of the gas. |
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Answer» Correct Answer - 64 L This is adiabatic irreversible process, so for this process `PV^(gamma)=` Constant, is not applicable `W= -P_(ext)(V_(2)-V_(1))` But for adiabatic process `W=dU=((P_(2)V_(2)-P_(1)V_(1))/(gamma-1))` `PV=nRT " "rArr10xx10=nxx0.082xx273rArrn=4.47` moles `-P_(ext)(V_(2)-V_(1))=((P_(2)V_(2)-P_(1)V_(1))/(gamma-1)) " "rArr-1xx(V_(2)-10)=(1xxV_(2)-10xx10)/(1.67-1)` `rArr (10-V_(2))=(V_(2)-100)/(0.67) " "6.7-0.67 V_(2)=V_(2)-100` `rArr 106.7V_(2) " " V_(2)=64` |
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| 3676. |
10 litre of a non linear polyatomic ideal gas at `127^(@)C` and 2 atm pressure is suddenly released to 1 atm pressure and the gas expanded adiabatically against constant external pressure. The final temperature and volume of the gas respectively are.A. T=350K,V = 17.5LB. T = 300 K ,V = 15 LC. T = 250 K, V = 12.5 LD. None of these |
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Answer» Correct Answer - A `DeltaU=w` `nxx(R )/(gamma-1)(T_(2)-T_(1))=-P_("ext")((nRT_(2))/(P_(2))-(nRT_(1))/(P_(1)))` `3(T_(2)-T_(1))=-1((T_(2))/(1)-(T_(1))/(2)]` `3T_(2)-3T_(1)=-T_(2)+(T_(1))/(2)` `4T_(2)=(7T_(1))/(2)` `T_(2)=(7T_(1))/(8)=(7xx400)/(8)=350 K` `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))implies(2xx10)/(400)=(1xxV_(2))/(350)impliesV_(2)=17.5 L` |
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| 3677. |
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=DeltaU+PDeltaV+VDeltaP` when P and V both changes |
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Answer» Correct Answer - A::B::C `DeltaH=DeltaU+P.DeltaU+V.DeltaP+DeltaP.DeltaV` is correct relation. |
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| 3678. |
Which of the following conditions may lead to a non-spontaneous change?A. `DeltaH " and "DeltaS " both "+ve`B. `DeltaH=-ve,DeltaS=+ve`C. `DeltaH=+ve,DeltaS=-ve`D. `DeltaH=-ve,DeltaS=-ve` |
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Answer» Correct Answer - A::C::D `DeltaG=DeltaH-TDeltaS` `(a) DeltaH=+ve,DeltaS=+ve` at low temperature `DeltaG=+ve` (b) `DeltaH=-ve,DeltaS=+ve, DeltaG=-ve` at any temperature. `(c )DeltaH=+ve, DeltaS=-ve, DeltaG=+ve` at any temperature (d) `DeltaH=-ve, DeltaS=-ve, DeltaG=+ve` at high temperature |
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| 3679. |
Calculate average molar heat capacity at constant volume of gaseous mixture contained 2 mole of each of two ideal gases `A(C_(v,m)=(3)/(2)R)` and `B(C_(v,m)=(5)/(2)R) :`A. RB. 2RC. 3RD. 8R |
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Answer» Correct Answer - B Average `C_(v,m)=(n_(1)C_(v,m_(1))+n_(2)C_(v,m_(2)))/(n_(1)+n_(2))` `=(2xx(3)/(2)R+2xx(5)/(2)R)/(2+2)=2R` |
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| 3680. |
0.5 mole each of two ideal gases`A (C_(v,m)=(5)/(2)R)` and `B (C_(v,m)=3R)` are taken in a container and expanded reversibly and adiabatically, during this process temperature of gaseous mixture decreased from 350 K to 250 K. Find `DeltaH` (in cal/mol) for the process :A. `-100 R`B. `-137.5 R`C. `-375 R`D. None of these |
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Answer» Correct Answer - C `DeltaH=(n_(1)C_(p,m_(1))+n_(2)C_(p,m_(2)))DeltaT` `=(0.5 xx(7)/(2)R+0.5xx4R)(-100)` `=-375 R` |
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| 3681. |
If `DeltaQ`and ` represent the heat supplied to the system and the work done on the system respectively, then the first law of thermodynamics can be written asA. `DeltaQ=DeltaU+DeltaW`B. `DeltaQ=DeltaU-DeltaW`C. `DeltaQ=DeltaW-DeltaU`D. `DeltaQ=DeltaW+DeltaU` |
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Answer» Correct Answer - B |
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| 3682. |
`DeltaHltDeltaU` for the reaction(s) :A. `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`B. `Ag_(2)O(s)rarr2Ag(s)+(1)/(2)O_(2)(g)`C. `CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g)`D. `C(s)+O_(2)(g)rarrCO_(2)(g)` |
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Answer» Correct Answer - A::C `DeltaH=DeltaU+Deltan_(g)RT` `(a) Deltan_(g)=-2" "DeltaUgtDeltaH` `(b) Deltan_(g)=+(1)/(2)" "DeltaUgtDeltaH` `(c )Deltan_(g)=-(1)/(2)" "DeltaU gt DeltaH` `(d) Deltan_(g)=0 " "DeltaU=DeltaH` |
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| 3683. |
A scientist says that the efficiency of his heat engine which operates at source temperature `127^(@)C` and sink temperature `27^(@)C is 26%`, thenA. It is impossibleB. It is possible but less probableC. It is quite probableD. Data are incomplete |
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Answer» Correct Answer - B |
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| 3684. |
The triple point of water is standard fixed point in modern thermometry. why? what is wrong in taking the melting point of ice and the boiling point of water as standard fixed points as was originally done in the Celsius scale? |
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Answer» It is because of the fact that triple point of substance unique, i.e., it occurs at one particular set of values of pressure and temperature. The melting point of ice and boiling point of water are not unique. They change with the change in value of pressure of due to the presence of impurities in water. |
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| 3685. |
The different in the work done when one mole of `Al_(4)C_(3)(s)` reacts with water in a closed vesselat `27^(@)C` against atmospheric pressure and that in an open vessel under the same conditions isA. greater in the open vessel by 600 calB. greater in the closed vessel by 600 calC. greater in the open vessel by 1800 calD. greater in the closed vessel by 1800cal |
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Answer» Correct Answer - C Reaction occuringis `:` `underset(1 mol)(Al_(4)C_(3) (s))+ 12 H_(2)O(l) rarr 4Al(OH)_(3) +underset(3 mol)(3CH_(4))` When the vessel is closed,volume `=` constant , i.e., `DeltaV = 0`.Hence,`w=- PDeltaV =0` When vessle is open, initial volume, `V_(1) =0` ( as no gas is present) Final volume , `V_(2) = `Volume of3 mole of `CH_(4)` `PV_(2) = nRT` `:.V_(2)= (nRT)/(P)` `:. w = - P(V_(2)-V_(1))= -PV_(2)` `= - P xx(nRT)/(P) = - nRT` `= - 3 xx 2 xx 300 = - 1800 cal` i.e., work isdone by the system. |
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| 3686. |
`DeltaU^(@)` of combustion of methane is `-X kJmol^(-1)`. The value of `DeltaH^(@)` is `:` (i) `= DeltaU^(@)` (ii) ` gt DeltaU^(@)` (iii) `lt DeltaU^(@)` (iv)` = 0` |
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Answer» The balanced equation for combustion of methane will be `CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l)` Thus, `Deltan_(g)= (n_(p) -n_(r))_(g)= 1 -3= -2` `DeltaH^(@) lt DeltaU^(@) + Deltan_(g) RT = - X - 2RT` Thus, `DeltaH^(@) lt DeltaU^(@)` . Hence, (iii) is the correct answer. |
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| 3687. |
It w is the amount of work by the system and q is the amount of heat supplied to the system, identify the type of the system.A. isolated systemB. closed systemC. open systemD. system with thermally conducting walls |
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Answer» Correct Answer - b When a system absorbs heat and does work, it is closed system. |
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| 3688. |
Changes in a system from an initial state to the final state were made by a different manner that `DeltaH` remains same but q changes becauseA. `DeltaH` is a path function but q is astate functionB. `DeltaH` is a state function and q is a path functionC. Both `DeltaH` andq are state functionD. Both `DeltaH` and q are path function |
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Answer» Correct Answer - b As `DeltaH` remains same by different paths,itisa state function. As q changes by different paths, it is a path function. |
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| 3689. |
If the enthalpy of combustion of diamond and graphite are – 395.4 mol-1 and -393.6kJ mol-1. What is the enthalpy change for the C (graphite) → C (diamond)? |
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Answer» C(diamond) + O2(g) → CO2(g) ΔH = -395.4 kJmol-1 …(1) C(graphite) + O2(g) → CO2(g) ΔH = -393.6 kJmol-1 … (2) C(graphite) → C(diamond ) substracting (1) from (2), we get C(graphite) → C(diamond) ΔH = -393.6 kJ – (-395.4 kJ), ΔH = +1.8 kJ mol-1 |
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| 3690. |
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are `- 890 .3 kJmol^(-1), -393.5kJ mol^(-1)` and `- 28.5 kJ mol^(-1)` respectively. Enthalpy of formation of `CH_(4)(g)` will be `:` (i) `-74.8 kJ mol^(-1)` (ii) `-52.27 kJ mol^(-1)` (iii) `+74.8 kJ mol^(-1)` (iv) `+ 52 .26 kJ mol^(-1)` |
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Answer» Given `: ` (i) `CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l), DeltaH = - 890 . 3 kJ mol^(-1)` (ii) ` C(s) + O_(2)(g)rarr CO_(2)(g) , DeltaH= -393.5 kJ mol^(-1)` (iii) `H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)O(l), DeltaH = - 285.8 kJ mol^(-1)` Aim `: C(s) + 2H_(2)(g) rarr CH_(4)(g) , DeltaH = ?` Eqn. (ii) `+ 2 xx` Eqn. (iii) - Eqn. (i) gives the required equation with `DeltaH = - 393.5 +2 ( - 285.8) - ( -890.3 ) kJ mol^(-1)` `= - 74.8 kJ mol^(-1)` Hence, (i) is the correct answer. |
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| 3691. |
If the enthalpy of combustion of diamond and graphite are `-395.4 kJ mol^(-1)` and `-393.6 kJ mol^(-1)`, what is enthalpy change for the `C("graphite") rarr C("diamond")`? |
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Answer» `C("diamond") +O_(2)(g) rarr CO_(2)(g) DeltaH =- 395.4 kJ mol^(-1)………(i)` `C("graphite") +O_(2)(g) rarr CO_(2)(g) DeltaH =- 393.6 kJ mol^(-1) …..(ii)` `C("graphite") rarr C ("diamond")` Subtracting equation (i) from equation (ii), we get `C("graphite") rarr C("diamond")` `DeltaH = -393.6 kJ -(-395.4 kJ)` `DeltaH = +1.8 kJ mol^(-1)`. |
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| 3692. |
Two carnot engines A and B are operated in series. The first one A receives heat at `900K` and rejects it to a reservoir at `T K`. The second engine B receives the heat rejected by the first engine and rejects it to a heat reservoir at `400K`. Calculate the value of `T`, when the efficiency ot two engines is the same. |
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Answer» Correct Answer - `600K` Here, `eta=(1-T/(900))=(1-(400)/T)` `:. T/(900)=(400)/T or T=sqrt(36xx10^(4))=600K` |
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| 3693. |
Two Carnot engines are operated in succession. The first engine receives heat from a source at `T=800K` and rejects to sink at `T_(2)K`. The second engine receives heat rejected by the first engine and rejects to another sink at `T_(3)= 300K`. If work outputs of the two engines are equal, then find the value of `T_(2)`.A. `100K`B. `300K`C. `550K`D. `700K` |
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Answer» Correct Answer - C `eta_(A)=(T_(1)-T_(2))/(T_(1))=(W_(A))/(Q_(1)),eta_(B)=(T_(2)-T_(3))/(T_(3))=(W_(B))/(Q_(2))` `(Q_(1))/(Q_(2))=(T_(1))/(T_(2))xx(T_(2)-T_(3))/(T_(1)-T_(2))=(T_(1))/(T_(2))" "( :.W_(A)=W_(B))` `T_(2)=(T_(1)+T_(3))/(2) ` |
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| 3694. |
Two Carnot engines are operated in succession. The first engine receives heat from a source at `T=800K` and rejects to sink at `T_(2)K`. The second engine receives heat rejected by the first engine and rejects to another sink at `T_(3)= 300K`. If work outputs of the two engines are equal, then find the value of `T_(2)`. |
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Answer» For the first engine, `Q_(1)-Q_(2)= W and (Q_(2))/(Q_(1))= (T_(2))/(T_(1))`….(i) For the second engine, `Q_(2)-Q_(3)= W and (Q_(3))/(Q_(2))=(T_(3))/(T_(2))`....(ii) `(Q_(3))/(Q_(2))xx(Q_(2))/(Q_(1))=(T_(2))/(T_(1))xx(T_(3))/(T_(2))` `(Q_(3))/(Q_(1))=(T_(3))/(T_(1))=(300)/(800)=3/8` Also, from (i) and (ii) `Q_(1)-Q_(2)= Q_(2)-Q_(3)` `Q_(1)+Q_(3)=2 Q_(2)` `1+(Q_(3))/(Q_(1))= 2(Q_(2))/(Q_(1))=(2T_(2))/(T_(1))` `1+3/8= (2T_(2))/(T_(1))= 11/8` `T_(2)= (11)/(16)xxT_(1)=(11)/(16)xx800= 550K` |
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| 3695. |
Enthalpy of combustion of carbon to `CO_(2)` is `-393.5 kJ mol^(-1)`. Calculate the heat released upon formation of `35.2 g` of `CO_(2)` from carbon and dioxygen gas.A. `-315 KJ`B. `+315KJ`C. `-630 KJ`D. `-3.15 KJ` |
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Answer» Correct Answer - A Given, `C(s) + O_(2)(g) to CO_(2)(g)` , `Delta_(f)H=-393.5 KJ "mol"^(-1)` Heat released on formation of 44 g or 1 mole `CO_(2) =- 395.5 KJ "mol"^(-1)` `because ` Heat released on formation of 35.2 g of `CO_(2)` `=(-393.5 KJ "mol"^(-1))/(44 g) xx35.2 g =- 315 KJ "mol"^(-1)` |
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| 3696. |
The standard heat of combustion of graphite carbon is `-393.5 kJ mol^(-1)`. The standard enthalpy of `CO_(2)` isA. `+393.5 kJ mol^(-1)`B. `-393.5 kJ mol^(-1)`C. `+196.75 kJ mol^(-1)`D. `-196.75 kJ mol^(-1)` |
| Answer» Correct Answer - B | |
| 3697. |
One of the most efficient engines ever developed operated between 2100K and 700K. Its actual efficiency is 40%. What percentage of its maximum possible efficiency is this?A. `40%`B. `60%`C. `66.67%`D. `33.37%` |
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Answer» Correct Answer - B `eta=(1-(T_(2))/(T_(1)))xx100` |
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| 3698. |
Two Carnot engines are operated in succession. The first engine receives heat from a source at `T=800K` and rejects to sink at `T_(2)K`. The second engine receives heat rejected by the first engine and rejects to another sink at `T_(3)= 300K`. If work outputs of the two engines are equal, then find the value of `T_(2)`.A. 100 kB. 300 kC. 550 kD. 700 k |
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Answer» Correct Answer - C |
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| 3699. |
A sample of `4` mole `He` is originally confined in `20 L` at `270 K` and then undergoes adiabatic expansion against a constant pressure of untill the volume has increased by factor of `3`. Calculate `Delta T` during process: |
| Answer» Correct Answer - 6 | |
| 3700. |
The value on a cyclinder containing initially `1` litres of an ideal gas at `7 atm` and `25^(@)C` is opened to the atmosphere, Whose the pressure is `760` torr and the temperature is `25^(@)C`. Assuming that the process is isothermal, how much work ( in `L atm`) is done on hte atmosphere by the action of expansion ? |
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Answer» Correct Answer - 6 `T="constant"` `P_(1)V_(1)= P_(2)V_(2)` `W= -P(DeltaV)` `=1xx(7-1)` `= -6 L. atm` `W= -6 L. atm` |
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