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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.
| 3951. |
A bulb of `100` watt is switched on in a room of dimensions `5xx4xx3 m^(3)`. What will be the increases in temperature of room after `15` minute, if specific heat of iar at room temperature and `1` atm is `0.71 J g^(-1) K^(-1)` and heat capacity of four walls and the roof is `50xx10^(3) JK^(-1)`? (Density of air `1.22xx10^(-6) kg mL^(-1)`) |
| Answer» Correct Answer - `0.88 K` | |
| 3952. |
Calculate the resonance energy of gaseous benzene form the following data. `BE(C-H) = 416.3 kJ mol^(-1)` `BE(C-C) = 331.4 kJ mol^(-1)` `BE(C=C) = 591.1 kJ mol^(-1)` `Delta_("sub")H^(Theta)(C,"graphite") = 718.4 kJ mol^(-1)` `Delta_("diss")H^(Theta)(H_(2),g) = 435.9 kJ mol^(-1)` `Delta_(f)H^(Theta) ("benzene", g) = 82.9 kJ mol^(-1)` |
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Answer» To compute resonance enegry, we compare the calaculated value of `Delta_(f)H^(Theta)` (benzene, g) with the given one. To calculate `Delta_(f)H^(Theta)` (benzene, g) we add the following reactions. (a). `6C(g) +6H(g) rarr C_(6)H_(6)(g), DeltaH^(Theta) =- (3BE_(C-C) +3BE_(C=C) +6BE_(C-H))` (b) `6C ("graphite") rarr 6C(g) DeltaH^(Theta) = 6 xx 718.4 kJ mol^(-1)` (c) `3H_(2)(g) rarr 6H(g) DeltaH^(Theta) = 3 xx 435.9 kJ mol^(-1)` `ulbar(6C("graphite")+3H_(2)(g)rarr C_(6)H_(6)(g))` The corresponding enthalpy change is `Delta_(f)H^(Theta) =- (3BE_(C-C)+ 3BE_(C=C) +6BE_(C-H)) +[6 xx 718.4 +3 xx 435.9] kJ mol^(-1)` `=[-(3xx331.4 xx 591.1 +6 xx 718.4 +3 xx 435.9] kJ mol^(-1)` The given `Delta_(f)H^(Theta)` is `Delta_(f)H^(Theta)` (benzene, g) `= 82.9 kJ mol^(-1)` This means benzene become more stable by `(352.8 -82.9) kJ mol^(-1)`, i.e., `269.7 kJ mol^(-1)` This is its resonance enegry. |
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| 3953. |
Consider the reactions `:` `C_((s))+2H_(2(g)) rarr CH_(4(g)), DeltaH=-X kcal` `C_((g))+4H_((g)) rarr CH_(4(g)),DeltaH-X_(1)kcal` `CH_(4(g))rarr CH_(3(g))+H_((g)), DeltaH=+Ykcal` The average bond energy of `C-H` bond is `:`A. `(X)/(4)` K cal. `mol^(-1)`B. Y K cal. `mol^(-1s)`C. `(X_(1))/(4)` K cal. `mol^(-1)`D. `X_(1)` K cal. `mol^(-1)` |
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Answer» Correct Answer - C `CH_(4(g))rarr C_((g))+4H_((g)) , Delta H= + x_(1) KCal` `therefore` B.E. of C-H bond `=(x_(1))/(4) KCal mol^(-1)` |
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| 3954. |
Calculate the work done in open vessel at `300K`, when `92g Na` reacts with water. Assume ideal gas neture. |
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Answer» `2Na +2H_(2)O rarr 2NaOH +H_(2)` Moles of `Na = (92)/(23) = 4` Mole of `H_(2)` formed `= (1)/(2) xx` Mole of `Na` used `= (1)/(2) xx 4 = 2` Work is done in giving out `2mol H_(2)`. Thus, `W =- P xx V_(H_(2)) = n_(H_(2))RT =- 2 xx 8.314 xx 300` `=- 4988.4J` The `H_(2)` liberated pushes the atmospheric gas block and thus, does work in driving back the atmosphere. Note that in the case of closed vessel, `DeltaV = 0`. Therefore, `W = 0`. |
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| 3955. |
Calculate the work done when a system raises a colume of water of radius `5.0mm` through `10cm`. |
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Answer» Work is given: `W =- mgh` where, `m = Volume xx Density` `pir^(2)l xxd = (22)/(7) xx (0.5)^(2) xx 10 xx1` `g = 980 cm s^(-2)` `h = (10)/(2) = 5cm` [The centre of mass of colume lies half way along its length `l`, therefore the centre of mass is raised to a height `h = (1//2)l`] `:.W =- (22)/(7) xx (0.5)^(2) xx 10 xx 980 xx 5xx1` `=- 38.5 xx 10^(3) erg =- 3.85 xx 10^(-3)J` The nagative sign shows that systwm shows a decrease in internal enegry. |
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| 3956. |
If values of `Delta_(f)H^(Theta)` of `ICI(g),CI(g)`, and `I(g)` are, respectively, `17.57,121,34`, and `106.96 J mol^(-1)`. The value of `I-CI` (bond energy) in `J mol^(-1)` isA. `17.57`B. `210.73`C. `35.15`D. `106.96` |
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Answer» The process will be `I(g) +CI(g) rarr I-CI(g),DeltaH^(Theta) = 17.57J mol^(-1)` `DeltaH^(Theta) =` Heat of atomisation of `I(g)` and `CI(g)`-Bond enegry of `I-CI` bond `17.51 = 121.34 +106.96-x` `x = 210.73 J mol^(-1)` |
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| 3957. |
If values of `Delta_(f)H^(Theta)` of `ICI(g),CI(g)`, and `I(g)` are, respectively, `17.57,121.34`, and `106.96 J mol^(-1)`. The value of `I-CI` (bond energy) in `J mol^(-1)` isA. `35.15 "J mol"^(-1)`B. `106.69 "J mol"^(-1)`C. `210.73 "J mol"^(-1)`D. `420.9 "J mol"^(-1)` |
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Answer» Correct Answer - C `ICl_((g))rarr I_((g))+Cl_((g)) , Delta H_("Bond")= ?` `Delta H=Sigma (Delta H_(f))_(P) -Sigma (Delta H_(f))_(R )` |
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| 3958. |
One mole of a liquid ( 1 bar, 100 ml) is taken in an adiabatic container and the pressure increases steeply to 100 bar. Then at constant pressure of 100 bar, volume decrases by 1 ml.Find `DeltaU` and`DeltaH`. |
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Answer» As the process is carriedout under adiabatic condition, `q=0` . But `DeltaU = q + w`. Hence `Delta U =w`.Further, at constant pressure of100 bar, volume has decreased by 1 ml, therefore, work of contraction . `= P Delta V = 100 ` bar ` xx 1 `ml. `= ( 100 xx10^(5) Nm^(-2)) ( 10^(-6)m^(3)) = 10J` Hence, `Delta U = 10 J` `Delta H =Delta U + Delta( PV)` `Delta ( PV) = (P_92)V_(2) - P_(1)V_(1))` But `P_(1) =1 "bar", V+(1) = 100 ml` `P_(2) = 100 "bar", V_(2) = 99 ml` `:. Delta ( PV) = ( 100 "bar" xx 99 ml) - ( 1 "bar" xx 100 ml) `= ( 9900 - 100 ) "bar" ml = 9800 "bar ml ) = 980 J ` `:. Delta H = 10J + 980 J = 990 J` |
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| 3959. |
There is 1 mol liquid (molar volume 100 ml) in an adiabatic container initial, pressure being 1 bar. Now the pressure is steeply increased to 100 bar ,and the volume decreased by 1 mL under constant pressure of 100 bar. Calculate `DeltaH "and" DeltaE`. [Given 1 bar = `10^(5) N//m^(2)`]A. `DeltaE=0 J, DeltaHne0 J`B. `DeltaH=990 J, DeltaE=10 J`C. `DeltaE=20 J, DeltaH=890 J`D. `DeltaE=0 J, DeltaH=10 J` |
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Answer» Correct Answer - B |
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| 3960. |
There is `1 mol` liquid (molar volume `100 ml`) in an adiabatic container initial, pressure being `1` bar Now the pressure is steeply increased to `100` bar, and the volume decreased by `1 ml` under constant pressure of `100` bar. Calculate `Delta H` and `Delta E`. [Given `1 "bar"=10^(5)N//m^(2)`] |
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Answer» Correct Answer - `Delta H=990, Delta E=19 J` `Delta U= q+W` for adiabatic process `q=0`, hence `Delta U=W` and `W= -p(Delta V)= -P(V_(2)-V_(1))` Now `Delta H= Delta U+Delta (PV)` Here `Delta U` already calculated abobe and `Delta PV=(P_(2)V_(2)-P_(1)V_(1))` So, `Delta H=100+(100xx99-1xx100)=9900 ` bar `mL=990J` |
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| 3961. |
Identify the state quantiy among the followingA. qB. `q xx W`C. q/WD. `q + W` |
| Answer» Correct Answer - D | |
| 3962. |
A mercury barometer is defective. When an accurate barometer reads `770` mn, the defective one reads `760` mn. When the accurate one reads `750` mn, the defective one reads `742` mn. Then The reading of the accurate barometer, when defective are reads `752`mm isA. `760mm`B. `764mm`C. `758mm`D. `761mm` |
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Answer» Correct Answer - D `(P_(0)-752)(80)=10xx72impliesP_(0)=761mm` |
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| 3963. |
The saturated water vapour is inclosed in a cylinder under a piston and occupies a volume `V_0 = 5.0 1` at the temperature `t = 100 ^@C`. Find the mass of the liquid phase formed after the volume under the piston decreased isothermally to `V = 1.6 1`. The saturated vapour is assumed to be ideal. |
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Answer» The volume of the condensed vapour was originally `V_0 - V` at temperature `T = 373 K`. Its mass will be given by `p(V_0 - V) = (m)/(M) RT` or `m = (Mp(V_0 - V))/(RT) = 2 gm` where `p = `atmospheric pressure. |
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| 3964. |
Assertion:A constant volume gas thermometer, reads temperature in terms of pressure. Reason : In this case a plot of ressure verus temperature gives a straight line.A. If both assetion and reason are true and reaasons is the correct expanation of assetionB. If both assetion and reason are tur but reason is not the correct explanation of assetionC. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - A `PVpropT` A gas is used to measure temperature in a constant volume gas thermometer `PpropT`(V constant)`, therefore, a constant volume gas thermometer reads tempterature in terms of pressure. A plot of pressure versus temperature gives a straight line. |
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| 3965. |
`DeltaU^(@)` of combustion of `CH_(4(g))` at certain temperature is `-"393 kJ mol"^(-1)`. The value of `DeltaH^(@)` isA. zeroB. `lt DeltaH^(@)`C. `gt DeltaU^(@)`D. equal to `DeltaU^(@)` |
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Answer» Correct Answer - B The balanced equation for combustion of methane is `CH_(4(G))+2O_(2(g))rarrCO_(2(g))+2H_(2)O_((l))` Here, `Deltan_(g)=1-3=-2` `DeltaH^(@)=DeltaU^(@_+Deltan_(g)RT` `DeltaH^(@)=-393-2RT` `therefore" "DeltaH^(@) lt DeltaU^(@)` |
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| 3966. |
7 g of `N_(2)g` at `27^(@)C` is expanded slowly and isothermally form pressuure of `0.5 M Pa` to a final pressure `0.1 Mpa`. Determine approximate work done `[In 5=1.6, R=8.3J mol^(-1)K^(-1)]`A. `-996J`B. `+125J`C. `-95J`D. `+200J` |
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Answer» Correct Answer - A |
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| 3967. |
A system performs work `DeltaW` when an amount of heat is `DeltaQ` added to the system, the corresponding change in the internal energy is `DeltaU` . A unique function of the initial and final states (irrespective of the mode of change) isA. `DeltaQ`B. `DeltaW`C. `DeltaU`and `DeltaU`D. `DeltaU` |
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Answer» Correct Answer - D |
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| 3968. |
Although graphite is thermodynamically most stable allotrope of Carbon but it has more entropy than diamond Because:A. It has layer structure which slides on each other .B. In diamond Carbon is bonded with covalent bond.C. In graphite covalent bond only present within sheet and weak vanderwaal froce present between sheets which allow sliding easily.D. Two types of bond lengths are present in graphite. |
| Answer» Correct Answer - C | |
| 3969. |
At `60^(@)C`, dinitrogen tetroxide is 50 per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere. |
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Answer» `N_(2)O_(4)(g)hArr2NO_(2)(g)` If `N_(2)O_(4 )`is 50% dissociated, the mole fraction of both the substances is given by `X_(N_(2O_(4)))=(1.05)/(1+0.5):x_(No_(2))=(2xx0.5)/(1+0.5)` `P_(N_(2O_(4)))=(0.5)/(1.5)xx1atm,P_(NO_(2))=(1)/(1.5)xx1atm`The equilibrium constant `K_(p)` is given by `K_(p)=((P_No_(2))^(2))/(P_(N_(2O_(4))))=(1.5)/((1.5)^(2)(0.5))` `=1.33 atm`. Since `Delta_(r)G^(Ө)=-RTlnK_(p)` `Delta_(r)G^(Ө)=(-8.314JK^(-1)"mol"^(-1))xx(333K) xx(2.303)xx(0.1239)` `=-763.8 "mol"^(-1)` |
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| 3970. |
For oxidation of iron. `4Fe(s)+3O_(2)(g)rarr2Fe_(2)O_(3)(s)` entropy change is`- 549.4 JK^(-1)mol^(-1)` at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous? `(Delta_(r)H^(Ө)` for this reaction is `-1648xx10^(3)J"mol"^(-1)`) |
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Answer» One decides the spontaneity of a reaction by considering `DeltaS_("total")(DeltaS_("sys")+DeltaS_(surr")).` For calculating `DeltaS_("surr")` have to consider the heat absorbed by the surroundings which is equal to `-Delta_(r)H^(Ө)` . At temperature T, entropy change of the surroundings is `DeltaS_("surr")=-(Delta_(r)H^(Ө))/(T)` (at constant pressure) `=-((-1648xx19^(3)J "mol"^(1)))/(298K)` `5530 JK^(-1)"mol"^(-1)` Thus, total entropy change for this reaction `Delta_(r)S_("total")=5530JK^(-1)"mol"^(-1)+` `" " (-549.4JK^(-1)"mol"^(-1))` `=4980.6 JK^(-1)"mol"^(-1)` This shows that the above reaction is spontaneous. |
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| 3971. |
For oxidation of iron,, `4Fe(s) + 3O_(2)(g) rarr 2Fe_(2)O_(3)(s)`, entropy change is `- 549 .4 JK^(-1) mol^(-1)` at 298 K . Inspite of the negative entropy change of this reaction , why is the reaction spontaneous ? `( Delta H ^(@)` for this reaction is `- 1648 xx 10^(3) J mol^(-1))` |
| Answer» Proceed as in Solved Problem on page `6 // 87` by either method (Prove that `DeltaG^(@) ` is `-ve` or `DeltaS_("total") ` is `+`ve). | |
| 3972. |
Calculate the entropy change for the conservation of following: (a) `1g` ice to water at `273K , DeltaH_f` for `ice=6.025kJmol^(-1)`. (b) `36g` water to vapour at `373K, DeltaH_v`for `H_2O=40.63kJmol(-1)` |
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Answer» `DeltaS_("fusion")=(DeltaH_(f))/T=(6.025xx10^(3))/273` `=22.1 JK^(-1) mol^(-1)` Now, `18 g` ice melts showing a change in entropy =`22.1` `:. 1 g` ice melts showing a change in entropy `22.1/18xx1=1.227 JK^(-1)` (b) `DeltaS_(V)=(DeltaH_(V))/T=(40.63xx10^(3))/373` `=108.9 KJ^(-1)mol^(-1)` `18 g H_(2)O` vaporises to show a change in entropy `=108.9` `:. 36 g H_(2)O` vaporises to show a change in entropy `=(108.9xx36)/18=217.85 JK^(-1)` |
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| 3973. |
A heat engine operating n between ` 227^(@)` and ` 27^(@) C` absorbs 2 K a cal of heat from the ` 227^(@)` reservoir reversibly per cycle . The amount of work done in one cycle isA. 0. 4 calB. 0.8 calC. 4 KcalD. 87 Kcal |
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Answer» Correct Answer - B `eta = 1 - (300)/(500) = 1 -0.6 -0.4` `%eta= 40 =|W|/(q)xx 100` `w = (40)/(100)xx2= 0.8 Kcal` |
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| 3974. |
The entropy change when two of ideal monoatomic gas is heated form `200 to 300^(@)C` reversiblity and isochorically?A. `(3)/(2)R ln.((300)/(200))`B. `(5)/(2)R ln.((573)/(273))`C. `3R ln.((573)/(473))`D. `(3)/(2)R ln.((573)/(473))` |
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Answer» Correct Answer - C `DeltaS = nC_(V) In (T_(2))/(T_(1))` `= 2 xx (3)/(2) R In (573)/(473)` |
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| 3975. |
A carnot engine absorbs `2000J` of heat from the source of heat engine at `227^(@)C` and rejects `1200J` of heat to the sink during each cycle. Calculate (i) Temp.of sink (ii) efficiency of engine (iii) amount of work done during each cycle. |
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Answer» Correct Answer - `27^(@)C ; 40% ; 800J` Here, `Q_(1)= 2000J, T_(1)=227^(@)C=(227+273)K= 500K` `Q_(2)= 1200J, T_(2)=? Eta=? W=?` As `(Q_(2))/(Q_(1))=(T_(2))/(T_(1)) :. T_(2)=(Q_(2))/(Q_(1))xxT_(1)` `=(1200)/(2000)xx500= 300K` `=(300-273)^(@)C` `27^(@)C` (ii) `eta=1 -(Q_(2))/(Q_(1))= 1- (1200)/(2000)` `=2/5=2/5xx100%= 40%` (iii) `W=Q_(1)-Q_(2)` `=2000-1200= 800J` |
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| 3976. |
A reversible heat engine A(based on carnot cycle ) absorbs heat from a reservoir at 1000 K and rejects heat to a reservoir at `T_(2)` . A second reserviour at `T_(2)` and rejects energy to a reservoir at 360 K . If the efficiencies of engines A and B are the same then the temperature `T_(2)`is .A. 680 KB. 640 KC. 600 KD. None |
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Answer» Correct Answer - C For A `eta_(1)= 1- (T_(2))/(1000)` For B `eta_(2) = 1 -(360)/(T_(2))` `1- (T_(2))/(1000)= 1 -(360)/(T_(2))` `T_(2)= 600K` |
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| 3977. |
A reversible heat engine A(based on carnot cycle ) absorbs heat from a reservoir at 1000 K and rejects heat to a reservoir at `T_(2)`. A second reversible engine B absorbs, the same amount of heat as rejected by the engine A, from the reservoir at `T_(2)` and rejects energy to a reservoir at 360 K. If the efficienices of engines A and B are the same then the temperature `T_(2)` is .A. 680 KB. 640 KC. 600 KD. none |
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Answer» Correct Answer - D For heat engines A: `eta= (1000- T_(2))/(1000) rArreta_(B) = (T_(2) - 360)/(T_(2))` `eta_(A) = eta_(B)` `So(1000-T_(2))/(1000)=(T_(2) - 360)/(T_(2)) = T_(2) = sqrt(360 xx 1000)=600 K` |
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| 3978. |
A Carnot engine absorbs 750 J of heat energy from a reservoir at `137^(@)C` and rejects 500 J of heat during each cycle then the temperature of sink isA. `0.25^(@)C`B. `0.34^(@)C`C. `0.44^(@)C`D. `0.54^(@)C` |
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Answer» Correct Answer - B Here `T_(1)=273+173=410, Q_(1)=750 J , Q_(2)=500 J` As, `(Q_(2))/(Q_(1))=(T_(2))/(T_(1))` `rArr T_(2) =(Q_(2))/(Q_(1))(T_(1))=(500)/(750)xx410` =273.34 K = 237.34-273 = `0.34^(@)C` |
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| 3979. |
A carnot engine absorbs `500J` of heat from a reservoir at `227^(@)C` and rejects of seek (ii) efficiency of engine (iii) amount of useful work done per cycle. |
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Answer» Here, `Q_(1)=500J` `T_(1)= 227^(@)C=227+273= 500K` `Q_(2)= 300J, T_(2)=?, eta=?, W=?` As `(T_(2))/(T_(1))= (Q_(2))/(Q_(1))= (300)/(500)`, `T_(2)= 6/(10)T_(1)= 6/(10)xx500= 300K` `=(300-273)^(@)C= -27^(@)C` `eta=1 - (T_(2))/(T_(1))=1 - (6)/(10)= 4/(10)= 40%` `W=Q_(1)-Q_(2)= 500-300= 200J` |
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| 3980. |
Two moles of a certain gas at a temperature `T_0=300K` were cooled isochorically so that the pressure of the gas got reduced 2 times. Then as a result of isobaric process, the gas is allowed to expand till its temperature got back to the initial value. Find the total amount of heat absorbed by gas in this process. |
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Answer» Let `v = 2` moles of the gas. In the first phase, uder isochoric process, `A_1 = 0`, therefore from gas law if pressure is reduced `n` times so that temperature i.e., new temperature becomes `T_0//n`. Now from first law of thermodynamics `Q_1 = Delta U_1 = (v R Delta T)/(gamma - 1)` =`(v R)/(gamma - 1)((T_o)/(n) -T_0) = (v RT_0 (1 - n))/(n(gamma - 1))` During the second phase (under isobaric process), `A_2 = p Delta V = v R Delta T` Thus from law of thermodynamics : `Q_2 = Delta U_2 + A_2 = (v R Delta T)/(gamma - 1)+ v R Delta T` =`(v R(T_0 - (T_0)/(n))gamma)/(gamma - 1) = (vRT_0(n - 1)gamma)/(n(gamma - 1))` Hence the total amount of heat absorbed `A = Q_1 + Q_2 = (v RT_0(1 -n))/(n(gamma - 1))+(v RT_0(n - 1)gamma)/(n(gamma - 1))` =`(vRT_0(n - 1)gamma)/(n(gamma - 1))(-1 + gamma) = V RT_0(1 - (1)/(n))`. |
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| 3981. |
A gas absorbs `100 Calorie` of heat and performs `150J` of work. Increase energy of the gas in the process isA. `420J`B. `570J`C. `270J`D. none of these |
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Answer» Correct Answer - C Volume, pressure |
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| 3982. |
If `300 ml` of a gas at `27^(@)` is cooled to `7^(@)` at constant pressure, then its final volume will beA. 540 mlB. 350 mlC. 280 mlD. 135 ml |
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Answer» Correct Answer - C |
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| 3983. |
Why is conversion of heat into work not possible without a sink at lower temperature? |
| Answer» For converting heat energy into work continuosly, a part of heat energy absorbed from the source has to be rejected. The heat energy can be rejected only if there is a body, whose temperature is less than that of the source. This body at lower temperature is the sink. | |
| 3984. |
Consider a `P-V` diagram in which the path followed by one mole of perfect gas in a cyclinderical container is shown in (figure) (a) Find the work done when the gas is taken from state 1 to state 2. (b) What is the ratio of temperatures `T_(1)//T_(2), if V_(2)= 2V_(1)`? (c ) Given the internal energy for one mole of gas at temperature `T is (3//2) RT`, find the heat supplied to the gas when it is taken from state 1 to 2, with `V_(2)= 2V_(1)`. |
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Answer» Let `pV^(1//2)` = Constant = `K, p=(K)/(sqrt(V))` (a) Work done for the process 1 to 2, ltBrgt `W=int_(V_(1))^(V_(2))pdV=Kint_(V_(1))^(V_(2))(dV)/(sqrt(V))=K[(sqrt(V))/(1//2)]_(V_(1))^(V_(2))=2K(sqrt(V_(2))-sqrt(V_(1)))` `=2p_(1)V_(1)^(1//2)(sqrt(V_(2))-sqrt(V_(1)))=2p_(2)V_(2)^(1//2)(sqrtV_(2)-sqrt(V_(1)))` (b) From ideal gas equation, `pV=nRTrArrT=(pV)/(nR)=(psqrt(V)sqrt(V))/(nR)` `rArr" "T=(Ksqrt(V))/(nR)" "("As",psqrt(V)=K)` Hence, `" "T_(1)=(Ksqrt(V_(1)))/(nR)rArrT_(2)=(Ksqrt(V_(2)))/(nR)` `rArr" "(T_(1))/(T_(2))=((Ksqrt(V_(1)))/(nR))/((Ksqrt(V_(2)))/(nR))=sqrt(V_(1)/(V_(2)))=sqrt(V_(1)/(2V_(1)))=(1)/(sqrt(2))" "(because V_(2)=2V_(1))` (c ) Given, internal energy of the gas = `U=((3)/(2))RT` `" "DeltaU=U_(2)-U_(1)=(3)/(2)R(T_(2)-T_(1))` `" "=(3)/(2)RT_(1)(sqrt(V)-1)" "[becauseT_(2)=sqrt(2)T_(1)"from"(b)]` ltbgt `" "DeltaW=2p_(1)V_(1)^(1//2)(sqrt(V_(2))-sqrt(V_(1)))` ltBrgt `" "=2p_(1)V_(1)^(1//2)(sqrt(2)xxsqrt(V_(1))-sqrt(V_(1)))` `" "=2p_(1)V_(1)(sqrt(2)-1)=2RT_(1)(sqrt(2)-1)` `because " " DeltaQ=DeltaU+DeltaW` `" "=(3)/(2)RT_(1)(sqrt(2)-1)+2RT_(1)(sqrt(2)-1)` `" "=(sqrt(2)-1)RT_(1)(2+3//2)` `" "=((7)/(2))RT_(1)(sqrt(2)-1)` This is the amount of heat supplied. |
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| 3985. |
Calculate the standard molar entropy change for the formation of gaseous propane `(C_(3)H_(8))` at `293K`. `3C("graphite") +4H_(2)(g) rarr C_(3)H_(8)(g)` Standard molar entropies `S_(m)^(Theta) (JK^(-1)mol^(-1))`are: `C("graphite") = 5.7, H_(2)(g) = 130.7,C_(3)H_(5)(g) = 270.2` |
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Answer» Standard molar entropy change, `Delta_(r)S_(m)^(Theta) = S_(m)^(Theta) (C_(3)H_(8))-[3S_(m)^(Theta) (C ) +4S_(m)^(Theta) (H_(2))]` `= 270.2 -[3 xx 5.7 +4 xx 130.7]` `= 270.2 - 539.9 =- 269.7 J K^(-1)` |
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| 3986. |
Calculate the enthalpy of vaporisation per mole for ethanol. Given `DeltaS = 109.8 J K^(-1) mol^(-1)` and boiling point of ethanol is `78.5^(@)`. |
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Answer» The entropy change for vaporisation, `Delta_(vap)S^(Theta) = (Delta_(vap)H)/(T_(b))` or `Delta_(vap)H^(Theta) = Delta_(vap)S^(Theta)T_(b)` `DeltaS^(Theta) = 109.8 J K^(-1)mol^(-1), T_(b) = 273 +78.5 = 351.5K` `:. DeltaH^(Theta) = (109.8 xx 351.5) = 38.595 kJ mol^(-1)` |
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| 3987. |
The correct match isA. `{:(A,B,C,D),(3,4,2,1):}`B. `{:(A,B,C,D),(2,3,1,4):}`C. `{:(A,B,C,D),(1,3,4,2):}`D. `{:(A,B,C,D),(1,3,4,2):}` |
| Answer» Correct Answer - A | |
| 3988. |
The change in Gibbs free energy of the system alone provides a criterion for the spontaneity of a process at constant temperature and pressure. A change in the free energy of a system at constant temperature and pressure will be : `DeltaG_("system")=DeltaH_("system")-T DeltaS_(system")` For a system in equilibrium, `DeltaG=0`, under conditions of contant....A. Temperature and pressureB. Temperature and volumeC. Pressure and volumeD. Energy and volume |
| Answer» Correct Answer - A | |
| 3989. |
The change in Gibbs free energy of the system alone provides a criterion for the spontaneity of a process at constant temperature and pressure. A change in the free energy of a system at constant temperature and pressure will be : `DeltaG_("system")=DeltaH_("system")-T DeltaS_(system")` The free energy for a reaction having `DeltaH=31400 cal, DeltaS=32 cal L^(-1) mol^(-1)` at `1000^(@)C`A. `-9336 cal`B. `-7386 cal`C. `-1936 cal`D. `+9336 cal` |
| Answer» Correct Answer - A | |
| 3990. |
Which of the following are false about Gibbs free energy?A. `DeltaG=DeltaH-TDeltaS`B. `DeltaGgt 0` indicates that process is spontaneousC. `DeltaG^(@)lt 0` implies `K_(eg)lt1`D. `DeltaG=0 ` implies`K_(eq)=1` |
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Answer» Correct Answer - b,c,d |
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| 3991. |
`DeltaH` (vap) for water is `40.7 kJ mol^(-1)`. The entropy of vapourisation of water isA. `-40.7 kJ mol^(-1)`B. `407 J mol^(-1) K^(-1)`C. `109 J mol^(-1) k^(-1)`D. `722 J mol^(-1)` |
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Answer» Correct Answer - C `DeltaS=(DeltaH)/(T)` |
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| 3992. |
The boiling point of chorofrom ,`CHCl_(3), is 61.7^(@)`C and its enthalpy of vaporization is `31.4 KJ-mol^(-1)` Calculate the molar entropy of vaporization for chlorofrom :A. `10.7Jxxmol^(-1).K^(-1)`B. `93.8Jxxmol^(-1).K^(-1)`C. `301J xxmol^(-1).K^(-1)`D. `509J xxmol ^(-1).K^(-1)` |
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Answer» Correct Answer - b |
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| 3993. |
The enthalpy of vapourisation of liquid water using the data : `H_(2)(g)+1//2O_(2)(g)rarr H_(2)O(l) , Delta =-285.77 KJ mol^(-1)` `H_(2)(g)+1//2O_(2)(g)rarr H_(2)O(g) , Delta H=-241.84 KJ mol^(-1)`A. `+43.93` KJ `mol^(-1)`B. `-43.93` KJ `mol^(-1)`C. `+527.61` KJ `mol^(-1)`D. `-527.61` KJ `mol^(-1)` |
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Answer» Correct Answer - A `H_(2)O_((l))rarr H_(2)O_((g)) , Delta H_("Vap.")= ?` Desired equation = eq(ii) - eq(i) `Delta H_("Vap.") =(-241.84)-(-285.77)` `Delta H_("Vap.")=+43.93 KJ mol^(-1)` |
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| 3994. |
Predict `DeltaH gt DeltaU or DeltaH lt DeltaU`. a. `C("graphite") +O_(2)(g) rarr CO_(2)(g)` b. `PCI_(5)(g) rarr PCI_(3)(g) +CI_(2)(g)` |
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Answer» a. `DeltaH = DeltaU` because `Deltan = 0`. b. `DeltaH gt DeltaU` because `Deltan = 1` and `DeltaH = DeltaU +DeltanRt` |
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| 3995. |
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` ltbrlt The value of Delta_(f)H^(Theta)` of `PCI_(5)` isA. `454.5 kJ mol^(-1)`B. `-454.5 kJ mol^(-1)`C. `-772 kJ mol^(-1)`D. `-498 kJ mol^(-1)` |
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Answer» `(1)/(4)P_(4)(s)+(3)/(2)CI_(2)(g)rarr PCI_(5)(l),DeltaH^(Theta) =- (635)/(2)kJ` `PCI_(3)(l)+CI_(2)(g)rarr PCI_(5)(s),DeltaH^(Theta) =- 137kJ` `ulbar({:(,"On adding,",,),(,"(1)/(4)P_(4)(s)+(5)/(2)C1_(2)(g)rarrPC1_(5)(s),DeltaH^(Theta)=-454.5kJ,,):})` |
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| 3996. |
The thermite reaction used ofr welding of metals involves the reaction `2Al(s) + Fe_(2)O_(3)(s) rarr Al_(2)O_(3)(s) + 2Fe(s)` What is `Delta H^(@) ` at `25^(@)C` for this reaction ? Given that the standard heats of formation of `Al_(2)O_(3)` and `Fe_(2)O_(3)` are - 1675.7 kJ and - 828.4 kJ `mol^(-1)` respectively. |
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Answer» We are given (i) `2Al(s) + (3)/(2) O_(2)(g) rarr Al_(2)O_(3)(s) , Delta _(r) H^(@) = - 1675.7 k J mol^(-1)` (ii) `2Fe(s) + (3)/(2) O_(2)(g) rarr Fe_(2)O_(3)(s) , Delta _(r) H^(@) = - 828 .4kJ mol^(-1)` We aim at `2Al(s) + Fe_(2)O_(3) (s), rarr Al_(2)O_(3)(s) + 2Fe(s), Delta_(r) H^(@) = ?` Equation (i) - Equation (ii) gives `2Al(s) + Fe_(2)O_(3)(s) rarr Al_(2)O_(3)(s)+2Fe(s), Delta_(r)H= - 1675.7 - ( 828.4)` `= - 847.3 kJ mol^(-1)` Alternative Method We aim at `2Al(s) + Fe_(2)O_(3)(s) rarr Al_(2)O_(3) +2Fe(s)lt Delta_(r) H^(@) = ?` `Delta _(r) H = ` Sum of `Delta_(r) H^(@)` of products `-` Sum of `Delta_(r) H^(@) =?` `= [ Delta_(f) H^(@) (Al_(2)O_(3))+ 2 xx Delta_(f ) H^(@) (Fe)] - [ 2 xx Delta _(f)H^(@) (Al)+Delta_(f) H^(@) ( Fe_(2)O_(3))]` `= [ 1675.7 +0] - [ 0+ ( -828.4)]` `= -847.3 kJ mol^(-1)` |
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| 3997. |
An ideal gas is enclosed in a thermally insulated container. Container is covered by an insulating piston which is connected with spring. Small displacement is given to piston as shown in figure. Which of the following statements is correct? Neglect friction everywhere. A. Motion of piston is oscillatoryB. Motion of piston is S.H.M. for small displacementC. Motion of piston can not be S.H.M.D. Piston will not oscillate at all |
| Answer» Correct Answer - A::B | |
| 3998. |
One end of a rod, enclosed in a thermally insulating sheath, is kept at a temperature `T_1` while the other, at `T_2`. The rod is composed of two sections whose lengths are `l_1` and `1_2` and heat conductivity coefficients `x_1` and `x_2`. Find the temperature of the interface. |
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Answer» Let `T` = temperature of the interface Then heat flowing from left = heat flowing into right in equilibrium. Thus `k_1 (T_1 - T)/(t_1) = k_2 (T - T_2)/(l_2)` or `T = (((k_1 T_1) /(l_1) + (k_2 T_2)/(l_2)))/(((k_1)/(l_1) + (k_2)/(l_2)))`. |
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| 3999. |
For which reaction from the following , will be maximum entropy change:A. `Ca(s) +(1)/(2)O_(2)(g) rarr CaO(s)`B. `CaCO_(3)(s) rarr CaO(s) + CO_(2)(g)`C. `C(s)+O_(2)(g)rarrCO_(2)(g)`D. `N_(2)(g)+ O_(2)(g) rarr 2NO(g)` |
| Answer» Correct Answer - B | |
| 4000. |
A `10g` piece of iron `( C =0.45 J//g^(@)C )` at `100^(@)C` is dropped into `25g` of water `( C =4.2 J//g^(@)C)` at `27^(@)C`. Find temperature of the rion and water system at thermal equilibrium .A. `30^(@)C`B. `33^(@)C`C. `40^(@)C`D. None of these |
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Answer» Correct Answer - A Heat lost by iron + Heat gained by water = 0 `10 xx0.45 (T -373)+25xx4.2 xx(T-300)=0` `" "T=303" K or " 30^(@)C` |
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