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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.
| 601. |
The pressure of a gas is increased by `50%` at constant temperature. The decrease in volume will be nearest toA. `66%`B. `33%`C. `17%`D. `8%` |
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Answer» Correct Answer - B Let initial pressure, `P_(1)= P, V_(1)=V` Final pressure, `P_(2)=P+50%P` `P=3/2P, V_(2)=?` As temperature is constant `P_(1)V_(1)= P_(2)V_(2)` or `V_(2)= (P_(1)V_(1))/(P_(2))= (PV)/(3P//2)=2/3V` `:.` Decrease in volume `= V-2/3V=V/3` `%` decrease in volume `=(V//3)/Vxx100` `=33.3%~~ 33%` |
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| 602. |
A thermodynamic system undergoes cyclic process `ABCDA` as shown in figure. The work done by the system is A. `P_(0)V_(0)`B. `2P_(0)V_(0)`C. `(P_(0)V_(0))/2`D. Zero |
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Answer» Correct Answer - D As is clear from (figure). `W=area BCE+ area EDA` `=(P_(0)xxV_(0))/2-(P_(0)xxV_(0))/2=Zero` Work done in expansion is positive and work done in compression is negative. |
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| 603. |
During adiabatic changes, `V prop 1//T^(2)`. How will pressure of the gas vary with temperature? |
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Answer» As `V prop 1/(T^(2))` or `V=("constant")/(T^(2))` But `(PV)/T= "constant"` `:. P/Txx("constant")/(T^(2))="constant"` `:. P prop T^(3)` |
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| 604. |
What is a perfect black body? |
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Answer» A perfect black body absorbs heat radiations of all wavelength and can emit the whole set of radiations when heated. |
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| 605. |
What is a diathermic material? |
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Answer» A material which does not allow the flow of heat is called diathermic material. |
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| 606. |
What is the S.I. unit of coefficient of expansion? |
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Answer» S.I. unit of α, β and γ is K-1 . |
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| 607. |
Give a common unit of heat and give its relation with the S.I. unit of heat. |
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Answer» Common unit of heat is 1 calorie. 1 calories = 4.2 joule. |
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| 608. |
What is temperature of gas? |
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Answer» Temperature of gas is measure of the mean kinetic energy per molecule of the gas. \(\frac{E}{N}=\frac{3}{T}kT\) or, \(E=\frac{3}{2}RT\) |
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| 609. |
What is an ideal gas? |
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Answer» An ideal gas is one in which the inter molecular forces are absent. Moreover, ideal gas strictly obey the gas laws. The molecules of an ideal gas considered as point masses, i.e., size of the molecules of an ideal gas is negligible |
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| 610. |
Give the range of a platinum resistance thermometer. |
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Answer» It Scan measure the temperature between −260°C to 1200°C. |
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| 611. |
Which is greater Cp or Cv? |
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Answer» Cp is greater than Cv (Cp > Cv). |
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| 612. |
What is an isothermal process? |
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Answer» Isothermal process is the process in which temperature variation does not exist. Such process are to be carried in (i) conducting cylinder, (ii) at a slow pace. |
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| 613. |
What is triple point? |
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Answer» The combination of pressure and temperature at which all the three states substance, i.e., solid, liquid and gas coexist is known as triple point. For what triple point is 273.15 K or 0.01°C at a pressure of 4 torr. |
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| 614. |
Find out the heat evolved in combustion if 112 litre (at 1 atm, 273 K) of water gas (mixture of equal volume of `H_(2)(g)` and CO(g)) is combusted with excess oxygen. `H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(g), Delta=-241.8 kJ` `CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g), Delta=-283 kJ`A. `241.8` KJB. 283 KJC. 1321 KJD. 1586 KJ |
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Answer» Correct Answer - C 112 litre water gas contains 56 litre CO and 56 litre `H_(2)`. `H_(2(g))+(1)/(2)O_(2(g))rarr H_(2)O_((g)) , Delta H =-241.8 kJ` In the combustion of mole. `H_(2)=22.4` litre `H_(2)` the evolved heat = 241.8 kJ `therefore` In the combustion of 56 litre `H_(2)` the `Delta H = (241.8)/(22.4)xx56` `therefore Delta H = 604.5 kJ` `CO_((g))+(1)/(2)O_(2(g))rarr CO_(2(g)) , Delta H=-283 kJ` By the combustion of 1 mole. CO = 22.4 litre CO teh evolved heat = 283 kJ `therefore` By the combustion of 56 litre CO the `Delta H=(283)/(22.4)xx56 =707.5 kJ` So, the total heat evolved = 604.5+707.5=1312 kJ . |
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| 615. |
When a certain amount of ethylene was combusted, 6226 kJ heat was evolved. If heat of combustion of ethylene is 1411 kJ, the volume of `O_(2)` (at STP) that entered into the reaction is :A. 296.5 litresB. 300.3 litresC. `6226xx22.7` litresD. 22.7 litres |
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Answer» Correct Answer - B |
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| 616. |
Given `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g),Delta_(r)H^(Ө)= -92.4 kJ mol^(-1)` What is the standard enthalpy of formation of `NH_(3)` gas? |
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Answer» Correct Answer - `-46.2 kJ mol^(-1)` Standard enthalpy of formation of a compound is the change in ethalpy that takes place during the formation of 1 mole of a substance in its standrad from its constituent elements in their standrad state. Re-writing the given equation for 1 mole of `MH_(3(g))` `(1)/(2)N_(2(g))+(3)/(2)H_(2(g))rarrNH_(3(g))` `:.` Standard enthalpy of formation of `NH_(3(g))` `=1//2Delta_(r)H^(theta)` `1//2 (-92.4 kJ "mol"^(-1))` `=-46.2 kJ "mol"^(-1)` |
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| 617. |
When a certain amount of ethylene was burnt 622 kJ heat was evolved. If heat of combustion of ethylene is 1411 kJ, the volume of `O_(2)` (at NTP) that entered into the reaction isA. 296.5 mlB. 29.62 litreC. `6226xx22.4` litreD. 22.4 litre |
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Answer» Correct Answer - B `C_(2)H_(4)+3O_(2) rarr 2CO_(2) +2H_(2)O` `28 gm" "rarr 1411 kJ` `? larr" "622 kJ` |
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| 618. |
Given `: N_(2)(g) + 3H_(2)(g)rarr2NH_(3)(g), Delta_(r)H^(@) +- 92.4 kJ mol^(-1)`. What is the standard enthalpy of formation of `NH_(3)` gas ? |
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Answer» Reaction for the enthalpy of `NH_(3)(g) ` is `: (1)/(2) N_(2)(g) +(3)/(2) H_(2)(g) rarr NH_(3)(g)` `:. Delta_(f)H^(@) = - 92.4//2 = - 46.2 kJ mol^(-1)` |
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| 619. |
The value of `Delta_(f) H^(Θ)` for `NH_(3)` is `-91.8 kJ mol^(-1)`. Calculate enthalpy change for the following reaction. `2NH_(3) (g) rarr N_(2) (g) + 3H_(2) (g)` |
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Answer» Given, `(1)/(2) N_(2) (g) + (3)/(2) H_(2) (g) rarr NH_(3) (g), Delta(f) H^(Θ) = - 91.8 kJ mol^(-1)` (`Delta_(f) H^(Θ)` means enthalpy of formation of 1 mole of `NH_(3)`) `:.` Enthalpy change for the formation of 2 moles of `NH_(3)` `N_(2) (g) + 3H_(2) (g) rarr 2NH_(3) (g), Delta_(f) H^(Θ) = 2 xx -91.8 = - 183.6 kJ mol^(-1)` And for the reverse reaction, `2NH_(3) (g) rarr N_(2) (g) + 3H_(2) (g), Delta_(f) H^(Θ) = + 183.6 kJ mol^(-1)` Hence, the value of `Delta_(f) H^(Θ)` for `NH_(3)` is `+ 183.6 kJ mol^(-1)` |
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| 620. |
The cycle is shown in figure is made of one mole of perfect gas in a cylinder with a piston . The processes A to B and C to D are isochoric wheras process B to C and D to A are adiabatic, the work done in one cycle is `(V_(A)=V_(B)=V,V_(C)=V_(D) =2V "and " gamma=5//3)` A. `[1-4^(3)/(2)](P_(B)-P_(A))V`B. `(3)/(2)[1-3^(2)/(3)](P_(B)-P_(A))V`C. `(3)/(2)[1-2^(-2)/(3)](P_(B)-P_(A))V`D. `(5)/(2)[1-2^(-2)/(3)](P_(B)-P_(A))V` |
| Answer» Correct Answer - C | |
| 621. |
Which of the following thermodynamic relation is correct ?A. `dG =V d p -SdT`B. `dE = P dV + T dS`C. `dH = - V dP +T d S`D. `Dg= vDp+ S dT` |
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Answer» Correct Answer - a `G= H - TS ` But `H =U + PV` `:. G= U + PV - TS` Differentiating it completely ,we get `dG=dU +PdV+ VdP - TdS - SdT` …(i) But `dU =dq+dw` and `w= -PdV` `:. dU = d-PdV` or `dq = dU+ PdV` …(ii) Further, for averversible process, `dS= ( dq)/(T)` or `dq=TdS` ....(iii) From (ii) and (iii), `dU +PdV=T DeltaS` Substituting this in (i),weget`dG =VdP - SdT` |
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| 622. |
The enthalpy change for the combustion of `N_(2)H_(4)(l)` (Hydrazine) is `-622.2 kJ mol^(-1)`. The products are `N_(2)(g)` and `H_(2)O(l)`. If `Delta_(f)H^(Theta)` for `H_(2)O(l) is -285.8 kJ mol^(-1)`. The `Delta_(f)H^(Theta)` for hydrazine isA. `-336.4 kJ mol^(-1)`B. `+50.6 kJ mol^(-1)`C. `-622.2kJ mol^(-1)`D. `+336.4 kJ mol^(-1)` |
| Answer» `DeltaH = DeltaH_(P) - DeltaH_(R)` | |
| 623. |
During an isothermal expansion, a confined ideal gas does `-150 J` of work aginst its surroundings. This implies thatA. 150 J of heat has been removed from the gsB. 300 J of heat has been added to the gasC. no heat is transferred because the process is isothermalD. 150 J of heat has been added to the gas |
| Answer» Correct Answer - D | |
| 624. |
Calculate the enthalpy of formation of `Delta_(f)H` for `C_(2)H_(5)OH` from tabulated data and its heat of combustion as represented by the following equaitons: i. `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g), DeltaH^(Theta) =- 241.8 kJ mol^(-1)` ii. `C(s) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 393.5kJ mol^(-1)` iii. `C_(2)H_(5)OH (l) +3O_(2)(g) rarr 3H_(2)O(g) + 2CO_(2)(g), DeltaH^(Theta) =- 1234.7kJ mol^(-1)` a. `-2747.1 kJ mol^(-1)` b. `-277.7 kJ mol^(-1)` c. `277.7 kJ mol^(-1)` d. `2747.1 kJ mol^(-1)`A. `-2747.1 kJ mol^(-)`B. `277.7 kJ mol^(-)`C. `-277.7 kJ mol^(-)`D. `2747.1 kJ mol^(-1)` |
| Answer» Correct Answer - C | |
| 625. |
Calculate the enthalpy of formation of `Delta_(f)H` for `C_(2)H_(5)OH` from tabulated data and its heat of combustion as represented by the following equaitons: i. `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g), DeltaH^(Theta) =- 241.8 kJ mol^(-1)` ii. `C(s) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 393.5kJ mol^(-1)` iii. `C_(2)H_(5)OH (l) +3O_(2)(g) rarr 3H_(2)O(g) + 2CO_(2)(g), DeltaH^(Theta) =- 1234.7kJ mol^(-1)` a. `-2747.1 kJ mol^(-1)` b. `-277.7 kJ mol^(-1)` c. `277.7 kJ mol^(-1)` d. `2747.1 kJ mol^(-1)` |
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Answer» Required equation: `2C(s) +3H_(2)(g) +(1)/(2)O_(2)(g) rarr C_(2)H_(5)OH(l)` `2C(s) +2O_(2)(g) rarr 2CO_(2)(g), DeltaH^(Theta) = - 2xx 393.5 kJ` `3H_(2)(g) +O_(2)(g)rarr 3H_(2)O(g), DeltaH^(Theta) =- 3 xx 241.8 kJ` `3H_(2)O(l) +2CO_(2)(g) rarr C_(2)H_(5)OH(l) +3O_(2)(g),DeltaH^(Theta) =+ 1234.7 kJ` `ulbar({:(,"On adding,",,),(,"2C(s)+3H_(2)(g)+(1)/(2)O_(2)(g)rarrC_(2)H_(5)OH(l),,,),(,,DeltaH^(Theta)=-277.7 kJ,):})` |
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| 626. |
Calculate the enthalpy of formation of `Delta_(f)H` for `C_(2)H_(5)OH` from tabulated data and its heat of combustion as represented by the following equaitons: i. `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g), DeltaH^(Theta) =- 241.8 kJ mol^(-1)` ii. `C(s) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 393.5kJ mol^(-1)` iii. `C_(2)H_(5)OH (l) +3O_(2)(g) rarr 3H_(2)O(g) + 2CO_(2)(g), DeltaH^(Theta) =- 1234.7kJ mol^(-1)` a. `-2747.1 kJ mol^(-1)` b. `-277.7 kJ mol^(-1)` c. `277.7 kJ mol^(-1)` d. `2747.1 kJ mol^(-1)`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 By definition, the enthalpy of formation of `H_(2) O (1)` is the enthalpy change when one mole of `H_(2) O (1)` is synthesized from its elements in their standard state: `H_(2) (g) + (1)/(2) O_(2) (g) = H_(2) O (1)`, `Delta_(f) H^(@) = - x_(2) kJ mol^(-1)` |
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| 627. |
In an isothermal expansion of an ideal gas. Select wrong statement:A. there is no change in the temperature of the gasB. there is no change in the internal energy of the gasC. the work done by the gas is equal to the heat supplied to the gasD. the work done by the gas is equal to the changes in its internal energy |
| Answer» Correct Answer - D | |
| 628. |
A system undergoes a process which absorbed `5 KJ` of heat and undergoing an expansion againt external pressure of `1 atm`, during the process change in internal energy is `300 J`. Then predict the change in volume (lit)A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B `Delta U= q+W= q-P_(ext)(V_(2)-V_(1))` `300=500-1(Delta V)xx100` `Delta V=(500-300)/(100)=2 lit` |
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| 629. |
Comment on the following statements: (a) An exothermic reaction is always thermodynamically spontaneous. Reaction with `DeltaG lt O` always have an equilibrium constant greater than `1` |
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Answer» Correct Answer - (a) When `DeltaH=-ve "and" TDeltaS`=- ve also `T DeltaS gt DeltaH` then `DeltaG`=+ ve in such case, the reaction will not be spontaneous. (b) `-DeltaG^(@) =2.303 RT "log"K_(c)` If `DeltaG_(@)` is negative (not to say that `DeltaG^(@) lt 0`) then `K_(c)` will be +ve and will have value greater than one. When `DeltaH =-ve` and `T DeltaS =-ve `,also `TDeltaS gt DeltaH " then " DeltaG =+ ve` in such case, the reaction will not be spontaneous. (b) `-DeltaG^(@) =2.303"RT log"K_(c)` If `DeltaG^(@) ` is negative (not to say that `DeltaG^(@) lt O)` then `K_(c)` will be `+ve` and will have value greather than one. |
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| 630. |
For the equilibrium, `PCI_(5)(g) hArr PCI_(3)(g) +CI_(2)(g) at 25^(@)C K_(c) =1.8 xx 10^(-7)` Calculate `DeltaG^(Theta)` for the reaction `(R = 8.314 J K^(-1) mol^(-1))`. |
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Answer» We know, `DeltaG^(Theta) =- 2.303 RT log_(10)K_(c)` `=- 2.303 xx 8.314 xx 298 log (1.8 xx 10^(-7))` `= 38484 J mol^(-1) = 38.484 kJ mol^(-1)` |
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| 631. |
For the followinf set of balanced reactions, `N_(2)O_(5)O_(2)rarr2NO_(2)+O_(3)" "DeltaH=200kJ` `NO_(2)+O_(2)rarr2NO+O_(3)" "DeltaH=20kJ` `10` moles of `N_(2)O_(5)` and 30 moles of `O_(2)` where taken in a chamber to cause complete conversion of `N_(2)O_(5)` to `NO_(2^.)NO_(2)` partially reacts with remaining oxygen such that volume percenfagr of `O_(3)` is `50%` Calculated overall enthaolpy change kJ. |
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Answer» Correct Answer - 1700 |
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| 632. |
In which of the following entrophy is creasing?A. `2SO_(2)(g)+O_(2)(g) to2SO_(3)(g)`B. `Ca^(2+)(aq.)+2F^(-)(aq)toCaF_(2)(s)`C. `N_(2)(g,1atm,300K) to N_(2), (g, 2atm, 300K)`D. `O_(2)(g,1 L,300K) to O_(2), (g, 2L, 300K)` |
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Answer» Correct Answer - D |
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| 633. |
Calcuate entrophy change in each step for an ideal gas (monoatomic) `underset((" 1 atm, 22.4 l, 273 K"))(State(A)) to underset(("1 atm, 33.6 l, 409.5 K"))(State(B)) to underset(("2 atm, 33.6 l, 819K"))(State(C))` |
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Answer» `DeltaS_(A - B)= .^(n)C_(p) ln.(T_(2))/(T_(2))+nRln .(P_(1))/(P_(2)) = nC_(p) ln .(409.5)/(273) = C_(p)ln .(3)/(2)` `DeltaS_(A - B)= .^(n)C_(v) ln.(T_(2))/(T_(2))+nRln .(V_(1))/(V_(2)) = C_(v) ln .(546)/(273) = C_(v)ln 2` |
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| 634. |
In an isothermal reversible expansion, if the volume of 96 gm of oxygen at `27^(@)C` is increased from 70 litres to 140 litres , then the work done by the gas will beA. `300Rlog_(10)2`B. `81Rlog_(e)2`C. `900Rlog_(10)2`D. `2.3xx900Rlog_(10)2` |
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Answer» Correct Answer - D |
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| 635. |
The change in entrophy of 2 moles of ideal gas upon isothermal expansion at `243.6 K` from 20 liter until the pressure becomes 1 atm, is :A. 1.385 cal/KB. `-1.2 cal//K`C. `-30 J//mol.K`D. None of these |
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Answer» Correct Answer - C `DeltaS= nRT In (V_(2))/(V_(1))` `V_(2) = 40` `V_(1) = 20` `DeltaS = 2 xx 2 In 2` ` = 4 In 2 = 2.77 cal` |
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| 636. |
How many of the following physical properties are extensive: (i) Free energy (ii) vapour pressure (iii) mole (iv) kinetic energy (v) Entropy (vi) Internal energy (vii) Enthalpy (viii) specific heat capacity (ix) Coefficient of viscosity |
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Answer» Correct Answer - 7 `1^(st)` process is isobaric volume become double so, `T` is also become double So, for adiabatic `T_(i)=2xx300=600` `V_(i)=40` `T_(f)=?, V_(f)=113` for adiabatic `P_(1)V_(1)^(gamma)="constant"=TV^(t-1)="constant"` `T_(i)V_(i)^(5//3-1)=T_(f)V_(f)^(2//3)` `600xx(40)^(2//3)=T_(f)xx(40)^(2//3)=T_(f)xx(110)^(2//3)` `T_(f)=300 K` `(300)/(100)=3` |
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| 637. |
Out of molar entropy (I), specific volume (II), heat capacity (III), volume (IV), extensive properties are :A. I, IIB. I, II, IVC. II, IIID. III, IV |
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Answer» Correct Answer - D Heat capacity and volume are extensive properties. |
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| 638. |
How much of the following are extensive properties ? Reasistance, Electromotive force, Heat enthalpy , Dipole moment, Heat capacity, Specific Heat, Denisity specific volume |
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Answer» Correct Answer - 7 Extensive properties arre: Enthalpy, Heat capacity, Resistance |
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| 639. |
How much of the following are intensive properties ? Vapour pressure, Molarity, Refractive index, Dielectric constant, Osmotic pressure, Molarity, specific gravity, Molar volume |
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Answer» Correct Answer - 0.694 atm Intensive properties are: Vapour pressure, molarity, Refractive index, Dielectric constant, molarity, specific gravity, molar volume |
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| 640. |
16g of `O_(2)` at `28^(@)C` is compressed to half of its initial volume under reversible isothermal conditions. If this gas behaves ideally, the workdone and change in internal energy in this process areA. `Delta U = 0` and w = -1251.2 ln 2 JB. `Delta U = 0` and w = +1251.2 ln 2JC. `w = Delta U = +1251.2 ln 2J`D. `Delta U = -1251.2 ln 2J` and `w = +1251.2 ln 2J` |
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Answer» Correct Answer - B `Delta U = Q+W` |
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| 641. |
a geyser heats water flowing at the rate of 4 litrre per minute from `30^(@) "to" 85^(2) C`. If the geyser operates on a gas burner then the amount of heat used per minute isA. `9.24xx10^(5)J`B. `6.24xx10^(7)J`C. `9.24xx10^(7)J`D. `6.24xx10^(5)J` |
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Answer» Correct Answer - A Here k,Volume of wster heated =4 litre `mi n^(-1)` 1 litre =1000g, then mass of water `=4xx1000g min ^(-1)=4000g mi n^(-1)` Rise in temperature `DeltaT=T_(2)-T_(1)=(85-30)^(@)C=55^(@)C` specific heat of water `S =4.2 J g^(-1)^(@)C^(-1)` The amount of heat used, `DeltaQ =msDelta` `=4000xx4.2xx55=924000` `=9.24xx10^(5) J mi n^(-1)` |
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| 642. |
The standard state Gibba free energies of formations of C("graphite") and ("diamond") at ``T =298k are `DeltafG^(@) [ C("graphite") = " O "kJ mol^(-1)``DeltafG^(@) [ C("graphite") = " O "kJ mol6^(-1)` The standed state means that the presses should be 1 bar, and substance of graphite [C(graphite)] to diamond [C(diamond )] reduces its volume by `2xx 10^(10) m^(-1)` If C(graphite) is converted to C(diamond ) isothemally at T= 298k the pressure at which C( graphite) is in equilibrium with C(diamond) ,is [ useful infromation : `1 J - Kgm^(2) s^(-2) ` ` 1pa = 1kg m^(-1) s^(-2) : 1^(-) = 10 ^(5) pa]`A. 58001 barB. 1450 barC. 14501 barD. 29001 |
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Answer» Correct Answer - c |
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| 643. |
`deltaU` is equal toA. Isochoric workB. isobaric workC. adiabatic workD. isothermal work |
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Answer» Correct Answer - c |
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| 644. |
The standard enthalpy of formation `(Delta_(f)H^(@))` at `298K ` for methane `(CH_(4(g)))` is `-74.8kJ mol^(-1)`. The additional information required to determine the average energy for `C-H` bond formation would be `:`A. the dissociation energy of `H_(2)` and enthelpy of sublimation of carbon.B. latent heat of vaporisation of methane.C. the first four ionization energies of carbon and electron gain enthlpy of hydrogen.D. the dissociation energy of hydrogen molecule, `H_(2)` |
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Answer» Correct Answer - A |
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| 645. |
If a `298K` the bond energies of `C-H,C-C,C=C` and `H-H` bonds are respectivly `414, 347, 615KJmol^(-1)` , the vlaue of enthalpy change for the reaction `H_(2)C=CH_(2)(g)+H_(2)(g)+H_(2)(g)rarrH_(3)C-CH_(3)(g)` at `298K` will beA. `-250 kJ`B. `+125 kJ`C. `-125 kJ`D. `+250 kJ` |
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Answer» Correct Answer - C `DeltaH=H_(R)-H_(P)` |
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| 646. |
Heat of combustion of `CH_(4),C_(2)H_(6), C_(2)H_(4)` and `C_(2)H_(2)` gases are -212.8, -212.8, -373.0, -337.0 and -310.5 Kcal respectively at the same temperature. The best fuel among these gases is :A. `CH_(4)`B. `C_(2)H_(6)`C. `C_(2)H_(4)`D. `C_(2)H_(2)` |
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Answer» Correct Answer - A Higher the calorific value, better is the fuel. Calorific value `=(Delta H_("Comb."))/("Molecular weight")` |
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| 647. |
Assertion: When a real gas is allowed to expand adiabatically through a fine hole from a region of high presssure to a region of low pressure, the temperature of gas falls in a completely insulated container. Reason: Work is done at the cost of internal energy of the gas.A. If both (A) and (R ) are correct, and (R ) is the correct explanation for (A).B. If the both (A) and (R ) are correct, but(R ) is not a correct explanation for (A).C. If (A) is correct, but (R ) is incorrect.D. If (A) is incorrect, but (R ) is correct. |
| Answer» In adiabatic process, `q = 0` hence is expansion process temperature of the system decreases. | |
| 648. |
The enthalpy of formation for `C_(2)H_(4)(g),CO_(2)(g)` and `H_(2)O(l)` at `25^(@)C` and 1 atm. Pressure are 52, -394 and -286 KJ `mol^(-1)` respectively. The entahlpy of combustion of `C_(2)H_(4)` will be :-A. `+1412 "KJ mole"^(-1)`B. `-1412 "KJ mole"^(-1)`C. `+142.2 "KJ mole"^(-1)`D. `-141.2 "KJ mole"^(-1)` |
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Answer» Correct Answer - B `C_(2)H_(2(g))+3O_(2(g))rarr 2CO_(2(g))+2H_(2)O_((l)) , Delta H_("Comb.") = ?` `Delta H = S?igma (Delta H_(f))_(P)-Sigma (Delta H_(f))_(R )` |
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| 649. |
An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If `T_i` is the initial temperature and `T_f` is the final temperature, which of the following statement is correct ?A. `(T_(f))_(irrev)gt(T_(f))_(rev)`B. `T_(f)gtT_(i)` for both reversible process but `T_(f)=T_(i)` for irreversible processC. `(T_(f))_(rev)=(T_(f))_(rev)`D. `T_(f)=T_(i)` for both reversible and irreversible process |
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Answer» Correct Answer - a Work done in reversible process is maximum. Thus `T_(2)lt lt T_(1)` in reversible process or `T_(2 rev)lt lt T_(2 irrev)` |
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| 650. |
The work done in adiabatic compression of `2` mole of an ideal monoatomic gas by constant external pressure of `2 atm` starting from initial pressure of `1 atm` and initial temperature of `30 K(R=2 cal//"mol-degree")` |
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Answer» Correct Answer - 72 `q=0, DeltaU=W` `nC_(1)(T_(2)-T_(2))= -P_(ext)(V_(2)-V_(1))` `n(3)/(2)R(T_(2)-30)= -2 ((nRT_(2))/(2)-(nRxx30)/(1))` `rArr (3)/(2)(T_(2)-30)=(60-T_(2))rArrT_(2)=42K` `W=nC_(v)(T_(2)-T_(1))=2xx(3)/(2)xx2(42-30)=72 cal` |
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