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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.
| 1. |
At room temperature `(300K)`, the rms speed of the molecules of a certain diatomic gas is found to be `1930 m//s`. Can you gusess name of the gas? Find the temperature at which the rms speed is double of the speed in part one `(R = 25//3 J// mol -K)` |
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Answer» Correct Answer - `H_(2), 1200K` `V_(rms) = sqrt((3RT)/(M)) rArr 1930` `M = (25 xx3)/(193 xx193) kg//mol = (25 xx 3)/(193xx193) xx 1000g//mol` `= 2.01348 g//mol ~~ 2g//mol` Hence, gas is `H_(2)`. Let `T_(1)` be temperature at which `V_(rms)` is doubled. `2V_(rms) = sqrt((3RT_(1))/(M))` from, equation (i) and (ii), `sqrt((T_(1))/(T)) = 2` `rArr T_(1) = 4T =4 (300)K, T_(1) = 1200K` `V_(rms) =sqrt((3RT)/(M)) rArr 1930 = sqrt((3((25)/(3))300)/(M))` `M = (25 xx3)/(193 xx 193) kg//mol` `=(25 xx3)/(193xx193) xx 100 g//mol, =2.01348 g//mol` `~~ 2g//mol, 2V_(rms) = sqrt((3RT_(1))/(M)) sqrt((T_(1))/(T)) =2` `rArr T_(1) = 4T = 4 (300)K T_(1) = 1200K` |
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| 2. |
Comprehension-1 Two closed identical conducting containers are found in the laboratory of an old scientist. For the verification of the gas some experiments are performed on the two boxes and the results are noted. Experiment 1. When the two containers are weighed `W_(A) = 225 g, W_(B) = 160 g` and mass of evacuated container `W_(C) = 100 g`. Experiment 2. When the two containers are given same amount of heat same temperature rise is recorded. The pressure changes found are `Deltap_(A) = 2.5 atm, Deltap_(B) = 1.5 atm`. Required data for unknown gas: `|(underset("molar mass")(Mono)",He,Ne,Ar,Kr,Xe,Rd),(,4g,20g,40g,84g,131g,222g),(underset(molar mass)(Dia),H_(2),F_(2),N_(2),O_(2),CI_(2),),(,2g,19g,28g,32g,71g,)|` Identify the type of gas filled in container `A` and `B` respectively.A. Mono, MonoB. Dia, DiaC. Mono, DiaD. Dia, Mono |
| Answer» Correct Answer - C | |
| 3. |
Two containers of equal volume enclose equal masses of hydrogen and helium at `300K`. Find (i) the ratio of pressure and (ii) the ratio of total `K.E.` of all the molecules of the two gases. |
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Answer» Correct Answer - `P_(H_(2)) = 2P_(He), (kE_(H_(2)))/(kE_(He)) = (10)/(3)` `V_(H2) = V_(He)` `m_(H2) = m_(He), T = 300 K` (i) `P_(H2) V_(H2) = n_(H2) RT ………(i)` `P_(He). V_(He) = n_(He) = RT ……..(ii)` `(P_(H_(2)))/(P_(He)) = ((m_(H_(2))//2))/(m_(He)//4)` `rArr P_(H2) = 2P_(He)` (ii) `K.E. = (f)/(2)nRT` `(KE_(H_(2)))/(KE_(He)) =(f_(H_(2).n_(H_(2))))/(f_(He).n_(He)) =(5((m_(H_(2)))/(2)))/(3((m_(He))/(4))) =(10)/(3)` |
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| 4. |
Find the average momentum of molecules of hydrogen gas in a container at temperature `300K`. |
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Answer» Correct Answer - Zero Average momentum of molecule of gas `vecP_(av) = MvecV_(av)` `[vecV_(av) = 0` Average velocity of gas molecules at any temperature is always zero] Hence, `vecP_(av) = 0` |
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| 5. |
Comprehension-1 Two closed identical conducting containers are found in the laboratory of an old scientist. For the vertification of the gas some experiments are performed on the two boxes and the results are noted. Experimenet 1. When the two containers are weighed `W_(A) = 225tg, W_(B) = 160 g` and mass of evacuated container `W_(C) = 100g`. Experiment 2. When the two containers are given same amount of heat same temperature rise is recorded. The pressure changes found are `DeltaP_(A) = 2.5 atm, Deltap_(B) = 1.5 atm`. Required data for unknown gas: `|(underset("molar mass")(Mono)",He,Ne,Ar,Kr,Xe,Rd),(,4g,20g,40g,84g,131g,222g),(underset(molar mass)(Dia),H_(2),F_(2),N_(2),O_(2),CI_(2),),(,2g,19g,28g,32g,71g,)|` Identify the gas filled in the container `A` and `B`.A. `N_(2), Ne`B. `He, H_(2)`C. `O_(2),Ar`D. `Ar,O_(2)` |
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Answer» Correct Answer - D Since `n_(1) = (5)/(3) n_(2)` Therefore `(125)/(M_(A)) = (5)/(3) ((60)/(M_(B)))` (From experiment `1 : W_(A) = 125 gm & W_(B) = 60 gm)` `rArr 5M_(B) = 4M_(A)` The above relation holds for the pair -Gas `A : Ar` and Gas `B : O_(2)` |
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| 6. |
A gaseous mixture enclosed in a vessel of volume V consists of one mole of a gas A with `gamma=(C_p//C_v)=5//3` and another gas B with `gamma=7//5` at a certain temperature T. The relative molar masses of the gasses A and B are 4 and 32, respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation `PV^(19//13)=constant`, in adiabatic processes. (a) Find the number of moles of the gas B in the gaseous mixture. (b) Compute the speed of sound in the gaseous mixture at `T=300K`. (c) If T is raised by 1K from 300K, find the `%` change in the speed of sound in the gaseous mixture. (d) The mixtrue is compressed adiabatically to `1//5` of its initial volume V. Find the change in its adaibatic compressibility in terms of the given quantities. |
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Answer» (a) As for ideal gas `C_(P) - C_(V) = R`and `gamma = (C_(P)//C_(V))`, So, `gamma-1 =(R )/(C_(V))` or `C_(V) = (R )/((gamma-1))` `(C_(V))_(1) = (R )/((5//3)-1) = (3)/(2)R`, `(C_(V))_(2) = (R )/((7//5)-1) = (5)/(2)R` and `(C_(V))_(mix) = (R )/((19//13)-1) = (13)/(6)R` Now from conservation of energy, i.e., `DeltaU = DeltaU_(1) +DeltaU_(2)`, `(n_(1)+n_(2)) (C_(V))_(mix) DeltaT = [n_(1)(C_(V))_(1) +n_(2) (C_(V))_(2)] DeltaT` i.e., `(C_(V))_(mix) = (n_(1)(C_(V))_(1)+n_(2)(C_(V))_(2))/(n_(1)+n_(2))` We have `(13)/(6)R = (1xx(3)/(2)R+nxx(5)/(2)R)/(1+n)=((3+5n))/(2(1+n))` or, `13 + 13n = 9 +15n`, (b) Molecular weight of the mixture will be given by `M = (n_(A)M_(A)+n_(B)M_(B))/(n_(A)+n_(B)) = ((1)(4)+2(32))/(1+2)` `M = 22.67` Speed of sound in a gas is given by `v = sqrt((gamma RT)/(M))` Therefore, in the mixture of the gas `v = sqrt(((19//13)(8.31)(300))/(22.67 xx 10^(-3)))m//s` `v ~~ 401 m//s` (c) `v prop sqrt(T)` or `v = KT^(1//2).....(2)` `rArr (dv)/(dT) = (1)/(2) KT^(-1//2)` `rArr dv = K ((dT)/(2sqrt(T)))` `rArr (dv)/(v) = (k)/(v) ((dT)/(2sqrt(T)))` `rArr (dv)/(v) = (1)/(sqrt(T)) ((dT)/(2sqrt(T))) = (1)/(2) ((dT)/(T))` `rArr (dv)/(v) xx 100 =(1)/(2) ((dT)/(T)) xx 100 = (1)/(2) ((1)/(300)) xx 100 = 0.167 = (1)/(6)` Therefore, percentage change in speed is `0.167%`. |
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| 7. |
A gaseous mixture consists of 16g of helium and 16 g of oxygen. The ratio `(C_p)/(C_v)` of the mixture isA. `1.59`B. `1.62`C. `1.4`D. `1.54` |
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Answer» Correct Answer - B `C_(v) =(n_(1)C_(v_(1))+n_(2)C_(v_(2)))/(n_(1)+n_(2))` For helium, `n_(1) = (16)/(4) = 4` and `gamma_(1) = (5)/(3)` For oxygen `n_(2) = (16)/(32) = (1)/(6)` and `gamma_(2) = (7)/(5)` `C_(v_(1)) = (R )/(gamma_(1)-1) = (R )/((5)/(3)-1) -(3)/(2)R` `C_(v_(2)) = (R )/(gamma_(2)-1) = (R)/((7)/(5)-1)=(5)/(2)` `:. C_(v) = (4xx(3)/(2)R+(1)/(2),(5)/(2)R)/(4+(1)/(2)) = (6R+(5)/(4)R)/((9)/(2))` `= (29 R xx 2)/(9xx4) = (29R)/(18), C_(P) = C_(V) +R` `rArr (C_(P))/(C_(V)) = (C_(V)+R)/(C_(V)) = 1+(R)/(C_(V))` `rArr (C_(P))/(C_(V)) = (R)/((29)/(18)R) +1 rArr (C_(P))/(C_(V)) = (18)/(19) +1` `= (18 +19)/(29) = 1.62` |
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| 8. |
A cylinder containing an ideal gas ( see figure ) and closed by a movable pistong is submerged in an ice- water mixture. The pistone is quickly pushed down from position (1) to position (2)( process `AB)`. The piston is held at position `(2)` until the gas is again at `0^(@)C` ( processs `BC)`. Then the pistone is slowly raised back to position (1) ( process `CA)` Which of the following `P-V` diagram will correctly represent the processes `AB, BC` and `CA` and the cycle `ABCA?`A. B. C. D. |
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Answer» Correct Answer - A Correct graph is shown in option `(A)` Process `1-2` adiabatic process, Process `2-3` Isohoric process, process `3-1` Isothermal process. |
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| 9. |
In a mixture of nitrogen and helium kept at room temperarture. As compared to a helium molecule nitrogen molecule hits the wallA. With greater average speedB. with smaller average speedC. with greater average kinetic energyD. with smaller average kinetic energy |
| Answer» Correct Answer - B::C | |
| 10. |
The average translational kinetic energy of nitrogen gas molecule is `0.02eV (1eV - 1.6 xx 10^(-19)J)`. Calculate the temperatuire of the gas. Boltzmann constant `k = 1.38 xx 10^(-23) J//K`. |
| Answer» Correct Answer - `(32000)/(207)K` | |
| 11. |
The cylinder shown in the figure has conducting walls and temperature of the surrounding is T, the piston is initially in equilibrium, the cylinder contains n moles of a gas, Now the piston is displaced slowly by an external agent to make the volume double of the initial. Find work done by external agent in term of n, R, T |
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Answer» `1^(st)` Method: Work done by external agent is positive, because `F_(ext)` and displacement are in the same direction. Applying equilibrium condition when pressure of the gas is `P` `PA +F_(ext) =P_(atm)A` `F_(ext) = P_(atm)A - PA` `W_(ext) =int_(0)^(d) F_(ext) dx` `= int_(0)^(d) P_(atm) Adx - int_(0)^(d) PA dx = P_(atm) A int_(0)^(d) dx - int_(V)^(2V) (nRT)/(V) dV` `= P_(atm) Ad - nRT In2` `= P_(atm). V_(0) - nRT In2 = nRT (1-In2)` `2^(nd)` Method Applying work energy theorem on the piston As `W_(all) = DeltaK.E` `DeltaK.E = 0` (given) `W_(gas) +W_(atm) +W_(ext) = 0` `nRT In (V_(f))/(V_(i)) - nRT +W_(ext) = 0` `W_(ext) = nRT (1- In2)` |
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| 12. |
If `Q` amount of heat is given to a diatomic ideal gas in a process in which the gas perform a work `(2Q)/(3)` on its surrounding. Find the molar heat capacity (in terms of `R`) for the process. |
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Answer» Correct Answer - `7.5R` `Q = nCDeltaT` As `Q = nC DeltaT` ………..(i) Also `DeltaU = Q - (2Q)/(3) = (Q)/(3)` `nC_(v)DeltaT = (Q)/(3), (n5RDeltaT)/(2) = (Q)/(3)` `rArr nDeltaT = (2Q)/(15R) ………(ii)` From (1) and (ii) `C = (Q15R)/(2Q) = 7.5R`. |
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| 13. |
When an ideal diatomic gas is heated at constant pressure fraction of the heat energy supplied which increases the internal energy of the gas isA. `(2)/(5)`B. `(3)/(5)`C. `(3)/(7)`D. `(5)/(7)` |
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Answer» Correct Answer - D `dQ = dW +dU , dQ = PdV +dU` `dQ = nRdT +dU , dQ = (2dU)/(f) +dU` `[dU = (nfRdT)/(2)]` `(dU)/(dQ) = (1)/(((2)/(f)+1)), (dU)/(dQ) = (5)/(7)` |
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| 14. |
Supposing the distance between the atoms of a diatomic gas to be constant, its specific heat at constant volume per mole (gram mole) isA. `(5)/(2)R`B. `(3)/(2)R`C. `R`D. `(7)/(2)R` |
| Answer» Correct Answer - A | |
| 15. |
Comprehension-3 An ideal gas initially at pressure `p_(0)` undergoes a free expansion (expansion against vacuum under adiabatic conditions) until its volume is `3` times its initial volume. The gas is next adiabatically compressed back to its original volume. The pressure after compression is `3^(2//3)p_(0)`. The gasA. is monoatomic.B. is diatomicC. is polyatomicD. type is not possible to decide from the given information. |
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Answer» Correct Answer - A For adiabatic compression, initial conditions are `(p_(0))/(3)` and `3v_(0)`. Find volume and pressure are `v_(0)` and `3^(2//3) p_(0)`. `(p_(0))/(3) (3v_(0))^(gamma) = 3^(2//3) p_(0)(v_(0))^(gamma) rArr 3^(gamma-1) = 3^(2//3)` or `gamma -1 =(2)/(3) rArr gamma - (5)/(3)` i.e. gas is monoatomic |
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| 16. |
Comprehension-3 An ideal gas initially at pressure `p_(0)` undergoes a free expansion (expansion against vacuum under adiabatic conditions) until its volume is `3` times its initial volume. The gas is next adiabatically compressed back to its original volume. The pressure after compression is `3^(2//3)p_(0)`. The pressure of the gas after the free expansion is:A. `(p_(0))/(3)`B. `p_(0)^(1//3)`C. `p_(0)`D. `3p_(0)` |
| Answer» Correct Answer - A | |
| 17. |
Find the amount of work done to increase the temperature of one mole of ideal gas by `30^(@)C` .if its is expanding under the condition `V prop R^(2//3) (R = 8.31 J//mol -K):`A. `16.62 J`B. `166.2J`C. `1662J`D. `1.662J` |
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Answer» Correct Answer - B `V = kT^(2//3)` `dV = (2)/(3) kT^(-(1)/(3)) dT , W = int PdV = int (nRT)/(V)` `= R int (T)/(V) dV = R (2)/(3) int (TKT^(-(1)/(3))dT)/(KT^((1)/(2)))` `= (2)/(3) R (T_(2) - T_(1)) = (2)/(3) R (30) = 20 (8.31) = 166.2 J` |
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| 18. |
`V = k((P)/(T))^(0.33)` where k is constant. It is an,A. isothermal processB. adiabatic processC. isochoric processD. isobaric process |
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Answer» Correct Answer - C `V = k ((P)/(T))^(0.33)` where `k` is constant. It is an, `V = k ((P)/(T))^(0.33) , V = k ((nRT)/(VT))^(0.33)` `V^(1.33) = const , V = const` `:.` process is isochroic |
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| 19. |
When an ideal gas undergoes an adiabatic change causing a temperature change `DeltaT` (i) there is no heat ganied or lost by the gas (ii) the work done by the gas is equal to change in internal eenrgy (iii) the change in internal energy per mole of the gas is `C_(V)DeltaT`, where `C_(V)` is the molar heat capacity at constant volume.A. (i),(ii),(iii) correctB. (i),(ii) correctC. (i),(iii) correctD. (i)correct |
| Answer» Correct Answer - C | |
| 20. |
In given figure, A uniform cylindrical tube closed at one end, contains a pallet of mercury `5cm` long. When the tube is kept verticaly with the closed end downward, the length of the air column trapped is `30cm`. Now the tube is inverted so that the closed end goes up find the new length of the air column trapped. atmospheric pressure `=75cm` of mercury. (Assume temperature remains constant in this process) |
| Answer» Correct Answer - `(240)/(7)cm` | |
| 21. |
Two moles of an ideal monoatomic gas are contained in a vertical cylinder of cross sectional area `A` as shown in the figure. The piston is frictionless and has a mass m. At a certain instant a heater starts supplying heat to the gas at a constant rate `q J//s`. Find the steady velocity of the piston under isobaric condition. All the boundaries are thermally insulated. |
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Answer» Correct Answer - `(2q)/(5(mg+P_(0)A))` `dQ_(p) = nC_(p)dT` Let heat be supplied for time dt `:. qdt =2. (5R)/(2). ([(mg)/(A)+P_(0)]Adx)/(2R)` where `[dT = (Pdv)/(nR)]` `:. qdt = (5)/(2) (mg +P_(0)A) dx` `:. (dx)/(dt) = (2q)/(5(mg+P_(0)A))`. |
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| 22. |
Calorie is defined as the amount of heat required to raise temperature of 1 g of water by `1^@C` and it is defined under which of the following conditions?A. `13.5^(@)C to 14.5^(@)C at 76 mm of Hg`B. `14.5^(@)C to 15.5^(@)C at 760 mm of Hg`C. `6.5^(@)C at 75^(@)C at 76 mm of Hg`D. `98.5^(@)C to 99.5^(@)C at 760 mm of Hg` |
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Answer» Correct Answer - B Above statement is a definition of one calorie. |
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| 23. |
A thermodynamic process of one mole ideal monoatomic gas is shown in figure. The efficiency of cyclic process `ABCA` will be |
| Answer» Correct Answer - 13 | |
| 24. |
In figure, a sample of an ideal gas is taken through the cyclic process abca. `800J` of work is done by the gas during process ab. If gas absorb no heat in process ab, rejects `100J` of heat during bc and absord `500J` of heat during process `ca`. Then (a) find the internal energy of the gas at `b` and `c` if its is `1000J` at a (b). Also calculate the work by the gas during the part `bc`, |
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Answer» (a) In process ab `DeltaQ = DeltaU +W` `Q = U_(B) - 100 +800` `U_(B) = 200J` for Cyclic process `DeltaQ = DeltaU +W` `400 = 0 +800 +W_(BC)` `W_(BC) =- 400 J` for process bc `DeltaQ = DeltaU +W` `- 100 =- 400 +U_(c) - 200 :. U_(c) = 500J` |
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| 25. |
The figure below shows the variation of specific heat capacity (C) of a solid as a function of temperature (T). The temperature is increased continuously form 0 to 500K at a constant rate. Ignoring any volume change, the following statement (s) is (are) correct to a reasonable approximation. A. the rate at which heat is absorbed in the range `0-100K` varies linerly with temperature `T`.B. heat absorbed in increasing the temperature from `0 -100K` is less than the heat required for increasing the temperature form `400-500K`.C. there is no change in the rate of heat absorbtion in the range `400-500K`.D. the rate of heat absorption increase in the range `200-300K`. |
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Answer» Correct Answer - B::C::D `q = mCT, (dq)/(dt) = mc (dT)/(dt)` `R =` rate or absortion of heat `= (dq)/(dt) prop C` (i) in `0 - 100k C` increases, so `R` increases but not lineraly (ii) `Deltaq = mCDeltaT` as `C` is more in `(400k - 500k)` then `(0-100k)` so heat is increasing. (iii) `C` remains constant so there no change in `R` from `(400k - 500k)` (iv) `C` is increases so `R` is increases in range `(200 k - 300k)` |
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| 26. |
Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure Efficiency of this cycle is nearly: (Assume the gas to be close to ideal gas) A. `15.4%`B. `9.1%`C. `10.5%`D. `12.5%` |
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Answer» Correct Answer - A `eta = (p_(0)v_(0))/((f)/(2)(p_(0)v_(0))+(f)/(2)(2p_(0))v_(0)+2p_(0)v_(0)) = (1)/((3)/(2)+3+2) = (200)/(13) = 15.4%` |
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| 27. |
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio `C_P//C_V` for the gas isA. `4//3`B. `2`C. `5//3`D. `3//2` |
| Answer» Correct Answer - D | |
| 28. |
In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas release `20 J` of heat and `8 J` of work is done on the gas. If initial internal energy of the gas was `30 J`, what will be the final internal energy? |
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Answer» Correct Answer - `18J` `dU = dQ - dW, U_(2) - U_(1) =- 20 +8 =- 12` `U_(2) = 30 - 12 = 18J` |
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| 29. |
A solid body of constant heat capacity `1J//^@C` is being heated by keeping it contact with reservoirs in two ways: (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with 8 reservoir such that each reservoir supplies same amount of heat. In both the cases body is brought from initial temperature `100^C` to final temperature `200^@C`. Entropy change of the body in the tow cases respectively is :A. `ln 2, 4ln2`B. `ln2, ln 2`C. `ln2, 2ln, 2`D. `2ln 2n, 8ln2` |
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Answer» Correct Answer - B Since entropy is a state function, therefore change in entropy in both the processes should be same. Therefore correct option is (2) |
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| 30. |
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume `V_1` and contains ideal gas at pressure `P_1` and temperature `T_1`.The other chamber has volume `V_2` and contains ideal gas at pressure `P_2` and temperature `T_2`. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will beA. `(T_(1)T_(2)(p_(1)V_(1)+p_(2)V_(2)))/(p_(1)V_(1)T_(2)+p_(2)V_(2)T_(1))`B. `(p_(1)V_(1)T_(1)+p_(2)V_(2)T_(2))/(p_(1)V_(1)+p_(2)V_(2))`C. `(p_(1)V_(1)T_(2)+p_(2)V_(2)T_(1))/(p_(1)V_(1)+p_(2)V_(2))`D. `(T_(1)T_(2)(p_(1)V_(1)+p_(2)V_(2)))/(p_(1)V_(1)T_(1)+p_(2)V_(2)T_(2))` |
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Answer» Correct Answer - A As no, work is done and system is thermally insulated from surrounding, it mean sum of internal energy of gas in two partitions is constant i.e, `U = U_(1) +U_(2)` Assume both gases have same degree of freedom, then Assuming both gases have same degree of freedom, then `U = (f(n_(1)+n_(2))RT)/(2)` and `U_(1) = (fn_(1)RT_(1))/(2), U_(2) = (fn_(2)RT_(2))/(2)` `n_(1) = (P_(1)V_(1))/(RT_(1))` and `(P_(2)V_(2))/(RT_(2))` Solving we get `T = ((p_(1)V_(1)+p_(2)V_(2))T_(1)T_(2))/(p_(1)V_(1)T_(2)+p_(2)V_(2)T_(1))` |
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| 31. |
Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature `T_0`, while Box contains one mole of helium at temperature `(7/3)T_0`. The boxes are then put into thermal contact with each other, and heat flows between them until the gasses reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gasses, `T_f` in terms of `T_0` isA. `T_(f) = (3)/(7)T_(0)`B. `T_(f) = (7)/(3)T_(0)`C. `T_(f) = (3)/(2)T_(0)`D. `T_(f) =(5)/(2)T_(0)` |
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Answer» Correct Answer - C Equarting internal energy. `1 xx (5)/(2) RT_(0) +1 xx (3)/(2) R ((7)/(3)T_(0)) = 1 xx(3)/(2)RT_(f)+(5)/(2)RT_(f)` `rArr T_(f) = (3)/(2)T_(0)` |
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| 32. |
A nonconducting cylinder having volume `2 V_(0)` is partitioned by a fixed non conducting wall in two equal part. Partition is attached with a valve. Right side of the partition is a vacuum and left part is filled with a gas having pressure and temperature `P_(0)` and `T_(0)` respectively. If valve is opened find the final pressure and temperature of the two parts. |
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Answer» From the first law thermodynamics `DeltaQ = DeltaU +W` Since gas expands freely therefore `w = 0`, since no heat is given to gas `DeltaQ = 0` `rArr DeltaU = 0` and temperature remains constant. `T_("final") = T_(0)` Since the process is isothermal therefore `P_(0) xx V_(0) = P_("final") xx 2V_(0) rArr P_("final") = P_(0)//2` |
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| 33. |
A vessel of volume `2 xx 10^(-2) m^(3)` contains a mixture of hydrogen and helium at `47^(@)C` temperature and `4.15 xx 10^(5) N//m^(2)` Pressure. The mass of the mixture is `10^(-2) kg`. Calculate the masses of hydrogen and helium in the given mixture. |
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Answer» Let mass of `H_(2) is m_(1)` and `He` is `m_(2)` `:. M_(1) +m_(2) = 10^(-2) kg =10 xx 10^(-3) kg ……….(1)` Let `P_(1),P_(2)` are partial pressure of `H_(2)` and `He` `P_(1) +P_(2) = 4.15 xx 10^(5) N//m^(2)` for the mixture `(P_(1)+P_(2))V = ((m_(1))/(M_(1))+(m_(2))/(M_(2))) RT` `rArr 4.15 xx 10^(5) xx2xx 10^(-2) = ((m_(1))/(2xx10^(-3))+(m_(2))/(4xx10^(-3))) 8.31 xx 320` `rArr (m_(1))/(2) +(m_(2))/(4) = (4.15 xx2)/(8.31 xx 320) = 0.00312 = 3.12 xx 10^(-3)` `rArr 2m_(1) +m_(2) = 12.48 xx 10^(-3) kg ..........(2)` Solving (1) and (2) `m_(1) = 2.48 xx 10^(-3) kg ~= 2.5 xx 10^(-3) kg` and `m_(2) = 7.5 xx 10^(-3) kg`. |
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| 34. |
Which of the following quantities is the same for all ideal gases at the same temperature?A. the kinetic energy of equal moles of gasB. the kinetic energy of equal mass of gasC. the number of molecules of equal moles of gasD. the number of molecules of equal mass of gas |
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Answer» Correct Answer - C For an ideal gas, the no of molecules of equal moles of gas is same. |
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| 35. |
The pressure of an ideal gas is written as `p=(2E)/(3V)`.Here `E` refers toA. average translational kinetic energyB. rotational kinetic energyC. total kinetic energyD. None of these |
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Answer» Correct Answer - A As traslation `K.E`. Is `K.E. = (3)/(2)nRT E = (3)/(2)PV` where `E =` total translational `K.E`. |
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| 36. |
A vessel of volume `V = 5` litre contains `1.4 g` nitrogen and `0.4g` of He at `1500K`. If `30%` of the nitrogen molecules are disassociated into atoms then the gas pressure becomes `(N)/(8)xx10^(5)N//m^(2)`. Find `N` (Assume `T` constant). `[R = (25)/(3)J//molK]` |
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Answer» Correct Answer - 33 Given mass of `N_(2) = 1.4 g` mass of `He = 0.4g` `V = 5` litre `T = 1500 K =` const `R = (25)/(3) J//"mole" K` number of moles of `He = (0.4)/(4) = 0.1` mole number of moles of `N_(2) = (1.4 xx 70)/(28xx100) = 0.035` mole number of moles of `N` atoms `=(1.4 xx 30)/(14 xx 100) = 0.03` mole Pressure of gas is `P = (nRT)/(V)` `P = ((0.1+0.035+0.03)(25)/(3)xx1500)/(5xx10^(-3))` `= ((0.165)25xx1500)/(3xx5xx10^(-3))` `= 4.125 xx 10^(5) N//m^(2) = (33)/(8) xx 10^(5) N//m^(2)` |
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| 37. |
What are the different ways of increasing the number of molecules collisions per unit time against the walls of the vessel containing a gas ? |
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Answer» Correct Answer - Number of collisions per unit time can be increased by (i) increasing the number of molecules. (ii) increasing the temperature of the gas and (iii) decreasing the volume of the gas. |
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| 38. |
A balloon containing an ideal gas has a volume of `10 litre` and temperature of `17^(@)`. If it is heated slowly to `75^(@)C`, the work done by the gas inside the balloon is (neglect elasticity of the balloon and take atmospheric pressure as `10^(5)` Pa) |
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Answer» Correct Answer - 2 Since elasticity of ballon is negligible `:.` pressure inside ballon `~~` pressure outside ballon `= P_(atm)` `:. W = P_(atm) DeltaV` `V_("in") = 10 litre`. `(V_("in"))/(T_("in")) = (V_("fin"))/(T_("fin")) rArr V_("final") = ((V_("in")T_("final"))/(T_("in"))) litre`. `rArr W = P_(atm) V_("in") (T_("final")/(T_("in"))-1)` `rArr 10^(5) xx 10^(-2) ((58)/(290)) = 200J` |
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| 39. |
In a process the pressure of a gas is inversely proportional to the square of the volume. If temperature of the gas increases, then work done by the gas:A. is positiveB. is negativeC. is zeroD. may be positive |
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Answer» Correct Answer - B `P alpha (1)/(V^(2)) rArr P = (k)/(V^(2)) rArr PV^(2) = k` `rArr PV. V = k rArr nRTV = k rarr TV = k_(1)` Since temperature increases therefore volume decreases or work done by the gas is negative. |
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| 40. |
When the state of a system changes from `A` to `B` adiabatically the work done on the system is `322` joule. If the state of the same system is changed from `A` to `B` by another process, and heat required is `50` calories of heat is required then find work done on the system in this process? `(J = 4.2 J//cal)` |
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Answer» Correct Answer - `112` joule `dU_(1) = dU_(2), dQ_(1) - dW_(1) = dQ_(2) - dW_(2)` `0 - 322 = 50 xx 4.2- dW_(2), dW_(2) =- 112J`. |
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| 41. |
During the expansion process the volume of the gas changes form `4m^(3)` to `6m^(3)` while the pressure change according to `p = 30 V +100` where pressure is in `Pa` and volume is in `m^(3)`. The work done by gas in `N xx 10^(2)J`. Find `N`. |
| Answer» Correct Answer - 5 | |
| 42. |
Specific heat of a substance can beA. finiteB. infiniteC. zeroD. negative |
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Answer» Correct Answer - A::B::C::D Specific heat of a substance can be finite, infinite, zero and negative. |
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| 43. |
For an ideal gas :A. the change in internal in a constant pressure process from temperature `T_(1)` to `T_(2)` is equal to `nC_(v) (T_(2)-T_(1))`, where `C_(v)` is the molar specific heat at constant volume and n the number of moles of the gas.B. the change in internal energy of the gas and the work doen by the gas are equal in magnitude in an adiabatic process.C. the internal energy does not change in an isothermal process.D. no heat is added or removed in an adiabatic process. |
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Answer» Correct Answer - A::B::C::D For any process `DeltaU = nC_(v)DeltaT`, For isothermal `DeltaT = 0` or `U =` constant `DeltaQ = 0` (For adiabatic process) `:. DeltaU +W = 0 DeltaU =- W` |
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| 44. |
Find the maximum attainable temperature of ideal gas in each of the following process : (a) `p = p_0 - alpha V^2` , (b) `p = p_0 e^(- beta v)`, where `p_0, alpha` and `beta` are positive constants, and `V` is the volume of one mole of gas. |
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Answer» Correct Answer - (a) `T_(max) = (2)/(3)(p_(0)//R) sqrt(p_(0)//3 alpha)]` (b) `T_(max) = p_(0)//e betaR`. (a) `pV = RT` `V = (RT)/(p)` `p = p_(0) - alphaV^(2)` `p = p_(0) - alpha ((RT)/(p))^(2)` `T = (1)/(Rsqrt(alpha)) sqrt(p_(0)p^(2)-p^(3))` for maximum value `T, sqrt(p_(0)p^(2)-p^(3))` should be maximum `(d)/(dP) (p_(0)p^(2)-p^(3)) = 0` `p = (2p_(0))/(3)` `T = (1)/(Rsqrt(alpha)) sqrt(p_(0)p^(2)-p^(3))` `p =(2p_(0))/(3)` `T_(max)= (2)/(3) (p_(0))/(R ) sqrt((p_(0))/(3alpha))` (b) `p = p_(0)e^(-betaV)` or `p = p_(0) e^(-(betaRT)/(p))` for maximum value of `T` `(dT)/(dp) = 0` `p = p_(0)e^(-(betaRT)/(p))` `ln (p) = ln p_(0) -(betaRT)/(p)` `ln ((p)/(p_(0))) =- beta (RT)/(p)` `T =- (p)/(betaR) ln ((p)/(p_(0)))` for `T_(max)` After solving `p = (p_(0))/(e) rArr T_(max) = (p_(0))/(e betaR)`. |
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| 45. |
In above question, if `U_(a) = 40J`, value of `U_(b)` will beA. `-50 J`B. `100 J`C. `-120 J`D. `160 J` |
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Answer» Correct Answer - D `U_(b) = U_(B) = 120` `U_(b) = 120 +40 = 160 J` |
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| 46. |
For a solid with a small expansion coefficient,A. `C_(p) -C_(v) = R`B. `C_(p) - C_(v) = 2R`C. `C_(p)` is slightly greater than `C_(v)`D. `C_(p)` is slightly less than `C_(v)` |
| Answer» Correct Answer - C | |
| 47. |
`0.040g` of He is kept in a closed container initially at `100.0^(@)C.`The container is now heated. Neglecting the expansion of the container, Calculate the temperature at which the internal energy is increased by `12J`. |
| Answer» Correct Answer - `196^(@)C` | |
| 48. |
The molar heat capacity for the process shown in figure is A. `C = C_(V)`B. `C = C_(P)`C. `C gt C_(V)`D. `C lt C_(V)` |
| Answer» Correct Answer - D | |
| 49. |
Which of the following is correct for the molecules of a gas in thermal equilibrium ?A. All have the same speedB. All have different speeds which remains constantC. They have a certain constant average speedD. They do not collide with one another. |
| Answer» Correct Answer - C | |
| 50. |
One kg of a diatomic gas is at pressure of `8xx10^4N//m^2`. The density of the gas is `4kg//m^3`. What is the energy of the gas due to its thermal motion?A. `5 xx 10^(4)J`B. `6 xx 10^(4)J`C. `7 xx 10^(4)J`D. `3 xx 10^(4)J` |
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Answer» Correct Answer - A `V = (m)/(d) = (1)/(4)m^(3), KE =` for diatomic `, KE = (5)/(2)` `PV = (5)/(2) xx8 xx 10^(4) xx(1)/(4) = 5 xx 10^(4) J` |
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