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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.
| 701. |
The equation of state for a gas is given by `PV = eta RT + alpha V`, where `eta` is the number of moles and `alpha` a positive constant. The intinal pressure and temperature of 1 mol of the gas contained in a cylinder is `P_(0)` and `T_(0)`, respectively. The work done by the gas when its temperature doubles isobarically will beA. `(P_(0)T_(0)R)/(P_(0) - alpha)`B. `(P_(0)T_(0)R)/(P_(0) + alpha)`C. `P_(0)T_(0)R In 2`D. None of these |
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Answer» Correct Answer - A `DeltaW = PDeltaV [PV=mu RT+alphaV]` `P_(0)[(muRDeltaT)/(P_(0)-alpha)]=:. PDeltaT = mu RT+alphaDeltaT` or `DeltaV = (muRDeltaT)//(P-alpha)]` `i.e., DeltaW = (P_(0)RT_(0))/(P_(0)-alpha) [mu-1]`. |
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| 702. |
A copper block of mass 1kg slides down one rough inclined plane of inclination `37^(@)` at a constant speed. Find the increase in the temperature of the block as it slides down temperature of the block as it slides down through 60cm assuming that the loss in mechanical energy goes into the copper block as thermal energy. (specific heat of copper=`420Jkg^(-1)K^(-1),g=10ms^(-2)`)A. `6.6xx10^(-3).^@C`B. `7.6xx10^(-3).^(@)C`C. `8.6xx10^(-3).^(@)C`D. `9.6xx10^(-3).^(@)C` |
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Answer» Correct Answer - C `(1)/(2)mv^(2)=mSDeltaT`, But `(1)/(2)mv^(2)=mgh` `rArrV^(2)=2gl sin theta,(1)/(2)xx2gl sin theta=SDeltaT` |
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| 703. |
A Carnot engine takes `3xx10^6cal`. of heat from a reservoir at `62^@C`, and gives it to a sink at `27^@C`. The work done by the engine isA. `4.2xx10^(6)J`B. `8.4xx10^(6)J`C. `16.8xx10^(6)J`D. zero |
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Answer» Correct Answer - B `T_(1)V_(1)^(gamma-1)=T_(2)V_(2)^(gamma-1)` |
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| 704. |
A carnot engine working between `300 K and 600 K` has work output of `800 J` per cycle. What is amount of heat energy supplied to the engine from source per cycle?A. 1400 JB. 1600 JC. 1500 JD. 1700 J |
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Answer» Correct Answer - B `eta=(dW)/(dQ)=1-(T_(2))/(T_(1))=1-(300)/(600)=(1)/(2)` `(800)/(dQ)=(1)/(2)` `dQ=1600J` |
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| 705. |
The heat reservoir of an ideal carnot engine is at 800K and its sink is at 400K. The amount of heat taken in it in one second to produce useful mechanical work at the rate of 750J isA. 2250JB. 1125JC. 1500JD. 750J |
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Answer» Correct Answer - C `(W)/(Q)=1-(T_(2))/(T_(1))` |
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| 706. |
n moles of a diatomic gas has undergone a cyclic process ABC as shown in figure. Temperature at a is `T_(0)`. Find a. Volume at C ? b. Maximum temperature ? c. Total heat given to gas ? d. Is heat rejected by the gas, if yes how much heat is rejected ? e. Find out the efficiency |
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Answer» Since triangle `O A V_(0)` and OC V are similar therefore `(2P_(0))/(V) = (P_(0))/(V_(0)) implies V = 2 v_(0)` b. Since process AB is isochoric hence `(P_(A))/(T_(A)) = (P_(B))/(T_(B)) implies T_(B) = 2T_(0)` Since process BC is isobaric therefore `(T_(B))/(V_(B)) = (T_(C))/(V_(C ))` `implies T_(C ) = 2 T_(B) = 4 T_(0)` Since process is cyclic therefore `Delta Q = Delta w`= area under the cycle ` =(1)/(2) P_(0) V_(0)`. d. Since `Delta u` and `Delta W` both are negative in process CA `:. Delta Q` is negative in process ca and heat is rejected in process CA `Delta Q_(CA) = Delta W_(CA) + Delta U_(CA)` `= - (1)/(20) [P_(0) + 2 P_(0)] V_(0) - (5)/(2) nr (T_(c ) - T_(a))` `= -(1)/(2) [P_(0) + 2 P_(0)] V_(0) - (5)/(2) nR ((4 P_(0) V _(0))/(nR) - (P_(0) V_(0))/(nR))` `= - 9 P_(0) V_(0) =` Heat injected. e. `eta = ` efficiency of the cycle `= ("work done by the gas")/("heat injected") = eta = (P_(0) V_(0)//2)/(Q_("injected") xx 100` `Delta Q_("inj") = Delta W_(AB) + Delta U_(BC)` ` = [(5)/(2) nr (2T_(0) - T_(0))] + [(5)/(2) nRT (2 T_(0)) + 2P_(0) (2V_(0) - V_(0))]` `(19)/(2) P_(0) V_(0)` |
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| 707. |
T-V curve of cyclic process is shown below, number of moles of the gas are n find the total work done during the cycle. A. `(3)/(2) nRT_(0) In 2`B. `nRT_(0) In 2`C. `(nRT_(0) In 2)/(2)`D. `2 nRT_(0) In 2` |
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Answer» Correct Answer - B Since path `AB` and `CD` are isochoric therefore work done is zero during path `AB` and `CD`. Process `BC` and `DA` are isothermal therefore, `W_(BC) = nR2T_(0)1n(V_C)/(V_B) = 2nRT_(0)1n2` `W_(DA) = nRT_(0)1n(V_A)/(V_D) = -nRT_(0)1n2` Total work done `W_(BC)+W_(DA)=2nRT_(0)1n 2=nRT_(0) 1n 2`. |
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| 708. |
T-V curve of cyclic process is shown below, number of moles of the gas are n find the total work done during the cycle. |
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Answer» Since path AB and CD are isohroic therefore work done is zero during AB and CD process BC and DA are isothermal, therefore `W_(BC) = nR2T_(0) 1n (V_(C ))/(V_(B)) = 2 nRT_(0) 1n 2` `W_(DA) = nRT_(0) 1n (V_(A))/(V_(D)) = - nRT_(0) 1n 2` Total wor done `W_(BC) + W_(DA) = e nRT_(0) 1n 2 = nRT_(0) 1n 2` |
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| 709. |
P-T curve of a cyclic process is shown. Find out the works done by the gas in the given proces if number of moles of the gas are n. A. `nR(T_(1) + T_(3) - T_(4) + T_(2))`B. `nR(T_(1) - T_(3) - T_(4) + T_(2))`C. `nR(T_(1) + T_(3) + T_(4) - T_(2))`D. `nR(T_(1) + T_(3) - T_(4) - T_(2))` |
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Answer» Correct Answer - D Since paths `AB` and `CD` are isochoric, therefore work done during `AB` and `CD` is zero. Paths `BC` and `DA` are isobaric. Hence , `W_(BC) = nRTDeltaT= nR(T_(3)-T_(2))` `W_(DA) = nR(T_1-T_4)` Total work done = `W_(BC)+W_(DA) = nR(T_(1)+T_(3)-T_(4)-T_(2))`. |
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| 710. |
An ideal gas heat engine operates in a Carnot cycle between `27^(@)C` and `127^(@)C`. It absorbs `6 kcal` at the higher temperature. The amount of heat (in kcal) converted into work is equal toA. `3.5`B. `1.6`C. `1.2`D. `4.8` |
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Answer» Correct Answer - C Efficiency of a Carnot engine is given by `eta=1-(T_2)/(T_1)` or `W/Q = 1-(T_2)/(T_1)implies W/6=1-((273+27))/((273+127))` `implies W = 1.2 kcal`. |
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| 711. |
An ideal gas heat engine operates in Carnot cycle between `227^(@)C` and `127^(@)C`. It absorbs `6 x 10^(4) cals` of heat at higher temperature. Amount of heat converted to work isA. `2.4 xx 10^(4) cal`B. `6 xx 10^(4) cal`C. `1.2 xx 10^(4) cal`D. `4.8 xx 10^(4) cal` |
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Answer» Correct Answer - C We have `Q/T` = consta nt `implies (Q_2)/(Q_1)=(T_2)/(T_1)` Given, `Q_(1)=6xx10^(4) cal`, `T_(1)=227+273=500K` `T_(2) = 127+273 = 400K` `:. (Q_2)/(6xx10^4)= 400/500` `implies Q_(2) = 4/5xx6xx10^(4)=4.8xx10^(4)cal` Now heat converted to work `=Q_(1) - Q_(2) = 6.0xx10^(4)-4.8xx10^(4)=1.2xx10^(4)cal` Aliter: `eta = (T_1-T_2)/(T_1) = W/Q implies W = (Q(T_1-T_2))/(T_1)` `W = (6xx10^(4)[(227+273)-(273+127)])/((227+273))` `implies W = (6xx10^(4)xx100)/(500)=1.2xx10^(4)cal`. |
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| 712. |
Which of the following conditions of the Carnot ideal heat engine can be realised in practice?A. Infinite thermal capacity of the sourceB. infinite thermal capacity of the sinkC. Perfectly non conducting standD. Less than 100% efficiency |
| Answer» Correct Answer - D | |
| 713. |
A container of volume `1m^3` is divided into two equal parts by a partition. One part has an ideal gas at 300K and the other part is vacuum. The whole system is thermally isolated from the surroundings. When the partition is removed, the gas expands to occupy the whole volume. Its temperature will now be .......A. `300 K`B. `250 K`C. `200 K`D. `10 K` |
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Answer» Correct Answer - A a. It is free expansion, temperature will remain constant. |
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| 714. |
A sound wave passing volume `1 m^(3)` is divided into two equal compartments by a partition. One of these compartments contains an ideal gas at `300 K`. The other compartment is vacuum. The whole system is thermally isolated from its surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would beA. `8.97 xx 10^(-4) K`B. `8.97 xx 10^(-6) K`C. `8.97 xx 10^(-8) K`D. none of these |
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Answer» Correct Answer - C c. Differentiating `gamma T^(gamma - 1) dT P^(1 - gamma) + T^(gamma) (1 - gamma) P^(-gamma) dP = 0` or `dT = ((gamma - 1) T)/(gamma P) = dP` or, `dT = ((1.5 - 1)/(1.5)) ((273)/(76 xx 13.6 xx 981) xx 0.001)` `= 8.97 xx 10^(-8) K` |
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| 715. |
A sample of `0.1 g` of water of `100^(@)C` and normal pressure `(1.013 xx 10^(5) N m^(-2))` requires 54 cal of heat energy to convert to steam at `100^(@)C`. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample isA. `84.5J`B. `104.3J`C. `42.2J`D. `208.7J` |
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Answer» Correct Answer - D `Delta Q = 54cal = 54xx4.18joule = 225.72` joule `DeltaW = P[V_(steam)-V_(water)]` [for water 0.1 gram = 0.1 c c] `=1.013xx10^(5)[167xx10^(-6)-0.1xx10^(-6)]"joule"` `=1.013xx167xx10^(-1)=16.917 "joule"` By first law of thermodyamics `implies DeltaU=DeltaQ-DeltaW=225.72-16.917` `DeltaU = 208.8 "Joule"`. |
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| 716. |
In the above example , if the system moves from Bto A, what is the Change in internal energy ?A. 300 JB. `-300 J`C. 400 JD. `-400 J` |
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Answer» Correct Answer - B Consider a closed path that through the state A and B . Internal energy is a state function , So `DeltaU` is zero for a closed path. Thus, `DeltaU=DeltaU_(AB)+DeltaU_(BA)=0` or `DeltaU_(BA)=-DeltaU_(AB)=-300J` |
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| 717. |
A cylinder of length `42cm` is divided into chambers of equal volumes and each half contains a gas of equal mass at temperature `27^(@)C`. The separator is a frictionless piston of insulating material. Calculate the distance by which the piston will be displacement if the temperature of one half is increaded to `57^(@)C`. |
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Answer» Correct Answer - `1cm` |
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| 718. |
When two blocks of ice are pressed against each other then they stick together (coalesce) becauseA. cooling is produced qB. heat is producedC. increase in pressure will increase in melting pointD. increase in pressure will decrease in melting point |
| Answer» Correct Answer - D | |
| 719. |
A cubical box of side `1 m` contains helium gas (atomic weight 4) at a pressure of `100 N//m^2`. During an observation time of `1 second`, an atom travelling with the root - mean - square speed parallel to one of the edges of the cube, was found to make `500 hits` with a particular wall, without any collision with other atoms . Take `R = (25)/3 j //mol - K and k = 1.38 xx 10^-23 J//K`. (a) Evaluate the temperature of the gas. (b) Evaluate the average kinetic energy per atom. ( c) Evaluate the total mass of helium gas in the box.A. Evaluate the temperaturef of the gas,B. Evaluate the average kinetic energy per atomC. Evaluate the total mass of helium gas in the box.D. |
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Answer» Correct Answer - `160 K,3.3xx10^_21J,0.3gm` |
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| 720. |
A cubical box with porous walls containing an equal number of `O_(2)` and `H_(2)` molecules is placed in a larger evacuated chamber. The entire system is maintained at constant temperature `T`. The ratio of `v_(rms)` of `O_(2)` molecules to that of the `v_(rms)` of `H_(2)` molecules, found in the chamber outside the box after a short interval isA. `1/(2sqrt(2))`B. `(1)/(4)`C. `(1)/(sqrt(2))`D. `sqrt(2)` |
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Answer» Correct Answer - B `(v_(rms))_(O_(2))/(v_(rms))_(H_(2)) = sqrt((M_(0))_(H_(2))/((M_(0))_(H_(2)))) = sqrt((2)/(32)) = (1)/(4)` |
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| 721. |
A cubical box containing a gas with internal energy U is given velocity V, then the new internal energy of the gasA. less than UB. more than UC. UD. zero |
| Answer» Correct Answer - C | |
| 722. |
In a cubical box of volume V, there are N molecules of a gas moving randomly. If m is mss of ech molecule and`V^(2)` is the meansquare of x component of the velocity of molecules, then the pressure of the gas is-A. `P = 1/3(mNv^2)/(V)`B. `P=(mNv^2)/(V)`C.D. `P=mNv^2` |
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Answer» Correct Answer - B refer to derivation (asked only x - compnent) |
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| 723. |
`N_(2)` molecules is 14 times heavier than a `H_(2)` molecule. At what temperature will the rms speed of `H_(2)` molecules be equal to that of` N_(2)` molecule at `27^(@)C- `A. `50^(@) C`B. `2^(@)C`C. `21.4^(@)C`D. 21.4 K |
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Answer» Correct Answer - D `V_1 =V_2 rArr sqrt((3RT_1)/(M_1)) = sqrt((3RT_2)/(M_2))` |
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| 724. |
At a certain temperature, the rms velocity for `O_(2)` is `400 ms^(-1)`. At the same temperature, the rms velocity for `H_(2)` molecules will beA. `100 ms^(-1)`B. `25 ms^(-1)`C. `1600 ms^(-1)`D. `6400 ms^(-1)` |
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Answer» Correct Answer - C `(C_(H_2))/(C_(O_2)) = sqrt((32)/(2)) = 4`. or, `C_(H_2) = 4 xx C_(O_2)=4xx400ms^(-1) = 1600ms^(-1)`. |
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| 725. |
Which of the following statement is true?A. Absolute zero degree temperature is not zero enery temperature.B. Two different gases at the same temperature pressure have equal root mean square velocities.C. The rms speed of the molecules of different ideal gases, maintained at the same temperature are the same.D. Given sample of 1cc of hydrogen and 1cc of oxygen both at N.T.P.,oxygen sample has a large number of molecules. |
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Answer» Correct Answer - A |
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| 726. |
At a certain temperature, the rms velocity for `O_(2)` is `400 ms^(-1)`. At the same temperature, the rms velocity for `H_(2)` molecules will beA. 100m/sB. 25m/sC. 1600m/sD. 6400m/s |
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Answer» Correct Answer - C `v_(rms)=sqrt((3RT)/(M))rArrv_(rms)prop(1)/sqrt(M)` |
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| 727. |
At a certain temperature , the ratio of the rms velocity of `H_(2)` molecules to `O_(2)` molecule isA. `1:1`B. `1:4`C. `4:1`D. `16:1` |
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Answer» Correct Answer - C As velocity `c=sqrt((3RT)/(M)), "so" (C_(H))/(C_(O))=sqrt((M_(O))/(M_(H)))=sqrt((16)/(1))=(4)/(1)` |
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| 728. |
The temperature at which the velocity of oxygen will be half of hydrogen at NTP isA. `1092^(@)C`B. `1492^(@)C`C. `273^(@)C`D. `819^(@)C` |
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Answer» Given , `v_(O_(2))=(1)/(2)V_(H_(2))` `therefore" " sqrt((3RT)/(32))=(1)/(2)sqrt((3Rxx273)/(2))` `(T)/(32)=(273)/(8)` ` therefore" " T=4xx273` `T=1092K=1092-273` `T=819^(@)C` |
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| 729. |
The molecules of a given mass of a gas have rms velocity of `200 m//s at 27^(@)C and 1.0 xx 10^(5) N//m_(2)` pressure. When the temperature and pressure of the gas are respectively `127^(@)C and 0.05 xx 10^(5) Nm^(-2)`, the rms velocity of its molecules in `ms^(-1)` isA. `(100sqrt(2))/(3)`B. `100sqrt(2)`C. `(400)/(sqrt(3))`D. none of these |
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Answer» Correct Answer - C |
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| 730. |
If the molecular weight of two gases are `M_(1)` and `M_(2)` then at a temperature the ratio of rms velocity `C_(1) and C_(2)` will beA. `(M_(1)/M_(2))^(1//2)`B. `(M_(2)/M_(1))^(1//2)`C. `((M_(1)-M_(2))/(M_(1)+M_(2)))^(1//2)`D. `((M_(1)+M_(2))/(M_(1)-M_(2)))^(1//2)` |
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Answer» Correct Answer - B As `(1)/(2)Mc^(2)=(3)/(2)RT" or " c=((3RT)/(M))^(1//2)rArrcprop(1)/(sqrtM)` `"So"," " (c_(1))/(c_(2))=(M_(2)/(M_(1)))^(1//2)` |
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| 731. |
Which of the following gases has maximum rms speed at a given temperature?A. (a)hydrogenB. (b)nitrogenC. (c)oxygenD. (d)carbon dioxide |
| Answer» Correct Answer - A | |
| 732. |
The mean square speed of the molecules of a gas at absolute temoerature T is proportional toA. (a)(1)/(T)B. (b)sqrtTC. (c)TD. (d)T^(2) |
| Answer» Correct Answer - C | |
| 733. |
The mean square speed of the molecules of a gas at absolute temperature `T` is proportional toA. `T^(-1)`B. `sqrt(T)`C. `T`D. `T^(2)` |
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Answer» Correct Answer - C `bar(v^(2)) = v_(rms)^(2) = (3 RT)/(M_(0))` `bar(v^(2)) prop T` |
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| 734. |
The mean square speed of the molecules of a gas at absolute temperature `T` is proportional toA. 1/TB. `sqrt(T)`C. TD. `T^(2)` |
| Answer» Correct Answer - B | |
| 735. |
The following four gases are at the same temperature. In which gas do the molecules have the maximum root mean square speed?A. HydrogenB. OxygenC. NitrogenD. Carbon dioxide |
| Answer» Correct Answer - A | |
| 736. |
The gas equation `PV//T =` constant is true for a constant mass of an ideal gas undergoingA. isothermal change onlyB. adiabatic change onlyC. both isothermal & adiabatic changesD. neither isothermal nor adiabatic change. |
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Answer» Correct Answer - C As PV=nRT for n =constant : `((PV)/T)` =constant for all change hence (2) |
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| 737. |
Which of the following paramenters is the same for molecules of all gases at a given temperature?A. (a)massB. (b)SpeedC. (c)MomentumD. (d)Kinetic energy. |
| Answer» Correct Answer - D | |
| 738. |
The pressure of an ideal gas is weitten as `p=(2E)/(3V)`.Here E refers toA. (a)translational kinetic energyB. (b)rotational kinetic energyC. (c)vibrational kinetic energyD. (d)total kinetic energy. |
| Answer» Correct Answer - A | |
| 739. |
How much heat energy in joules must be supplied to 14gms of nitrogen at room temperature to rise its temperature by `40^(@)C` at constant pressure? (Mol. Wt. of `N_(2)=28gm`, R=constant)A. 50 RB. 60 RC. 70 RD. 80 R |
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Answer» Correct Answer - C `Here, " " DeltaT=40^(@)C` `C_(p)=(7)/(2)R` (for diatomic gases) Number of moles, `mu=(14)/(28)=(1)/(2)` Heat supplied at constant pressure , `Q=muC_(p)DeltaT` `rArr " " Q=(1)/(2)xx(7)/(2)Rxx40=70R` |
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| 740. |
The gas equation `PV//T =` constant is true for a constant mass of an ideal gas undergoingA. isothermal changeB. adiabatic changeC. isobaric changeD. any type of change |
| Answer» Correct Answer - D | |
| 741. |
A gas behaves more closely as an ideal gas atA. (a)low pressure and low temperatureB. (b)low pressure and high temperatureC. (c)high pressure and low temperatureD. (d)high pressure and high temperature. |
| Answer» Correct Answer - B | |
| 742. |
A gas behaves more closely as an ideal gas atA. low pressure and low temperatureB. low pressure and high temperatureC. high pressure and low temperatureD. high pressure and high temperature |
| Answer» Correct Answer - B | |
| 743. |
A real gas behaves like an ideal gas if itsA. Phase transitionB. temperatureC. pressureD. None of these |
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Answer» Correct Answer - C No attractive or repulsive force acts between gas molecules in an ideal gases. Molecules do not exert that force on the wall which they would have exerted in the absence of intermolecular force. Therefore, the observed pressure P of the gas will be less than that present in the absence of the gas will be less than that present in the absence of intermolecular force. Hence, the attractive becomes `(p+(a)/(V_(2)))` |
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| 744. |
A real gas behaves like an ideal gas if itsA. pressure and temperature are both highB. pressure and temperature are both lowC. pressure is high and temperature is lowD. pressure is low and temperature is high |
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Answer» Correct Answer - D For idela gas behaviour pressure should be low and temperature should be high. Hence correct option is |
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| 745. |
A real gas behaves like an ideal gas if itsA. pressure and temperatture are both highB. pressure is high and temperature is lowC. pressure is high and temperature is lowD. pressure is low and temperature is high |
| Answer» Correct Answer - D | |
| 746. |
A refrigerator, whose coefficient performance `beta` is 5, extracts heat from the collingcompartment at the rate of 250 J per cycle. (a) How much work per cycle is required to operate the refrigerator? (b) How much heat per cycle is discharges the room which acts as the high temperature revervoir? |
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Answer» (a) As coefficient of performance of a refrigeration is defined as `beta=Q_(1)//W`, So `W=(Q_(1))/(beta)=(250)/(5)=50` (b) As `Q_(B)=Q_(1)+W,` so `W_(B)=250+50=300J` |
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| 747. |
A carnot engine operating between temperatures `T_(1)` and `T_(2)` has efficiency. When `T_(2)` is lowered by 62K, its efficience increases to `(1)/(3)`. Then `T_(1)` and `T_(2)` are respectively: |
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Answer» `eta=1-(T_(2))/(T_(1))rArr(1)/(6)=1-(T_(2))/(T_(1))rArr(T_(2))/(T_(1))=(5)/(6)`.......(1) `eta_(2)=1-(T_(2)-62)/(T_(1))rArr(1)/(3)=1-(T_(2)-62)/(T_(1))` ....(2) On solving Equation (1) and (2) `T_(1)=372K` and `T_(2)=310K` |
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| 748. |
Find the efficeincy of the thermodynamic cycle shown in figure for an ideal diatomic gas. |
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Answer» Let n be the number of moles of the gas and the temperature be `T_(0)` in the state A. Now, work done during the cycle `W=(1)/(2)xx(2V_(0)-V_(0))(2P_(0)-P_(0))=(1)/(2)P_(0)V_(0)` For the heat `(DeltaQ_(1))` given during the process `ArarrB`, we have `DeltaQ_(1)=DeltaW_(AB)+DeltaU_(AB)` `DeltaW_(AB)=` area under the straight line AB `=(1)/(2)(P_(0)+2P_(0))(2V_(0)-V_(0))=(3P_(0)V_(0))/(2)` Applying equation of state for the gas in the state A `&` B. `(P_(0)V_(0))/(T_(0))=((2P_(0))(2V_(0)))/(T_(B))rArrT_(B)=4T_(0)` `thereforeU_(AB)=nC_(v)DeltaT=n((5R)/(2))(4T_(0)-T_(0))` `=(15nRT_(2))/(2)=(15P_(0)V_(0))/(2)` `therefore DeltaQ_(1)=(3)/(2)P_(0)V_(0)+(15)/(2)P_(0)V_(0)=9P_(0)V_(0)` Obviously, the processes `BrarrC` and `CrarrA` involve the abstraction of heat from the gas. `"Efficiency"=("Work done per cycle")/("Total heat supplied per cycle")` i.e., `eta=((1)/(2)P_(0)V_(0))/(9P_(0)V_(0))=(1)/(18)` |
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| 749. |
A Carnot engine, having an efficiency of `eta=1//10` as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is |
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Answer» Coefficient of performance of refrigerator `beta=(1-eta)/(eta)=(1-1//10)/(1//10)=9=("Heat extracted")/("workdone")` `therefore` Heat extracted `=9xx10=90J` |
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| 750. |
In (figure). shows two path that may be taken by a gas to go from a state A to state C In the process AB, `400 J` of heat is added to the system and in process Bc, `100 J` of heat is added to the system. The heat absorbed by the system in the process AC will beA. `380J`B. `500J`C. `460J`D. `300J` |
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Answer» Correct Answer - C In cyclic process `Q_(cyclic) = W_(cyclic)` `Q_(AB)+Q_(BC) + Q_(CA) =` area enclosed `400+100+Q_(CA) =` area enclosed `400+100+Q_(CA)=1/2xx(4xx10^(4)xx2xx10^(-3))` `Q_(AC) = 500-40J = 460J`. |
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