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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.
| 351. |
In a given process on an ideal gas, `dW=0 and dQlt0.` Then for the gasA. temperature-increasesB. volume-decreasesC. pressure-decreasesD. pressure-increases |
| Answer» Correct Answer - C | |
| 352. |
The temperature determines the direction of net change ofA. gross kinetic energyB. intermolecular kinetic energyC. gross potential enrgyD. intermolecular potential energy |
| Answer» Correct Answer - B | |
| 353. |
The thermal motion meansA. motion due to heat engineB. disorderly motion of the body as a wholeC. motion of the body that generate heatD. random motion of molecules |
| Answer» Correct Answer - D | |
| 354. |
If specific heat of a substance is infinite, it meansA. heat is given outB. heat is taken inC. no change in temperature whether heat is taken in (or) given outD. all of the above |
| Answer» Correct Answer - C | |
| 355. |
70 calories of heat required to raise the temperature of 2 moles of an ideal gas at constant pressure from `30^@Cto 35^@C.` The amount of heat required (in calories) to raise the temperature of the same gas through the same range `(30^@C to 35^@C)` at constant volume is:A. 28JB. 50CalC. 75JD. Zero |
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Answer» Correct Answer - B `DeltaQ_(P)=nC_(P)DeltaT,` `C_(P)-C_(v)=RrArrC_(v)=C_(p)-R` `DeltaU=nC_(v)DeltaT` |
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| 356. |
Heat required to raise the temperature of one gram of water through `1^@C` isA. 0.001 K calB. 0.01 K calC. 0.1 K calD. 1.0 K cal |
| Answer» Correct Answer - A | |
| 357. |
A vessel contains 2 mole of gas and the pressure is 40 cm of Hg. Under the same conditons, iif 1 gm of gas is introduced into the vessel, the pressure will beA. 40 cm of HgB. 20 cm of HgC. 76 cm of HgD. 60 cm of Hg |
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Answer» Correct Answer - D `P_(1) prop n_(1) and P_(2) propn_(2)(P_(2))/(P_(1))=(n_(2))/(n_(1))=(n_(1)+1)/(n_(1))` |
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| 358. |
It is required to double the pressure of helium gas, contained in a steel cylinder, by heating. If the initial temperature of helium be `27^(@)C` the temperature up to which it ought to be heated is –A. `327^(@)C`B. `273^(@)C`C. `108^(@)C`D. `54^(@)C` |
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Answer» Correct Answer - A `(T_(2))/(T_(1))=(P_(2))/(P_(1))` |
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| 359. |
The surface of a house hold radiator has an emissivity of 0.5 and an area of 1.5 `m^(2)`. The rate at which the radiation is emitted by then the radiator when its temperature is `47^(@)C` will be nearly (`sigma=5.7xx10^(-8)` Si units)A. 598 wattB. 299 wattC. 448 wattD. 346 watt |
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Answer» Correct Answer - C `E=sigma A rhoT^(4)` `=5.7xx1.5xx(1)/(2)xx1.5xx0.5xx[3.20xx10^(2)]` `=5.7xx1.5xx(1)/(2)xx10.24xx10.24` `=5.7xx0.75xx10.24xx10.24` =448 watt. |
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| 360. |
A spherical balloon of volume V contains helium at a pressure P. How many moles of helium are in the balloon if the average kinetic energy of the helium atoms is `vec(K)`?A. `(2PV)/vecKN_(A)`B. `3PV/KN_(A)`C. `5PV/3KN_(A)`D. `3PV/KN_(A)` |
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Answer» Correct Answer - D `P=(2N)/(3V) bar(K)` From the kinetic theory account for pressure. `N = 3/2 (PV)/(barK)` `n = (N)/(N_A) = 3/2 (PV)/(bar(K)N_A)`. |
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| 361. |
A spherical balloon of volume `4.00 xx 10^(3) cm^(3)` contains helium at a pressure of `1.20 xx 10^(5) Pa`. How many moles of helium are in the balloon if the average kinetic energy of the helium atoms is `3.60 xx 10^(-22) J`?A. `3.32 mol`B. `2.16 mol`C. `4.12 mol`D. `2.8 mol` |
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Answer» Correct Answer - A The gas temperature must be that implied by we know kinetic energy of a molecules is given by `1/2m_(0) bar(v^2) = 3/2 k_(B)T implies 2/3 ((1/2 m_(0)bar(V^2))/(k_B))` `implies T = 2/3 ((3.60xx10^(-22))/(1.38xx10^(-23)J//K)) = 17.4K` Now `PV = nRT implies n = (PV)/(RT)` `n = ((1.20xx10^(5)N//m^2)(4.00xx10^(-3)m^3))/((8.314 j//mol.K)(17.4K)) = 3.32mol`. |
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| 362. |
Aseertion: Thermodynamics process in nature are irreversible. Reason: Dissipactive effects cannot be eliminated.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If the assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false |
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Answer» Correct Answer - A In reversible process, there always occurs some loss of enrergy . This is because energy spent in working against the dissipative force not recovered back. Some irreversible process occur in nature such as friction where extra work to cancle the effect of friction. Salt dissolve in water but a salt does not separate by itself into pure salt and pure water. |
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| 363. |
310 J of heat is required to rise the temperature of 2 moles of an ideal gas at constant pressure from `25^(@)C` to `35^(@)C`. The amount of heat required to raise the temperature of the gas through the same range at constant volume, isA. 384 JB. 144 JC. 276 JD. 452 J |
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Answer» Correct Answer - B At constant pressure, Heat required `=nC_(p)DeltaT` `rArr" " 310=2xxC_(p)xx(35-25)` `rArr" " C_(p)=(310)/(20)=15.5J "mol" ^(-1)K^(-1)` Similarly ,at constant volume , Heat required `nC_(v)DeltaT` `=2(C_(p)-R)xx(35-25)" " [becauseC_(p)-C_(v)=R] ` `=2xx(15.5-8.3)xx10` `=2xx7.2xx10=144 J` |
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| 364. |
`C_(v)" for "O_(2)" is " (5)/(2)R` with increase in tempeature it becomes `(7)/(2)R` due to aA. translational motionB. rotational motionC. vibrational motionD. None of these |
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Answer» Correct Answer - C Oxygen is a diatomic gas and hence has 5 degree of freedom 3 trenslational and 2 rotational . As the heat is given to oxygen `C_(v)` increases from `(5R)/(2) "to" (7R)/(2)`. Hence , internal energy `(DeltaU=nC_(v)DeltaT)` increases. With the increases in internal energy the atoms within the molecule may also vibrate with respect to each other .In such cases the molecule will have an additional degree of freedom due to vibratinal motion. |
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| 365. |
A big container having ideal gas at pressure P and of volume `V_(0)` is being evacuated by using vacuum pump of cylinder volume V. Find pressure in the big container after n strokes . Assume the whole process to be isothermal. |
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Answer» For first stroke, pressure become `P _(1)` `PV_(0)=P_(1)(V_(0)+V)` `P_(1)=(PV_0)/((V_(0)+V))` for `2^(nd)`stroke, pressure become `P_(2)` `P_(1)V_(0)=P_(2)(V_(0)+V)` `P_(2)=(P_(1)V_(0))/((V_(0)+V))` `P_(2)=(PV_(0)^(2))/(V_(0)+V)^2` for `n^(tn)` Stroke, pressure becomes `P_(n)` `P_(n)=(PV_(0)^(n))/((V_(0)+V)^(n)` |
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| 366. |
The rms speed of oxygen at room temperature is about 500m/s.The rms speed of hudrogen at the same temperature is aboutA. (a)`125ms_(-1)`B. (b)`2000ms_(-1)`C. (c)`8000ms_(-1)`D. (d)`31ms_(-1)` |
| Answer» Correct Answer - B | |
| 367. |
A balloon contains `500m^(3)` of He at `27^(@)C` and 1 atm pressure. Then , the volume of He at `-3^(@)C` and 0.5 atm pressure will beA. `700m^(3)`B. `900m^(3)`C. `1000m^(3)`D. `500m^(3)` |
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Answer» Correct Answer - B Using standard gas equation, `(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2))orV_(2)=(p_(1)V_(1)T_(2))/(p_(2)T_(1))` `therefore ` Volume fo the `V_(2)=(1xx500xx(273-3))/(0.5xx(273+27))` `(1xx500xx270)/(0.5xx300)=900m^(3)` |
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| 368. |
The remperature at which the `rms` speed of hydrogen molecules is equal to escape velocity on earth surface, will beA. `1060 K`B. `5030 K`C. `8270 K`D. `10063 K` |
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Answer» Correct Answer - D `v_(rms) = v_(e)` `sqrt((3RT)/((M_(0))_(H_(2)))) = sqrt(2 g R_(e))` `T = ((2g R_(e))(M_(0))_(H_(2)))/(3R) = ((11.2 xx 10^(3))^(2) (2 xx 10^(-3)))/(3 xx 8.3)` `= 10075 K` |
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| 369. |
A monoatomic gas is supplied heat Q very slowly keeping the pressure constant. The work done by the gas isA. `(2)/(5) Q`B. `(3)/(5) Q`C. `(Q)/(5)`D. `(2)/(3) Q` |
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Answer» Correct Answer - D For monoatomic gas , `(Delta U)/(Q) = 1/3 or Delta U = Q/3` From the first law of thermodynamics `Q = Delta U +W :. W = (2//3)Q`. |
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| 370. |
Find the amount of work done to increase the temperature of one mole jof an ideal gas by `30^(@)C`, if it is expanding under condition `VooT^(2//3)`. |
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Answer» We have, `VaT^((3)/(2)),` PV=RT `PVaT, PVaV^((3)/(2))rArrPasqrt(V)` `therefore P=Ksqrt(V)rArrP=kV^(1//2)` work done `W=intpdV=kintV^(1//2)dV=(2k)/(3)V^(3//2)` |
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| 371. |
Find the amount of work done to increase the temperature of one mole jof an ideal gas by `30^(@)C`, if it is expanding under condition `VooT^(2//3)`.A. `10 R`B. `20 R`C. `30 R`D. `40 R` |
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Answer» Correct Answer - B `W = int PdV = int (RT)/(V) dV` Since `V = kT^(2//3)` Hence, `(dV)/(V) = 2/3KT^(-1//3) dt :. (dV)/(V)=2/3 *(dT)/(T)` Hence `W = int_(T_1)^(T_2) 2/3 (RT)/(T) dT` ` = 2/3 R(T_(2)-T_(1)) = 2/3 R(30) = 20R`. |
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| 372. |
In a cold storage, ice melts at the rate of `2 kg//h` when the external temperature is `20^(@)C`. Find the minimum power output of the motor used to drive the refrigerator which just prevents the ice from melting. Latent heat of fusion of `ice = 80 cal//g`A. `28.5 W`B. `9.75 W`C. `16.4 W`D. `13.6 W` |
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Answer» Correct Answer - D `omega = (T_(2))/(T_(1)-T_(2)) = (273)/(293-273) = (273)/(20)` ltbr gt Mass of ice melting per second `= (2xx1000)/(3600) = (5/9)g` To prevent the melting of ice, heat to be driven out per second i.e., `Q_(2) = (5/9)xx80=400/9 cal = 186J` As, `omega = (Q_2)/(W) , W=(Q_2)/(omega) = (186)/((273//20)) = 13.6J` Power of the motor = `13.6W`. |
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| 373. |
`1gm` water at `100^(@)C` is heated to convert into steam at `100^(@)C` at `1atm`. Find out chage in internal energy of water. It is given that volume of `1gm` water at `100^(@)C = 1c c`, volume of `1gm` steam at `100^(@)C = 167 1 c c`. Latent heat of vaporization `= 540 cal//g`. (Mechanical equivalent of heat `J = 4.2J//cal)` |
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Answer» 1 atmosphere `=1.013xx10^(5) Nm^(2)` : volume of 1 gm of water, `V_(1)=1 c c=10^(-6)m^(3)`, `(because rho =(m)/(V) & rho=1gm // c c)` Volume of steam `=1671 c c=1671xx10^(-6)m^(3)` External work doen `dW=P(V_(2)-V_(1))` `=1.013xx10^(5)(1671xx10^(-6)-1xx1-^(-6))` `=1.013xx10^(5)xx1670xx10^(-6)` `=1.013xx167=169.2J`. Latent heat of vaporisation of steam=540 cal/g So, heat supplied to convert 1g of water into steam, `DeltaQ=540xx4.2J=226.8J` By first law of thermodynamics `DeltaU=DeltaQ-DeltaW=2268-169.2=2098.8J` |
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| 374. |
The amount of heat energy required to raise the temperature of `1 g` of Helium at `NTP`, from `T_(1)K` to `T_(2)K` isA. `3/8 N_(a)k_(B)(T_(2)-T_(1))`B. `3/2 N_(a)k_(B)(T_(2)-T_(1))`C. `3/4 N_(a)k_(B)(T_(2)-T_(1))`D. `3/4 N_(a)k_(B)((T_2)/(T_1))` |
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Answer» Correct Answer - A The amount of heat energy required. `Q = f/2 nRDeltaT` `implies Q = 3/2xx1/4xxk_(B)N_(a)DeltaT = 3/8 N_(a)k_(B)(T_(2)-T_(1))`. |
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| 375. |
Calculate the external workdone by the system in Kcal, when 40 Kcal of heat is supplied to the system and internal energy rises by 8400J. |
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Answer» `dQ=dU+dW` `dU=8400J=(8400)/(4200)K Cal=2 K Cal` `therefore 40 K Cal =2 K Cal+` external work done The external work done `=40-2=38` K Cal |
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| 376. |
When heat energy of 1500J is supplied to a gas the external workdone by the gas is 525J what is the increase in its internal energy |
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Answer» Heat energy supplied `DeltaQ=1500J` External workdone `DeltaW=525J` By `1^(st)` law of thermodynamics `DeltaQ=DeltaU+DeltaW` `therefore DeltaU=DeltaQ-DeltaW=1500-525=975J`. |
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| 377. |
Which of the following methods will enable the volume of an ideal gas to be made four timesA. Quarter the pressure at constant temperatureB. Quarter the temperature at constant pressureC. Half the temperature, double the pressureD. Double the temperature, double the pressure |
| Answer» Correct Answer - A | |
| 378. |
The relation between rms velocity, `v_(rms)` and the most probable velocity, `v_(mp)`, of a gas isA. `v_(rms)=v_(mp)`B. `v_(rms)=sqrt((3)/(2))mp`C. `v_(rms)=(2)/(3)v_(mp)`D. `v_(rms)=(2)/(3)v_(mp)` |
| Answer» Correct Answer - B | |
| 379. |
When unit mass of water boils to become steam at `100^(@)C`, it absorbs Q amount of heat. The densities of water and steam at `100^(@)C` are `rho_(1)` and `rho_(2)` respectively and the atmospheric pressure is `P_(0)`. The increase in internal energy of the water isA. `Q`B. `(Q)+p_0((1)/(rho_1)-(1)/(rho_2))`C. `(Q) + p_0((1)/(rho_2)-(1)/(rho_1))`D. `Q-p_0((1)/(rho_1) + (1)/(rho_2))` |
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Answer» Correct Answer - B `Q = triangleU +W` |
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| 380. |
When unit mass of water boils to become steam at `100^(@)C`, it absorbs Q amount of heat. The densities of water and steam at `100^(@)C` are `rho_(1)` and `rho_(2)` respectively and the atmospheric pressure is `P_(0)`. The increase in internal energy of the water isA. QB. `Q+P_(0)((1)/(rho_(1))-(1)/(rho_(2)))`C. `Q+P_(0)((1)/(rho_(2))-(1)/(rho_(1)))`D. `Q-P_(0)((1)/(rho_(1))+(1)/(rho_(2)))` |
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Answer» Correct Answer - B `DeltaU=DeltaQ-W` `=Q-P_(0)[(1)/P_(2)-(1)/(P_(2))]` `=Q+P_(0)[(1)/(P_(2))-(1)/(P_(1))]` |
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| 381. |
In a heat engine, the temperature of the working substance at the end of the cycle isA. equal to that at the beginningB. more than that at the beginningC. less than that at the beginningD. determined by the amount of heat rejected to the sink. |
| Answer» Correct Answer - A | |
| 382. |
A gas is being compressed adiabatically. The specific heat of the gas during compression isA. zeroB. infiniteC. finite but non zeroD. undefined |
| Answer» Correct Answer - A | |
| 383. |
A vessel contains `1` mole of `O_(2)` and `1` mole of He. The value of `gamma` of the mixture isA. 1.4B. 1.5C. 1.53D. none of these |
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Answer» `C_(Vmix)=(3/4R+5/2R)/2=2R` `C_(Pmix)=2R+R=3R` `y_(mix)=C_(P)/C_(V)=3/2` |
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| 384. |
Consider the statement (A) and (B) and identify the carrect answers. A . First law of thermodynamics specifics the consitions under which a body can use its heat energy to produce the work. B. Second law of thermodynamics states that heat always flows from haot body to cold body the itself .A. Both A and B are trueB. Both A and B are falseC. A is true but B is falseD. A is false B is true |
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Answer» Correct Answer - A According to first law of thermodynamics. Amount of heat energy Q taken or given to a thermodynamic system is equal to the sum of work done by or the system W and change in its internal energy `DeltaU` . `Q=W+DeltaU` Second law of thermodynamics states that heat cannot be made to flow cold body to a hot without any external aid of machine but heat can flow from a hot body to a cold body by itself. |
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| 385. |
A piece of lead falls from a height of 100m on a fixed non-conducting slab which brings it to rest. If the specific heat of lead is 30.6 cal/kg `.^(@)C`, the increase in temperature of the slab immediately after collision isA. `6.72^(@)C`B. `7.62^(@)C`C. `5.62^(@)C`D. `8.72^(@)C` |
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Answer» Correct Answer - B `mgh=JmSDeltathetarArrDeltatheta=(gh)/(JS)` |
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| 386. |
Hailstones fall from a certain height. If only 1% of the hailstones melt on reaching the ground, find the height from which they fall. `(g=10ms^(-2),L=80"calorie//g` & `J=4.2J/"calorie")`A. 336mB. 236mC. 436mD. 536m |
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Answer» Correct Answer - A `mgh=(JmL_("ice"))/(100)` |
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| 387. |
How much work to be done in decreasing the volume of an ideal gas by an amount of `2.4xx10^(-4)m^(3)` at constant normal pressure of `1xx10^(5) N//m^(2)`?A. 28 jouleB. 27jouleC. 24jouleD. 25 joule |
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Answer» Correct Answer - C `W=JhrArr(1)/(2)mv^(2)=JmSDeltatheta` |
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| 388. |
A man of 60kg gains 1000 cal of heat by eating 5 mangoes. His efficiency is 56%. To what height he can jump by using this energy?A. 4mB. 20mC. 28mD. 0.2m |
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Answer» Correct Answer - A `mSDeltatheta=J(1)/(2)m(V_(2)^(2)-v_(1)^(2))rArr(theta_(1))/(theta_(2))=(v_(2)^(2)-v_(1)^(2))/(v_(1)^(t^(2))-v_(1)^(t^(3)))` |
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| 389. |
Assuming the expression for the pressure exerted by the gas on the walls of the container, it can be shown that pressure isA. `[(1)/(3)]^(rd)` kinetic energy per unit volume of a gasB. `[(2)/(3)]^(rd)` kinetic energy per unit volume of a gasC. `[(3)/(4)]^(rd)` kinetic energy per unit volume of a gasD. `(3)/(2)xx` kinetic energy per unit volume of a gas |
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Answer» Correct Answer - B According to Kinetic of gases , the pressure exerted by the gas on the walls of container is `p=p_(0)+p_(1)+p_(2)` `"i.e " p=p_(0)+(1)/(3)8v^(2)+3gh` for a container , `p_(1)=(1)/(3)rhov^(2)=(1)/(3)(m)/(v).v^(2)xx(2)/(2)` `(2)/(3)KE[KE=(1)/(2)mv^(2)]` |
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| 390. |
A vessel contains a mixture of `7 g` of nitrogen and `11 g` of carbon dioxide at temperature `T = 300 K`. If the pressure of the mixutre is 1 atm `(1 xx 10^(5) N//m^(2))`, its density is (gas constant `R = 25//3 J//mol K)`A. `0.72 kg//m^(3)`B. `1.44 kg//m^(3)`C. `2.88 kg//m^(3)`D. `5.16 kg//m^(3)` |
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Answer» Correct Answer - B b. Density or mixture, `rho_(mix) = (PM_(mix))/(RT)` Mass or nitrogen gas, `m_(N) = 7 g = 7 xx 10^(-3) kg` Mass of carbon dioxide, `m_(C0_(2)) = 11 g = 11 xx 10^(-3) kg` Molecular weight of nitrogen gas, `M_(N) = 28 xx 10^(-3) kg` Molecular weight of carbon dioxide, `M_(CO_(2)) = 44 xx 10^(-3) kg` `M_(mix) = (n_(N) M_(N))/(n_(N) + n_(CO_(2))) = ((m_(N))/(M_(N)) M_(N) + (m_(CO_(2)))/(M_(CO_(2))) M_(CO_(2)))/((m_(N))/(M_(N)) + (m_(CO_(2)))/(M_(CO_(2)))` ` = (m_(N) + m_(CO_(2)))/((m_(N))/(M_(N)) + (m_(CO_(2)))/(M_(CO_(2)))) = ((7 + 11) xx 10^(-3))/(((7)/(28) + (11)/(44)))` ` = (18 xx 10^(-3))/((1)/(4) + (1)/(4)) = 36 xx 10^(-3) kg` `rho = ((1 xx 10^(5))(36 xx 10^(-3)))/((25)/(3) xx 300) = 1.44 kg//m^(3)` |
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| 391. |
The temperature of a gas is raised while its volume remains constant, the pressure exerted by the gas on the walls of the container increases because its moleculesA. lose more kinetic energy to the wallB. are in contact with the wall for a shorter timeC. strike the wall more often with higher velocitiesD. collide with each other with less frequency. |
| Answer» Correct Answer - C | |
| 392. |
One mole of oxygen is heated at constant pressure starting at `0^(@)C`. How much heat energy must be added to the gas to double its volume ?A. `2.5xx273xxr`B. `3.5xx273xxR`C. `2.5xx546xxR`D. `3.5xx546xxR` |
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Answer» Correct Answer - B `(T_(2))/(T_(1))=(V_(2))/(V_(1))` , Heat to be supplied `=nC_(P)(T_(2)-T_(1))` |
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| 393. |
The equation of a certain gas can be written as `(T^(7//5))/(P^(2//5))=` cons `tan 1`. Its specific at constant volume will be.A. `(3)/(2)R`B. `(5)/(2)R`C. `(7)/(2)R`D. `2R` |
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Answer» Correct Answer - B `P^(1-gamma)T^(gamma)`= constnat , `C_(V)=(R)/(gamma-1)` |
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| 394. |
If the door of a refrigerator is kept open, then which of the following is trueA. Room is cooledB. Room is heatedC. Room is either cooled or heatedD. Room is neither cooled nor heated |
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Answer» Correct Answer - B In a refregerator the heat dissipated in the atmosphere is more than that taken from the cooling chamber, therefore the room is heated if the door of a refrigeratore is kept open. |
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| 395. |
For adiabatic processes `(gamma = (C_(p))/(C_(v)))`A. `P^(gamma) V = constant`B. `T^(gamma) V = constant`C. `T V^(gamma-1) = constant`D. `T V^(gamma) = constant` |
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Answer» Correct Answer - C In adiabatic process `PV^(gamma)=consta nt` `implies ((RT)/(V)) V^(gamma)` = constant `implies TV^(gamma-1) = consta nt` . |
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| 396. |
If `P` is the atmospheric pressure in the last problems find the percentage increase in tension of the string after heatingA. `(25)/((1 - P//P_(0)))`B. `(25)/((1 - P//P_(0)))`C. `25 (1 - P//P_(0))`D. `25 (1 - P//P_(0))` |
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Answer» Correct Answer - A a. Atmospheric pressure `= P` Tension before heating `= P_(0)A = PA` Tension after heating `= P_(2)A - PA` `= (5)/(4) P_(0) A - PA` Increase in tension `= P_(0)A // 4` % Increase = `((1//4) P_(0) A xx 100)/(P_(0) A - P.A) = (25)/((1 - P//P_(0)))` |
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| 397. |
Two identical containers `A` and `B` having volume of ideal gas at the same temperature have mass of the gas as `m_(A)` and `m_(B)` respectively. `2 m_(A) = 3 m_(B)`. The gas in each cylinder expand isothermally to double its volume. If the change in pressure in `A` is `Delta p`, find the change in pressure in `B`:A. `2 Delta p`B. `3 Delta p`C. `(2)/(3) Delta P`D. `(4)/(3) Delta P` |
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Answer» Correct Answer - C d. `2m_(A) = 3m_(g)` `p_(A) V = (m_(A))/(M) RT` `p_(B) V = (m_(B))/(m) RT = (2)/(3) (m_(A))/(M) RT` So, `(p_(A))/(p_(B)) = (3)/(2)` The expanion is isothermal, so pressure will reduce to half when volume is doubled. `p_(A) - (p_(A))/(2) = Delta P` or `p_(A) = 2 Delta p` `p_(B) - (p_(B))/(2) = (p_(B))/(2) = (1)/(2) . (1)/(3) p_(A) = (2)/(3) Delta p` So, `p_(B) = (4)/(3) Delta p` |
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| 398. |
The ratio of pressure of the same gas in two containers in `(n_(1) T_(1))/(n_(2) T_(2))` where `n_(1)` & `n_(2)` are the number of moles and `T_(1)` & `T_(2)` are respective temperatures. If the containers are now joined find the ratio of pressure to the pressure :A. `(P_(1) T_(2) + P_(2) T_(1))/(2 T_(1) T_(2))`B. `(P_(1) T_(1) + P_(2) T_(1))/(T_(1) T_(2))`C. `(P_(1) T_(1) + P_(2) T_(2))/(2 T_(1) T_(2))`D. none of these |
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Answer» Correct Answer - A a. Given `(P_(1))/(P_(2)) = (n_(1) T_(2))/(n_(2) T_(2))` `P prop nT`. It implies that volume of both containers is same. After mixing. `P (2V) = (n_(1) + n_(2)) RT = ((P_(1) V)/(RT_(1)) + (P_(2) V)/(RT_(2))) RT` `P//T = (1)/(2) ((P_(1))/(T_(1)) + (P_(2))/(T_(2))) = (1)/(2) ((P_(1) T_(2) + P_(2) T_(2))/(T_(1) T_(2)))` |
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| 399. |
In the given elliptical `P-V` diagram A. the work done is positiveB. the change in internal energy is non-zeroC. the work done ` = - (pi//4) (P_(2) - P_(1)) (V_(2) - V_(1))`D. the work done `= pi (V_(1) - V_(2))^(2) - pi (P_(1) - P_(2))^(2)` |
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Answer» Correct Answer - C c. The work done = area of `P - V` diagram `a = (V_(2) - V_(1))/(2), b = (P_(2) - P_(1))/(2)` `W = - pi ((V_(2) - V_(1))/(2)) ((P_(2) - P_(1))/(2))` But the cyclic process is anitclockwise, Hence, the wor done is negative |
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| 400. |
In the given elliptical `P-V` diagram A. The work done is positveB. The change in internal energy is non-zeroC. The work done `=-((pi)/(4))(P_(2)-P_(1))(V_(2)-V_(1))`D. The work done `=(pi)(V_(2)-V_(1))^(2)=pi(P_(2)P_(1))^(2)` |
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Answer» Correct Answer - C The work done =area of P-V diagram `a=(V_(2)-V_(1))/(2),b=(P_(2)-P_(1))/(2)` `W=-pi((V_(2)-V_(1))/(2))((P_(2)-P_(1))/(2))` But the cyclic process is anticlockwise. Hence, the work done is negative.. |
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