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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.
| 301. |
Three processes compose a thermodynamic cycle shown in the accompanying `P-V`, diagram of an ideal gas. Process `1 rarr 2` take place at constant temperature, during this process `60 J` of heat enters the system. Process `2 rarr 3` takes place at constant volume. During this process `40 J` of heat leaves the system. Process `3 rarr 1` is adiabatic. What is the change in internal energy of the during process `3 rarr 1`? A. `- 40 J`B. `- 20 J`C. `+ 20 J`D. `+ 40 J` |
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Answer» Correct Answer - D d. `1 rarr 2` : isothermal, `Delta U_(12) = 0` `2 rarr 2`: isochoric, `Delta W = 0` `implies Delta Q_(23) = Delta U_(23) implies - 40 = Delta U_(23)` For a cyclic process, `Delta U = 0` `Delta U_(12) + Delta U_(23) + Delta U_(31) = 0` `0 + (- 40) + Delta U_(31) = 0` `0 + (-40) + Delta U_(31) = 0` `Delta U_(31) = + 40 J` |
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| 302. |
A sealed container contains helium gas at `300 k`. If it is heated to `600 K`, the average kinetic energy of the helium atomsA. remains unchangedB. is doubledC. become `sqrt(2)` timesD. becomes `4` times |
| Answer» Correct Answer - B | |
| 303. |
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?A. 58 JB. 18 JC. 42 JD. 2 J |
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Answer» Correct Answer - B `dQ=-20J` `dW=-8J` (on the gas) `dU=dQ-dW` `=-20+8` =-12 `U_(r)=U_(i)+(dQ-dW)` `=30-12` =18 J |
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| 304. |
Assertion: When a bottle of cold carbonated drink is opened, a slight fog forms around the opening. Reason: Adiabatic expansion of the gas causes lowering of temperature and condersation of water vapours.A. If both assertion and reason are true and the reason is correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true, but the reason is false.D. If assertion is false but the reason is true |
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Answer» Correct Answer - A When a bottle of cold carbonate drink is opened. Then adiabatic expansion of gas evolved is done. Due to this temperature of gas decreases. It condenses the water vapour, which forms a slight fog around the opening. |
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| 305. |
When `1 kg` of ice at `0^(@)C` melts to water at `0^(@)C`, the resulting change in its entropy, taking latent heat of ice to be `80 cal//g` isA. `8xx10^(4)cal//K`B. `80cal//K`C. `293 cal//K`D. `273 cal//K` |
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Answer» Correct Answer - C change in entropy `DeltaS = (ml)/(T) = (1000xx80)/(273)=293 cal K^(-1)`. |
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| 306. |
An ideal gas expands in such a manner that its pressure and volume can be related by equation `PV^(2) = ` constant. During this process, the gas isA. heatedB. cooledC. neither heated nor cooledD. first heated and then cooled |
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Answer» Correct Answer - B b. For an isothermal process, `PV =` constant and for the given process `PV^(2) =` constant. Therefore has is cooled because volume expands by a greater exponent than is in an isothermal porcess. |
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| 307. |
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_1T_2((P_1V_1+P_2V_2))/(P_1V_1T_2+P_2V_2T_1)`B. `(P_1V_1T_1+P_2V_2T_2)/(P_1V_1+P_2v_2)`C. `P_1V_1T_2+P_2V_2T_1/(P_1V_1+P_2V_2)`D. `T_1T_2((P_1V_1+P_2V_2))/(P_1V_1T_1+P_2V_2T_2)` |
| Answer» Correct Answer - A | |
| 308. |
When a system is taken from state i to state f along the path iaf, it is found that `Q=50 cal and W=20 cal`. Along the path ibf `Q=36cal`. W along the path ibf is |
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Answer» From first law of thermodynamics, `dQ=dU+dW` For path iaf, `50=DeltaU+20rArrDeltaU=30` cal For path ibf, `dW=dQ-dU=36-30=6` cal |
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| 309. |
The pressure and density of a diatomic gas `(gamma=7//5)` change adiabatically from (p,d) to `(p^(1),d^(2))`. If `(d^(1))/(d)=32`, then `(P^(1))/(P)` should beA. 1/128B. 32C. 128D. none of the above |
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Answer» Correct Answer - C `W=nRT "log"_(e((V_(2))/(V_(1)))` |
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| 310. |
A solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature `120^(@)C`, thenA. Both of themwill cool down at the same rateB. the cube will cool down faster than the sphereC. the sphere will cool down faster than the cubeD. whichever of the two is heavier will cool down faster |
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Answer» Correct Answer - A When both have same are and same temperature, they will cool at the same rate. |
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| 311. |
Two spherical black bodies of radii `R_(1)` and `R_(2)` and with surface temperature `T_(1)` and `T_(2)` respectively radiate the same power. `R_(1)//R_(2)` must be equal toA. `((T_(2))/(T_(1)))^(2)`B. `((T_(2))/(T_(1)))^(4)`C. `((T_(1))/(T_(2)))^(2)`D. `((T_(1))/(T_(2)))^(4)` |
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Answer» Correct Answer - A Radius of Ist body = `r_(1)` radius of Iind body = `r^_(2)` Temperature of Ist body = `T_(1)` Temperature of Iind body = `T_(2)` The emissivity of a black body or radiated power is given `"by, "E=AsigmaT^(4)xxtpropAT^(4)` or `Aprop(E)/(T^(4))` (where,A is the surface area of the spherical black body) As in the condition of question , the power radiated by Ist and IInd body is same. `"Hence " Aprop(1)/(T^(4))` So , `(A_(1))/(A_(2))=((T_(2))/(T_(1)))^(4)rArr(4pir_(1)^(2))/(4pir_(2)^(2))=((T_(2))/(T_1))^(4)` `"Thus "(r_(1))/(r_(2))=((T_(2))/(T_(1)))^(2)` |
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| 312. |
Which of the following parameters is the same for molecules of all gases at a given temperature?A. MassB. SpeedC. MomentumD. Kinetic energy |
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Answer» Correct Answer - D `K.E. = (3)/(2)kT` |
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| 313. |
if quantity of heat 1163.4 J supplied to one mole of nitrogen gas , at room temperture at constant pressure , then the rise in temperature isA. 54 KB. 28 KC. 65 KD. 8 K |
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Answer» Correct Answer - D Heat given to the gas at room temperture and at constant temperature , `Q=nCpDeltaT` `therefore1163.4=1xx(7)/(2)RxxDeltaT" " (becauseC_(p)=(7)/(2)R " for diatomic gas")` `"or"" " DeltaT=(2xx1163.4)/(7xx8.31)" " (because R=8.31J"mol"^(-1)K^(-1))` `"or " DeltaT=40K` |
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| 314. |
The direction of flow of heat between two bodies is determined byA. average kinetic enrgyB. total energyC. internal energyD. potential energy |
| Answer» Correct Answer - A | |
| 315. |
Heat is absorbed by a body. But its temperature does not raised. Which of the following statement explains the phenomena?A. only K.E. of vibration increasesB. Only P.E. of inter molecular force changeC. No increases in internal energy takes placeD. Increase in K.E. is blanced by decrease in P.E. |
| Answer» Correct Answer - B | |
| 316. |
For a gas the differce between the two specific heat is `4150 J//kg K`. What is the specific heat at constant volume of gas if the ratio of sepcific heat is 1.4A. 8475B. 5186C. 1660D. 10375 |
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Answer» Correct Answer - D `C_(p)-C_(V)=R and C_(p)//C_(V)=1.4` `therefore C_(p)=1.4C_(V) " "therefore 1.4C_(V)-C_(V)=R` `therefore0.4C_(V)=R` `C_(V)=(4150)/(0.4)=(41500)/(4)` `=10375` |
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| 317. |
The temperature of argon, kept in a vessel, is raised by `1^(@)C` at a constant volume. The total heat supplied to the gas is a combination of translational and rotational energies. Their respective shares areA. `60%` and `40%`B. `40%` and `60%`C. `50%` and `50%`D. `100%` and `0%` |
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Answer» Correct Answer - D Rotational `K.E.` for monoatomic gas `= 0` |
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| 318. |
The temperature of argon, kept in a vessel, is raised by `1^(@)C` at a constant volume. The total heat supplied to the gas is a combination of translational and rotational energies. Their respective shares areA. 60%,40%B. 100%,0%C. 0%,100%D. 40%,60% |
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Answer» Correct Answer - B As argon is monatomic gas its atoms have no rotational motion. Therefore the entire energy is translational. |
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| 319. |
Carnot cycle (reversible) of a gas represented by a pressure volume curve is shown in the diagram Consider the following statement I Area ABCD = Work done on the gas II Area ABCD = Net heat absorbed III Change in the internal energy in cycle = `0` Which of these are correct? A. I onlyB. II onlyC. II and IIID. I, II and III |
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Answer» Correct Answer - C Work done by the gas (as cyclic process is clockwise ) `:. DeltaW = Area ABCD` so from the first law of thermodynamics `Delta Q` (net heat absorbed) = `DeltaW = Area ABCD` as change in internal energy in cycle `DeltaU = 0`. |
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| 320. |
A motor cycle engine delivers a power of `10 kW`, by consuming petrol at the rate of `2.4 kg//hour`. If the calorific value of petrol is `35.5 MJ//kg`, the rate of heat rejection by the exhaust byA. `5.5 kW`B. `13.7 kW`C. `11.2 kW`D. `9.7 kW` |
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Answer» Correct Answer - B Heat recived (or produce by the burning of petrol) in one hour will be `=(2.4 kg//hour) (3.5 MJ//kg)` `=85.2xx10^(6)J//hour` `:.` The rate at which heat is recevied ltbr. `=(85.2xx10^6 J)/((3600s)) = 2.37xx10^(4) J//s = 23.7kW` Evidently the rate of heat rejection =rate at which heat is produced - rate at which work is obtained = `23.7kW - 10kW = 13.7kW`. |
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| 321. |
An ideal gas is taken from state `1` to state `2` through optional path `A, B, C and D` as shown in the `PV` diagram. Let `Q, W and U` represent the heat supplied, work done and change in internal energy of the gas respectively. Then, A. `Q_(A) - Q(D) = W_(A) - W(D)`B. `Q_(B) - W(B) gt Q_(C) - W(C)`C. `W_(A) lt W(B) lt W_(C) - W(D)`D. `Q_(A) lt Q(B) lt Q_(C) - Q(D)` |
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Answer» Correct Answer - A Work done by gas in all four process is positive and in order `W_A gt W_B gt W_C gt W_D` `implies ` c is false The change in intial energy `U` is same for all process. `:. Q_(A) = U +W_A` ..(1) ` Q_(B) = U +W_B` ..(2) ` Q_(C) = U +W_C` ..(3) ` Q_(4) = U +W_A` ..(4) Hence `Q_(A) gtQ_B gt Q_C gtQ_D` `implies ` (d) is false From equation (1) and (4) ltbr. `Q_A - Q_C = W_B - W_C` `implies` (b) is false. |
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| 322. |
During adiabatic process pressure P versus density `roh` equation isA. `P(roh)^(gamma) = constant`B. `P(roh)^(-gamma) = constant`C. `P^(gamma)(roh)^(1+gamma) = constant`D. |
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Answer» Correct Answer - B In adiabatic process ltbr. `PV^(gamma) ` = constant..(1) Density `rho=m/V` or `rhooo V^(-1)` Hence equation (1) can be written as `prho^(-gamma) = constant`. |
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| 323. |
P-V diagram of a diatomic gas is a straight line passing through origin. The molar heat capacity of the gas in the process will beA. 4 RB. 3 RC. `4R//3`D. 2.5 R |
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Answer» Correct Answer - B As p- V diagram is a straight line passing through origin , therfore, `ppropV" or "pV^(_1)` = constant . In the process , `pV^(x)` = constant, molar heat capacity is given by , `C=(R)/(gamma-1)+(R)/(1-x)` Where , `x=-1` here and `gamma=1.4` for diatomic gas `C=(R)/(1.4-1)+(R)/(1-(-1))` `=(5)/(2)R+(R)/(2)=3R` |
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| 324. |
P-V diagram of a diatomic gas is a straight line passing through origin. The molar heat capacity of the gas in the process will beA. `4R`B. `2.5R`C. `3R`D. `4R/3` |
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Answer» Correct Answer - C `P-V` diagram of the gas is a straight line passing through origin. Hence `p prop V` Molar heat capacity in the process `PV^(x)`= constant is given by `C = R/(gamma-1) + R/(1-x)` Here, `gamma = 1.4` (for diatomic gas) `C = R/(1.4-1) + R/(1+1) = 3R` . |
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| 325. |
P-V diagram of a diatomic gas is a straight line passing through origin. The molar heat capacity of the gas in the process will be |
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Answer» P-V diagram of the gas is a straight line passing through origin. Hence, `PpropV` or `PV^(-1)=` constant `rArrx=-1` `C=(R)/(.-1)+(R)/(1-x)` Here, `gamma=1.4` (for diatomic gas) `therefore C=(R)/(1.4-1)+(R)/(1+1)` or `C=R` |
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| 326. |
A monoatomic gas undergoes a process given by `2dU+3dW=0`, then what is the process |
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Answer» `dQ=dU+dWrArrdQ=dU-(2dU)/(3)=(dU)/(3)` `=(1)/(3)nC,dT=(1)/(3)n(3)/(2)RdT=(nRdT)/(2)` `C=(1)/(n)(dQ)/(dT)=(R)/(2)` it is not isobaric as C is not equal to `(5R)/(2),` it is not adiabatic as `Ccancel=0` It is not isothermal as `Ccancel=oo` so it is a polytrophic process. |
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| 327. |
The volume of a gas is reduced adibatically to `(1//4)` of its volume at `27^(@)C`. if `gamma = 1.4` . The new temperature will beA. `350xx4^(0.4)K`B. `300xx4^(0.4)K`C. `150xx4^(0.4)K`D. None of these |
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Answer» Correct Answer - B Isothermal process `PdV+VdP=0` `(dP)/(P)=-(dV)/(V), (K)=(dP)/(-((dV)/(V)))=(dP)/(((dP)/(P)))=P` |
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| 328. |
A solid sphere cools at the rate of `2.8^(@)C` per minute, when its temperature is `127^(@)C`. Find the rate at which another solid copper sphere of twice the radius looses its temperature at `127^(@)C`, in both the cases, the room temperature is maintained at `27^(@)C` isA. `9.72^(@)C//`minB. `11.2^(@)C`/minC. `3.6^(@)C`/minD. `1.4^(@)C`/min |
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Answer» Correct Answer - B `(R_(2))/(R_(1))=(A_(2))/(A_(1))((T_(2)^(4)-T_(0)^(4)))/((T_(1)^(4)-T_(0)^(4)))=(4r_(1)^(2))/(r_(1)^(2))=4` `R_(2)=4R_(1)` `=4xx2.8` `=11.2` c/min. |
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| 329. |
A vessel contains a mixtrue consisting of `m_(1) = 7g` of nitrogen `M_(1) = 28 and m_(2) = 11 g` of carbon dioxide `(M_(2) 44)` at temperature` T = 300 K` and pressure `p_(0) = 1` atm. Find the density of the mixture. |
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Answer» Let V be the volume of the vessel. Then `rho_(mis)=(m_(1) + m_(2))/(V)` Let `p_(1) and p_(2)` be the partial pressure Then `p_(1)V = (m_(1))/(M_(1))RT` and `P_(2)V = (m_(2))/(M_(2))RT, p_(0) = p_(1) + p_(2)` `therefore " "p_(0) = (m_(1)/M_(1) + m_(2)/M_(2))(RT)/(T)` `therefore " "rho_(mix)=((m_(1) + m_(2))M_(1)M_(2))/(M_(1)M_(2) + m_(2)M_(1))xx(p_(0))/(RT)` `=((7+11)xx28xx44xx10^(-3))/(7xx44+11xx28) xx (10^(5))/(8.3xx300)` `=1.446 kg m^-3 = 1.446` per litre |
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| 330. |
8 gm of oxygen and x gm of hydrogen posses same pressure, volume and temperature. Then x=A. `1//2`B. `16`C. 32D. 2 |
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Answer» Correct Answer - A `(n_(1)P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))n_(2)n_(1)=n_(2)(8)/(32)=(x)/(2)x=(1)/(2)` |
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| 331. |
The internal energy of a monatomic ideal gas is `1.5 nRT`.One mole of helium is kept in a cylinder of cross section `8.5 cm^(2)`. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of `42J` heat is given to the gas. If the temperature rise through `2^(@)C`, find the distance moved by the piston. atmospheric pressure `=100 kPa.`A. `10 cm`B. `15 cm`C. `20 cm`D. `5 cm` |
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Answer» Correct Answer - C The change in internal energy of the gas is `DeltaU=1.5nR(DeltaT)` `=1.5(1mol) (8.3J//mol-K) = 24.9 J` The heat given to the gas = `42J` The work done by the gas is `DeltaW=DeltaQ-DeltaU = 42J-24.9J = 17.1J` If the distance moved by the piston is `x`, the work done is `DeltaW = (100kPa)(8.5cm^(2))x` Thus, `(10^ N//m^(2))(8.5xx10^(-4)m^(2))x = 17.1J` or `x= 0.2m = 20 cm`. |
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| 332. |
In a p-V diagram for an ideal gas (Where , p is along y-axis and V is along X- axis) , the value of the ratio "slope pof adiabatic curve/ Slope of the isothermal curve"at any point will be (Where symbols have their usual meanings ).A. 1B. 2C. `C_(p)//C_(v)`D. `C_(v)//C_(p)` |
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Answer» Correct Answer - C As, it is known , `("Slope of isothermal curve")/("Slope of abiabatic curve")=gamma=(C_(p))/(C_(v))` |
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| 333. |
The temperature inside a refrigerator is `t_(2)^(@)C` . The amount of heat delivered to the room for each joule of electrical energy consumed ideally will beA. `(t_(2)+273)/(t_(1)-t_(2))`B. `(t_(1)+t_(2))/(t_(1)-273)`C. `(t_1)/(t_(1)-t_(2))`D. `(t_(1)+273)/(t_(1)-t_(2))` |
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Answer» Correct Answer - D Heat delivered = `Q_(1)` `COP_(beta)=(t_(2)+273)/(t_2-t_1)=(Q_2)/(W)=(Q_(1)-W)/(W) = (Q_1)/(W)-1` `(Q_1)/(W) = 1+(t_(2)+273)/(t_(1)-t_(2))= (t_(1)+273)/(t_(1)+t_(2))` |
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| 334. |
A refrigerator placed in a room at 300K has inside temperature 200K. How many caslories of heat shall be delivered to the room for each 2 KiloCal of energy consumed by the refrigerator ideally?A. 4K.calB. 2K.calC. 8K.calD. 6Kcal |
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Answer» Correct Answer - D `beta=(Q_(2))/(W)=(T_(2))/(T_(1)-T_(2))` |
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| 335. |
Which one of the following gases possesses the largest internal energyA. `2` moles of helium occupying `1 m^(3) at 300 K`B. `56` kg of nitrogen at `107 Nm^(-2) and 300 K`C. `8` grams of oxygen at `8 atm 300 K`D. `6 xx 10^(26)` molecules of argon occupying `40 m^(3) at 900 K` |
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Answer» Correct Answer - B `Delta U = mu C_(v) DeltaT=m/M C_v Delta T = N/(N_A) C_v Delta T` `implies (DeltaU)_(n) = (56xx10^3))/(14) xx 5/2 R xx 300` and `(DeltaU)_(A) = (6xx10^(26))/(6xx10^(23)) xx 3/2Rxx900implies (Delta U)_(N) gt (Delta U)_(A)`. |
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| 336. |
A given system undergoes a change in which the work done by the system equals to the decrease in its internal energy. The system must have undergone anA. isothermal changeB. adiabatic changeC. isobaric changeD. isochoric change |
| Answer» Correct Answer - B | |
| 337. |
A closed vessel contains some gas at a given temperature and pressure. If the vessel is given a very high velocity, then the temperature of the gasA. increasesB. decreasesC. may increase or decrease depending upon the nature of the gasD. does not change |
| Answer» Correct Answer - D | |
| 338. |
Unit mass of liquid of volume `V_(1)` completely turns into a gas of volume `V_(2)` at constant atmospheric pressure P and temperature T. The latent heat of vaporization is "L". Then the change in internal energy of the gas isA. LB. `L+P(V_(2)-V_(1))`C. `L-P(V_(2)-V_(1))`D. Zero |
| Answer» Correct Answer - C | |
| 339. |
The molecular kinetic energy of 1 g of helium (molecular weight 4) at `127^(@)C` is (Given , R `=8.31 J"mol"^(-1)K^(-1)` )A. 12.84 JB. 12.465 JC. 14.34D. 14.384 J |
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Answer» Correct Answer - B Given , T = 273 + 127 = 400 K `because` Helium is monoatmic gas. `therefore` Average kinetic energy per mole of helium = `(3)/(2)RT` Average kinetic energy of 1 g of helium = `(3)/(2)(RT)/(M)` `=(3xx8.31xx4.00)/(2xx4)=12.465J` |
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| 340. |
At a constant pressure of `10^(4)Nm^(-2)` , a gas expands by `0.25m^(3)` work done by the gas isA. 2500 JB. 250 JC. 25 JD. 2.5 J |
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Answer» Correct Answer - A Given pressure , `p=10^(4)Nm^(-2)` , volume , `V=0.25m^(3)` , work done , W= ? , we know that , `W=pdV=10^(4)xx0.25` =2500 J |
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| 341. |
A refrigerator transfers 250 J heat per second from `-23^(@)C` to `-25^(@)C` . Find the power consumed , assuming no loss of energy.A. `48 Js^(-1)`B. `50 Js^(-1)`C. `52 Js^(-1)`D. `53 Js^(-1)` |
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Answer» Correct Answer - A Here , `Q_(2)=250Js^(-1),T_(2)=-23+273=250K` `T_(1)=25^(@)C=25+273=298K` power = work per second We know that , `beta=(Q_(2))/(W)=(T_(2))/(T_(1)-T_(2))=(250(298-250))/(250)` `therefore W=(250xx48)/(250)=48Js^(-1)` |
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| 342. |
Starting with the same initial conditions, an ideal gas expands from volume `V_(1)` to `V_(2)`. The amount of work done by the gas is greatest when the expansion isA. isothermalB. isobaricC. adiabaticD. equal in all cases |
| Answer» Correct Answer - B | |
| 343. |
Which of the following laws of thermodynamics leads to the interference that it is difficult to convert whole of heat into work?A. ZerothB. SecondC. FirstD. both 1 `&` 2 |
| Answer» Correct Answer - B | |
| 344. |
At absolute zero temperature, the kinetic energy of the moleculesA. becomes zeroB. becomes maximumC. becomes minimumD. remains constant |
| Answer» Correct Answer - A | |
| 345. |
The average energy for molecules in one degree of reedom is :A. `(3)/(2)kT`B. `(kT)/(2)`C. `(3)/(2)kT`D. `kT` |
| Answer» Correct Answer - B | |
| 346. |
Two ballons are filled, one with pure He gas and other by air, repectively. If the pressure and temperature of these ballons are same then the number of molecules per unit volume is:A. more in the He filled balloonB. same in both balloonsC. more in air filled balloonD. in the ratio of 1:4 |
| Answer» Correct Answer - B | |
| 347. |
A large block of ice is placed on a table when the surroundings are at `0^@C`.A. ice melts at the sidesB. ice melts at the topC. ice melts at the bottomD. ice does not melt at all |
| Answer» Correct Answer - C | |
| 348. |
The temperature determines the direction of net change ofA. gross kinetic energyB. intermolecular kinetic enegyC. gross potential energyD. intermolecular potential energy |
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Answer» Correct Answer - A Temperature is proportional to the average molecular kinetic energy. |
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| 349. |
A gas is taken through the cycle `A rarrB rarr C rarr A`, as shown in figure, what is the net work done by the gas? A. `2000J`B. `1000J`C. ZeroD. `-2000J` |
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Answer» Correct Answer - B Work done in `P-V` diagram is equal to area enclosed in `P-V` curve `implies W = 1/2 xx 5 xx 10^(-3)xx4xx10^(5)=10xx10^(2)` ltbr. `=1000J`. |
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| 350. |
The maximum energy in thermal radiation from a source occurs at the wavelength 4000Å. The effective temperature of the sourceA. 7000 KB. 80000 KC. `10^(4)K`D. `10^(6)K` |
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Answer» Correct Answer - A `lamda_(m)=(b)/(T) " "therefore T=(b)/(lamda_(m))=(2.93xx10^(-3))/(4000xx10^(-10))=7325K`. |
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