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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.
| 201. |
Find the work done by a gas when it expands isothermally at `37^(@)C` to four times its initial volume.A. 3753JB. 3573JC. 7633JD. 5375J |
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Answer» Correct Answer - B `T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1)` |
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| 202. |
A monatomic gas expands at constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involed in the expansion isA. 100%,0B. 60%,40%C. 40%,60%D. 75%,25% |
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Answer» Correct Answer - B `W=(nR)/(gamma-1)(T_(1)-T_(2))` |
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| 203. |
Given molecular weight of hydrogen molecule is `M = 2.016 xx 10^(-3) kg//mol`. Calculate the root-mean-square speed of hydrogen molecules `(H_(2))` at `373.15 K (100^(@) C)`. |
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Answer» The mass of an `H_(2)` molecule may be calculated from the molecule weight as `m = (M)/(N_(0)) = (2.016 xx 10^(-3) kg//mol)/(6.02 xx 10^(23) mol^(-1)) = 3.35 xx 10^(-27) kg` Then `(1)/(2) mv_(rms)^(2) = (3)/(2) kT` `v_(rms) = sqrt((3 kT)/(m)) = sqrt((3(1.38 xx 10^(-23))(373.15))/(3.35 xx 10^(-27))) = 2.15 km//s` |
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| 204. |
The mass of hydrogen molecule is `3.23 xx 10^(23)` hydrogen molecules strike `2 cm^(2)` of a wall per second at an angle of `45^(@)` with the normal when moving with a speed of `10^(5) cm s^(-1)`, what pressure do they exert on the wall ? Assume collision to be elasitc. |
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Answer» Velocity of each molecule along the normal `= v cos theta` where `theta` is the angle made by the velocity of the molecule with the normal. Therefore, change in velocity `= v cos theta - (- v cos theta ) = 2 v cos theta` Change in momentum `= m xx 2v cos theta = 2 m v cos theta` There, rate of change of momentum `= 2 m v cos theta xx N` where N is the number, of molecules striking the wall each second. Force exerted by molecules `= 2 m v N cos theta` `Pressure = ("Force")/("Area")` `= (2 m v N cos theta)/(A) = (2 xx 3.32 xx 10^(-27) xx cos 45^(@))/(2 xx 10^(-4))` `= 2.35 xx 10^(3) N m^(-2)` |
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| 205. |
At `100^(@)C` the volume of `1 kg` of water is `10^(-3) m^(3)` and volume of `1 kg` of steam at normal pressure is `1.671 m^(3)`. The latent heat of steam is `2.3 xx 10^(6) J//kg` and the normal pressure is `10^(5) N//m^(2)`. If `5 kg` of water at `100^(@)C` is converted into steam, the increase in the internal energy of water in this process will beA. `8.35 xx 10^(5) J`B. `10.66 xx 10^(6) J`C. `11.5 xx 10^(5) J`D. zero |
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Answer» Correct Answer - B b. Heat required to convert 5 kg of water into steam `Delta Q = mL = 5 xx 2.3 xx 10^(6) = 11.10^(6) J` Work done in expanding volume, `Delta W = P Delta V` `= 5 xx 10^(5) (1.671 - 10^(-3)) = 0.835 xx 10^(6) J` Now by the first law of thermodynamics `Delta U = Delta Q - Delta W` `implies Delta U = 11.5 xx 10^(6) - 0.835 xx 10^(6) J` |
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| 206. |
Two vessels A and B having equal volume contain equal masses of hydrogen in A and helium in B at 300 K. Then, mark the correct statement?A. The pressure exerted by hydrogen is half that exerted by heliumB. The pressure exerted by hydrogen is equal to that exerted by heliumC. Average KE of the molecule of hydrogen is half the average KE of the molecules of heliumD. The pressure exerted by hydrogen is twice that exerted by helium |
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Answer» Correct Answer - D We have `pV=nRT` Number of moles of the gas pressure pf hydrogen `P_(H_(2))=(m)/(M_(H_(2))).(RT)/(V)rArr` Bessera of Helium `P_(He)=(m)/(M_(He)).(RT)/(V)` `therefore" " (p_(H_(2)))/(p_(H_(2)))=(M_(He))/(M_(H_(2)))=(4xx10^(-3))/(2xx10^(-3))=2` `rArr" " p_(H_(2))=2P_(He)` |
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| 207. |
In the two vessels of same volume, atomic hydrogen and helium at pressure 1 atm and 2 atm are filled. If temperature of both the sample is same then average speed of hydrogen atoms `ltC_(B)lt` will be related to that of helium `ltC_(He)gt` asA. `ltC_(H)gt=sqrt(2)ltC_(He)gt`B. `ltC_(H)gt = ltC_(He)gt`C. `ltC_(H)gt=2ltC_(He)gt`D. `ltC_(H)gt=(ltC_(He)gt)/(2)` |
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Answer» Correct Answer - C `ltCgt=sqrt((8RT)/(nM))rArrltCgtprop(1)/sqrt(M)` |
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| 208. |
Determine the gas temperature at which (a) the root mean square velocity of hydrogen molecules exceeds their most probable velocity by `Delta v = 400 m//s` , (b) the velocity distribution function `F (v)` for the oxygen molecules will have the maximum value at the velocity `v = 420 m//s`.A. 384KB. 342KC. 300KD. 280K |
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Answer» Correct Answer - A `v_(rms)-v_(P)=(sqrt(2)-sqrt(2))sqrt((RT)/(M))=Deltav` `thereforeT=(M)/(R)((Deltav)/(sqrt(3)-sqrt(2)^(2)))=384k` |
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| 209. |
If 294 joules of heat energy is required to raise the temperature of 2 moles of an ideal gas from `30^(@)C` to `35^(@)C` at constant pressure, then the specific heat at constant pressure will be (R=8.4J/mol `.^(@)K`)A. 27.4 J/mol KB. 28.4 J/mol KC. 29.4 J/mol KD. 30.4 J/mol K |
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Answer» Correct Answer - C `dQ_(p)=nC_(p)dT` `therefore C_(p)=(dQ_(p))/(ndT)=(294)/(2xx5)=29.4` |
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| 210. |
A system changes from the state `(P_(1),V_(1))` to `(P_(2)V_(2))` as shwon in the diagram. The workdone by the system is A. `12xx10^(4)J`B. `12xx10^(8)J`C. `12xx10^(5)J`D. `6xx10^(4)J` |
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Answer» Correct Answer - C W=Area of triangle+Area of rectangle |
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| 211. |
In the given (V-T) diagram, what is the relation between pressure `P_(1) and P_(2)` ? A. `P_(2)=P_(1)`B. `P_(2)gtP_(1)`C. `P_(2)ltP_(1)`D. Cannot be predicated |
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Answer» Correct Answer - C Slope of the graph `prop 1/(pressure)`. |
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| 212. |
A heat-conducting piston can freely move inside a closed thermally insulated cylinder with an ideal gas. In equilibrium the piston divides the cylinder into two equal parts, the gas temperature being equal to `T_0`. The piston is slowly displaced. Find the gas temperature as a function of the ratio `eta` of the volumes of the greater and smaller sections. The adiabatic exponent of the gas is equal to `gamma`. |
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Answer» Correct Answer - `T_0[(n+1)^(2)/(4n)]^(gamma-1)/(2)` Let d be the distance through Which the piston has to be moved to male the volume of one part n times that of the other part. Then n `=(V_(0)+Ad)/(V_(0)-Ad)` or `d =(n-1)/(n+1)xx(V_(0))/(A)` Now consider any displacement of the piston by x. Since the pistionis conducting and the process of compressionis slow, the temperature of gases on the two sides of the piston will be the same." `therefore (p_(0)v_(0))/(T_(0))=(p_(1)(V_(0)+Ax))/(T)=(P_(2)(V_(0)-Ax))/(T)` Since it is a quasi static process `p_(1)A+F_("agent")=p_(2)A or F_("agent")=(P_(2)-P_(1))A` Then dW, elementary work done by agent `= F_(agent)dx=(P_(2)-p_(1))Adx` or `dW=(m)/(M)RT((1)/(V_(0)-Ax)-(1)/(V_(0)+Ax))Adx` `=(m)/(M)RT(2A^(2)xdx)/(V_(0)^(2)-A^(2)x^(2))` Now `dU=(2m)/(M)C_(v)dT` By 1st law of thermodynamics `dQ=dU+dW` Here `dQ= 0``therefore dU=-dW` or `(2m)/(M)C_(v)dT=(m)/(M)RT(2A^(2)xdx)/(V_(0)^(2)-A^(2)x^(2))` or `(R)/(gamma-1)dT=RT(A^(2)xdx)/(V_(0)^(2)-A^(2)x^(2))` or `(1)/(gamma-1)overset(T)underset(T_(0))int(dT)/(T)=A^(2)overset(d)underset(0)int(xdx)/(V_(0)^(2)-A^(2)x^(2))` or`(1)/(gamma-1) In (T)/(T^(0)) = A^(2)overset(d)underset(0)int(xdx)/(V_(0)^(2)-A^(2)x^(2))` Put`( V_(0)^(2)-A^(2)x^(2))=z`. Then-`(2A^(2)xdx)=dz` `therefore int(xdx)/(V_(0)^(2)-A^(2)x^(2))=int-(1)/(2A^(2))(dz)/(z)=-(1)/(2A^(2))Inz=(1)/(2A^(2))In(V_(0)^(2)-A^(2)x^(2))` `therefore (1)/(gamma-1)in(T)/(T_(0))=A^(2)xx-(1)/(2A^(2))[In(V_(0)^(2)-A^(2)x^(2))]_(0)^(d)` `=(1)/(2)[In(V_(0)^(2)-A^(2)d^(d))-InV_(0)^(2)]` `=-(1)/(2)[In(1-(A^(2))/(V_(0)^(2)).((n-1)^(2)V_(0)^(2))/((n+1)^(2)A^(2)))]=(1)/(2)In(4n)/(n+1)^(2)` or `In(T)/(T_(0))=-(gamma-1)/(2)In(4n)/(n+1)^(2)` or `T = T_(0)[(n+1)^(2)/(4n)]^((gamma-1)/(2))` |
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| 213. |
A freely sliding massive piston is supported by a spring inside a vertical cylinder as shown. When all air is pumped out of the container, the piston remains in equilibrium with only a tiny gap between the piston and the bottom surface of the cylinder. An ideal gas at temperature `T_(0)` is slowly injected under the piston so that it rises to height `h_(0)` (see figure). Calculate the height of the piston from the bottom of the container if the temperature of the gas is slowly raised to `4 T_(0)`. |
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Answer» Correct Answer - `2h_(0)` |
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| 214. |
An ideal gas is inside a cylinder with a piston that can move freely. The walls of the cylinder and piston are non- conducting. The piston is being moved out of the cyl- inder at a constant speed u. (a) Consider a gas molecule of mass m moving with speed `v (gt gt u)`. It hits the piston elastically at an angle of incidence q. Calculate the loss in kinetic energy of the molecule. (b) If the area of piston is A and pressure of the gas is P, calculate the rate of decreases of molecular kinetic energy of the gas sample. (c) If` u gt gt` molecular velocities, at what rate will the gas lose its molecular kinetic energy |
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Answer» Correct Answer - (a) 2 mvu `cos theta` " " (b) Pau " " (c ) zero |
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| 215. |
Choose the correct statement . When the temperature of a gas is increasedA. the kinetic energy of its molecules increaseB. the potential energy of its molecules increaseC. the potential energy decreases and the kinetic energy increases, the total energy remaining unchanged.D. the potential energy increases, he kinetic energy decreases and the total energy remaining unchanged |
| Answer» Correct Answer - A | |
| 216. |
The number of molecules per unit volume (n) of a gas is given byA. `(P)/(kT)`B. `(kT)/(P)`C. `(P)/(kT)`D. `(RT)/(P)` |
| Answer» Correct Answer - A | |
| 217. |
The number of molecules of `N_(2)` and `O_(2)` in a vessel are same. If a fine hole is made in the vessel then which gas escapes out more rapidly?A. `N_(2)`B. `O_(2)`C. both equallyD. sometimes `N_(2)` and sometimes `O_(2)` |
| Answer» Correct Answer - A | |
| 218. |
The mean kinetic energy per unit volume of gas (E) is related to average pressure P, exerted by the gas isA. `(P)/(2)`B. PC. `(3P)/(2)`D. 2P |
| Answer» Correct Answer - C | |
| 219. |
Pressure exerted by a perfect gas equal toA. mean `K.E.` per unit volumeB. half of mean `K.E.` per unit volumeC. one-third of mean `K.E.` per unit volumeD. two-third of mean `K.E.` per unit volume |
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Answer» Correct Answer - D `P = (1)/(3) (M)/(V) v_(rms)^(2) = (2)/(3)(((1)/(2)Mv_(rms)^(2))/(V))` `= (2 K)/(3V), K :K.E`. |
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| 220. |
In an adiabatic change, the pressure and temperature of a monoatomic gas are related with relation as `P prop T^(C )`, Where `C` is equal to:A. `5/4`B. `5/3`C. `3/5`D. `5/2` |
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Answer» Correct Answer - D For adiabatic process `PV^(gamma)` constant ..(i) We have ideal gas relation , `PV=RT` `implies V = (RT)/(P) `...(ii) From equ. (i) and (ii) we get `P((RT)/(P))^(gamma)` = constant `implies (T^(gamma))/(P^(gamma-1)) = consta nt` ..(iii) where `gamma` is adiabatic consta nt of the gas Given `P prop T^(C)` ...(iv) On comparing with eq, (iii) we have `C = (gamma)/(gamma-1)` For a monoatomic gas `gamma=5/3` `:.` we have , `C = (5/3)/(5/3-1) = 5/2`. |
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| 221. |
According to the kinetic theory of gases (i) `P prop v_(rms)` (ii) `v_(rms) prop T` (iii) `v_(rms) prop T^(1//2)` (iv) `P prop v_(rms)^(2)`A. (i), (iii)B. (ii), (iii)C. (iii), (iv)D. (i), (iv) |
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Answer» Correct Answer - C `P = (1)/(3)(mN)/(V) bar(v^(2)) = (1)/(3) (M)/(V) v_(rms)^(2) rArr P prop v_(rms)^(2)` `v_(rms) = sqrt((3 R T)/(M_(0))) rArr v_(rms) = prop T^(1//2)` |
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| 222. |
The pressure P, Volume V and temperature T of a gas in the jar A and the other gas in the jar B at pressure `2P`, volume `V//4` and temperature `2 T`, then the ratio of the number of molecules in the jar A and B will beA. `1:1`B. `1:2`C. `2:1`D. `4:1` |
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Answer» Correct Answer - D |
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| 223. |
The pressure `P`, volume `V` and temperature `T` of a gas in the jqar `A` and other gas in the jar `B` as at pressure `2P`, volume `V//4` and temperature `2T`, then the ratio of the number of molecules in the jar `A` and `B` will beA. `1 : 1`B. `1 : 2`C. `2: 1`D. `4: 1` |
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Answer» Correct Answer - D `PV = n_(A)RT` (i) `2P.(V)/(4) = n_(B)R.2T` (ii) Number of moles `prop` number of molecules `(n_(A))/(n_(B)) = (4)/(1)` |
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| 224. |
At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be `1930 m//s`. The gas isA. `H_(2)`B. `F_(2)`C. `O_(2)`D. `CI_(2)` |
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Answer» Correct Answer - A `v_(rms) = sqrt((3 RT)/(M_(0)))` `M_(0) = (3 RT)/(v_(rms)^(2)) = (3 xx 8.3 xx 300)/((1930)^(2)) = 2 xx 10^(-3) kg = 2g` Molecular mass `= 2 g` i.e. gas is `H_(2)` |
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| 225. |
The rms speed of the molecules of a gas in a vessel is `400ms^(-1)`. If half of the gas leaks out at constant temperature, the rms speed of the ramaining molecules will be………….. |
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Answer» Correct Answer - A::D |
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| 226. |
The kinetic energy of 1 g molecule of a gas, at normal temperature and pressure, isA. `3.4xx10^(3)J`B. `1.7xx10^(3)J`C. `2.4xx10^(3)J`D. None of these |
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Answer» Correct Answer - A So, Kinetic theory of 1 g molecule of an ideal gas is `KE=(3)/(2)RT=(3)/(2)xx8.31xx273` `=3.4xx10^(3)J` |
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| 227. |
`S_(1)`: "All collisions between the molecules of the gas and walls of container are elastic". This assumption has validity only when temperature of both, the gas and the walls of the container, are same `S_(2)`: If `DeltaV=0` in a process then work done by the gas must be zero. (T/F) `S_(3)`: Internal energy of one kilogram of Helium is same as that of one kilogram of Argon at the same temperature.A. TFFB. TTFC. TTTD. TFT |
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Answer» Correct Answer - A `U=(f)/(2) nRT` for both gases f and T are same but n is different |
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| 228. |
Assertion: The isothermal curves intersect each other at a certain point. Reason: The isothermal changes takes place rapidly, so the isothermal curves have very little slope.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 - D To carry out isothermal process, a perfect gas is compressed or allowed to expand very slowly. Isothermal curves never intersect earth other as they have little lope. |
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| 229. |
Figure shows the pressure P versus volume V graphs for a certains mass of a gas at two constant temperature `T_(1) and T_(2)`. Which of the following interface is correct? A. `T_(1) = T_(2)`B. `T_(1) gt T_(2)`C. `T_(1) lt T_(2)`D. no inference can be drawn due to insufficient information |
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Answer» Correct Answer - C For a given pressure `V` is small for `T_(1)`. Since `V prop T` therefore, `T_(1) lt T_(2)`. |
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| 230. |
Which one of the following is not a thermodynamical coordinate ?A. `R`B. `V`C. `T`D. `P` |
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Answer» Correct Answer - A Each thermodynamic process is distinguised from other process in energetic character according to which parameters as temperature , pressure or volume. While `R` is a gas constant which is not a thermodynamical coordinate. |
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| 231. |
The internal energy change in a system that has absorbed `2 kcal` of heat and done `500J` of work isA. `8900J`B. `6400J`C. `5400J`D. `7900J` |
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Answer» Correct Answer - D Heat given to a system `(DeltaQ)` is equal to the sun of increases in the internal energy `(Deltau)` and the work done `(DeltaW)` by the system against the surrounding and `1 cal=4.2J`. According to first law of thermodynamics `DeltaU=Q-W` `=2xx4.2xx1000-500=7900J`. |
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| 232. |
If `DeltaU` and `Delta W` represent the increase in internal energy and work done by the system resectively in a thermodynamical process, which of the following is true?A. `DeltaU=-DeltaW`, in an adiabatic processB. `DeltaU=DeltaW`, in an isothermal processC. `DeltaU=DeltaW`, in an adiabatic processD. `DeltaU=-DeltaW`, in an isothermal process |
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Answer» Correct Answer - A From the first law of thermodynamics `Q=DeltaU+W` For adiabatic process `Q = 0` `implies DeltaU=-W`. |
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| 233. |
A Carnot engine has the same efficiency between `800 K` to `500 K` and `x K to 600 K`. The value of `x` isA. 100 KB. 960 KC. 846 KD. 754 K% |
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Answer» Correct Answer - B As, `eta=1-(T_(2))/(T_(1))=1-(500)/(800)=1-(600)/(x)` `therefore(3)/(8)=1-(600)/(x)rArr(600)/(x)=1-(3)/(8)=(5)/(8)` 5x=4800 `x=(4800)/(5)=960 K` |
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| 234. |
A Carnot engine has the same efficiency between `800 K` to `500 K` and `x K to 600 K`. The value of `x` isA. `1000 K`B. `960 K`C. `846 K`D. `754 K` |
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Answer» Correct Answer - B In first case, `(eta_1) = 1-500/800=3/8` and in second case, `(eta_2) = 1-600/x` or, `(600)/x = 1-3/8=5/8 or x = (600xx8)/(5) = 960 K`. |
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| 235. |
An ideal heat engine working between temperature `T_(1)` and `T_(2)` has an efficiency `eta`, the new efficiency if both the source and sink temperature are doubled, will beA. `eta/2`B. `eta`C. `2 eta`D. `3 eta` |
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Answer» Correct Answer - B In first case , `eta_(1) = (T_1-T_2)/(T_1)` in second case `eta_(2) = (2T_(1)-2T_(2))/(2T_1) = (T_1-T_2)/(T_1) = eta` . |
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| 236. |
Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same velocity V. The mass of the gas in A is `m_A,` and that in B is `m_B`. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be `DeltaP and 1.5 DeltaP` respectively. ThenA. `4m_(A)=9m_(B)`B. `2m_(A)=3m_(B)`C. `3m_(A)=2m_(B)`D. `9m_(A)=3m_(B)` |
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Answer» Correct Answer - C Process is isothermal , therefore `T`= constant `(P prop 1/V)` volume is increasing , therefore pressure will decreases. In chamber A: `Delta P = P_(i)-P_(f)=(mu_(A)RT)/(V)-(mu_(A)RT)/(2V)=(mu_(A)RT)/(2V)`...(i) In chamber B: `1.5DeltaP=P_(i)-P_(f)=(mu_(B)RT)/(V)-(mu_(B)RT)/(2V)=(mu_(B)RT)/(2V)`....(ii) From equation (i) and (ii) `(mu_A)/(mu_B) = 1/1.5 = 2/3` `implies (m_(A)//M)/(m_(B)//M) = 2/3implies 3m_(A) = 2m_(B)`. |
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| 237. |
Starting with the same initial conditions, an ideal gas expands from volume `V_1 to V_2` in three different ways. The work done by the gas is W_1 if the process is purely isothermal, `W_2`if purely isobaric and `W_3` if purely adiabatic. Then A. `W_(2)gtW_(1)gtW_(3)`B. `W_(2)gtW_(3)gtW_(1)`C. `W_(1)gtW_2)gtW_(3)`D. `W_(1)gtW_(3)gtW_(2)` |
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Answer» Correct Answer - A |
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| 238. |
P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should corresponds respectively to A. He and` O_(2)`B. `O_(2)`and HeC. He and ArD. `O_(2)` and `B_(2)` |
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Answer» Correct Answer - B |
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| 239. |
A vessel of volume , `V = 5.0` litre contains `1.4 g` of nitrogen at a temperature `T = 1800 K`. Find the pressure of the gas if `30%` of its molecules are dissociated into atoms at this temperature. |
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Answer» Correct Answer - `p=(n+1)mRT//MV=1.9atm` |
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| 240. |
The speed of sound in hydrogen is `1270 ms^(-1)` at temperature T. the speed at the same T in a mixture of oxygen and hydrogen mixed in a volume ratio 1:4 will be ?A. `317"ms"^(-1)`B. `635"ms"^(-1)`C. `830"ms"^(-1)`D. `950"ms"^(-1)` |
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Answer» Correct Answer - B As given , the volumes of hydrogen and oxygen in a mixture is `4:1` , so let V be the volume of oxygen . The volume of hydrogen will be 4 V .If `rho_(m)` be the density of mixture, then `rho_(m)=(4Vxx1+Vxx16)/(5V)=4` `As " " vprop((1)/(rho))^(1//2)` `therefore"Velocity in mixture"=(1270)/((4)^(1//2))=635ms^(-1)` |
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| 241. |
The volume of a monatomic ideal gas increases linearly with pressure, as shown in Fig. Calculate (a) increase in internal energy, (b) work done by the gas and (c ) heat supplied to the gas. |
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Answer» We know that for a perfect gas, internal energy is given by `U = (pV)/(gamma - 1)` `implies Delta U = (p_(2)V_(2) - p_(1)V_(1))/(gamma - 1)` `:. Delta U = (8 xx 10^(5) xx 0.5 - 4 xx 10^(5) xx 0.2)/(5//3 - 1) = 4.8 xx 10^(5) J` Now, `Delta W = int_(0.2)^(0.5) PdV` Here, p varies with V along a straight line. Therefore, one way say2 `p = kV + V_(0)` At `p = 4 xx 10^(5), V = 0.2 m^(2)` and `p = 8 xx 10^(5), V = 0.5 m^(3)` `4 xx 10^(5) = k xx 0.2 + V_(0)` and `8 xx 10^(5) = k xx 0.5 + V_(0)` Here, `k = (4)/(3) xx 10^(6)` and `V_(0) = (4)/(3) xx 10^(5)` `p = (4)/(3) xx 10^(5) (10 V + 1)` `Delta W = int_(0.2)^(0.5) (4)/(3) xx 10^(5) ((V^(2))/(2))_(0.2)^(0.5) + (4)/(3) xx 10^(5) (V)_(0.2)^(0.5)` `(4)/(3) xx 10^(6) xx (0.21)/(2) + (4)/(3) xx 10^(5) xx 0.3 = 1.8 xx 10^(5) J` By the first law of thermodynamics `Delta Q = Delta U + Delta W` `:. Delta Q = 4.8 10^(5) + 1.8 xx 10^(5) = 6.6 xx 10^(5) J` |
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| 242. |
Variation of internal energy with density of `1 "mole"` of monatomic gas is depicted in Fig. Corresponding variation of pressure with voluem can be depicted as (assume the curve is rectangular hyperbola) A. B. C. D. |
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Answer» Correct Answer - D d. Sincee `U rho` = constant, `(P)/(rho) = (RT)/(M)` `P =` constant since `rho` is increasing, therefore `V` is decreasing. |
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| 243. |
A helium balloon has its rubber envelope of weight `w_("Rubber")` and its is used to lift a weight of `w_("Load")`. The volume of fully inflated balloon is `V_(0)`. The atmospheric temperature is `T_(0)` throughout and the atmospheric pressure is known to change with height y according to equation `P = P_(0) e^( – ky)` where k is a positive constant and `P_(0)` is atmospheric pres- sure at ground level. The balloon is inflated with sufficient amount of helium so that the net upward force on is `F_(0)` at the ground level. Assume that pressure inside the balloon is always equal to the outside atmospheric pressure. Density of air and helium at ground level is `rho_(a)` and `rho_(He)` respectively. (a) Find the number of moles (n) of the helium in the balloon. For following two questions assume n to be a known quantity. (b) Find the height `(y(0))` at which the balloon is fully inflated. (c) Prove that the balloon will be able to rise to height `y_(0)` if `(nRT_(0)g)/(P_(0))(rho_(a)-rho_(Hg)) gt w_("Rubber"+w_("Load")` |
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Answer» Correct Answer - `(a) n=(P_(0)(F_(0)+w_("Rubber"+w_("Load"))))/(RT_(0)(rho_(a)-rho_(He))g)` (b) `y_(0)=(1)/(k)ln((P_(0)V_(0))/(nRT_(0)))` |
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| 244. |
Let A and B the two gases and given: `(T_(A))/(M_(A))=4.(T_(B))/(M_(B))`, where T is the temperature and M is the molecular mass. If `C_(A)` and `C_(B)` are the rms speed, then the ratio `(C_(A))/(C_(B))` will be equal to ……….. |
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Answer» Correct Answer - B |
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| 245. |
A flask contains `10^(-3)m^(3)` gas. At a temperature, the number of molecules of oxygen at `3.0xx10^(22)`. The mass of an oxygen molecule is `5.3xx10^(-26)` kg and at that temperature the rms velocity of molecules is 400m/s. The pressure in `N//m^(2)` of the gas in the flask is .............. |
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Answer» Correct Answer - A::B::D |
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| 246. |
When volume of system is increased two times and temperature is decreased half of its initial temperature, then pressure becomes …………. |
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Answer» Correct Answer - A::D |
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| 247. |
A metal ball of surface area `200 cm^(2)` and temperature `527^(@)C` is surrounded by a vessel at `27^(@)C` . If the emissivity of the metal is 0.4, then the rate of loss of heat from the ball is `(sigma = 5.67 xx10^(-8)J//m^(2)-s-k^(4))`A. 108 wattsB. 168 wattC. 182 wattD. 192 watt |
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Answer» Correct Answer - C `(dQ)/(dt)=sigmaAT^(4)rho` |
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| 248. |
One mole of a monoatomic ideal gas is mixed with one mole of a diatomic ideal gas . The molar specific heat of the mixture at constant volume isA. `(3//2)R`B. `(5//2)`C. 2 RD. 4 R |
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Answer» Correct Answer - C Molar specific heat of the mixture `C_(v)=(n_(1)C_(v_(1))+n_(2)C_(v_(2)))/(n_(1)+n_(2))` `=(1xx(3)/(2)R+1xx(5)/(2)R)/(1+1)=2R` |
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| 249. |
A body cools in 7 minutes from `60^(@)C` to `40^(@)C`. What will be its temperature after the next 7 minutes? The temperature of the surroundings is `10^(@)C`.A. `32^(@)C`B. `28^(@)C`C. `20^(@)C`D. `25^(@)C` |
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Answer» Correct Answer - B `(d theta)/(dt)=k(theta_(AV)-theta_(0))` `(20)/(7)=k(50-10)` . . (i) `(40-theta)/(7)=k((40+theta)/(2)-10)` . . . (ii) Dividing equation (ii) by (i) `(40-theta)/(20)=(40+theta-20)/(2xx40)` `4(40-theta)=theta+20` `5 theta=160-20` `theta=(140)/(5)=28^(@)C`. |
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| 250. |
The ratio of loss of heat from a metal sphere at `327^(@)C` to that of the same sphere at `527^(@)C`. When placed in the surrounding of temperature `127^(@)C` isA. `(1)/(2)`B. `(13)/(48)`C. `(22)/(48)`D. `(1)/(16)` |
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Answer» Correct Answer - B `(R_(1))/(R_(2))=(T_(1)^(4)-T_(0)^(4))/(T_(2)^(4)-T_(0)^(4))=(500^(4)-400^(4))/(800^(4)-400^(4))=(13)/(48)` |
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