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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The following graphs shows two isotherms for a fixed mass of an ideal gas. Find the ratio of r.m.s speed of the molecules at tempertaures `T_(1)` and `T_(2)`? |
| Answer» Correct Answer - `1:sqrt(2)` | |
| 2. |
Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram The work done on the gas in taking it from D to A is :A. `-414 R`B. `+414R`C. `-690 R`D. `+690R` |
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Answer» Correct Answer - B `D` to `A` is isothermal process Work done by the gas in `SD` to `A` is `W_(DA) = nRT In (V_(2))/(V_(1)) = nRT In (P_(1))/(P_(2))` `= (2)(R )(300)In (10^(5))/(2xx10^(5)) = (600R)[-In2]` `=- (600R) (0.693) =- 414 R` `W_(DA) = - 414R`, it is work done by the gas So work done on the gas is `+414 R` |
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| 3. |
In figure, `P-V` curve of an ideal gas is given. During the process, the cumulative work done by the gas A. continuously increasesB. continuously decreasesC. first increases then decreasesD. first decreases then increases |
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Answer» Correct Answer - A As volume increases `:. WD` continously increases |
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| 4. |
In a cyclic process shown on the `P -V` diargam the magnitude of the work done is : A. `pi ((P_(2)-P_(1))/(2))^(2)`B. `pi((V_(2)-V_(1))/(2))^(2)`C. `(pi)/(4)(P_(2)-P_(1)) (V_(2)-V_(1))`D. `pi(P_(2)V_(2)-P_(1)V_(1))` |
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Answer» Correct Answer - C `W.D = pi xx` Pressure Radius `xx` volume Radius (area of ellipse) `W = pi ((P_(2)-P_(1))/(2))((V_(2)-V_(1))/(2))` `= (pi)/(4) (P_(2) -P_(1)) (V_(2)-V_(1))` |
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| 5. |
Find the work done by gas going through a cyclic process shown in figure? |
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Answer» Correct Answer - `-100piJ` Work done = area under `P-V` curve =Area of ellipse `=pi xx` Pressure radius `xx` volume radius `=- pi ((30-10))/(2) xx 10^(3) xx ((30-10))/(2) xx 10^(-3) =- 100piJ`. |
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| 6. |
An ideal gas is compressed at constant pressure of `10^(5)Pa` until its volume is halved. If the initial volume of the gas as `3.0 xx 10^(-2)m^(3)`, find the work done on the gas? |
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Answer» Correct Answer - `1500J` `W = PdV [P = constant]` `= 10^(5) (3.0 xx 10^(-2) - 1.5 xx 10^(-2)), 10^(5) xx 1.5 xx 10^(-2)` `W = 1500J` |
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| 7. |
A monoatomic gas is enclosed in a nonconducting cylinder having a piston which can move freely. Suddenly gas is compressed to `1//8` of its initial volume. Find the final pressure and temperature if initial pressure and temperature are `P_(0)` and `T_(0)` respectively. |
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Answer» Since process is adiabatic therefore `P_(0)V^((5)/(3)) = P_("final") ((V)/(8))^(5//3) [gamma = (C_(P))/(C_(V)) =(5R)/(2)l(3R)/(2) =(5)/(3)]` `P_("final") = 32 P_(0)`. Since process is adiabatic therefore `T_(1)V_(1)^(gamma-1) = T_(2)V_(2)^(gamma-1) rArr T_(0) V_(0)^(2//3) =T_("final") ((V_(0))/(8))^(2//3)` `rArr T = 4T_(0)` |
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| 8. |
P-V curve of a diatomic gas is shown in the Fig. Find the total heat given to the gas in the process `A rarr B rarr C` |
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Answer» From first law of thermodynamics `DeltaQ_(ABC) = DeltaU_(ABC) +W_(ABC)` `W_(ABC) = W_(AB) +W_(BC) = 0 +nRT_(B) In (V_(C))/(V_(B)) = nRT_(B) In (2V_(0))/(V_(0))` `= nRT_(B) In 2 = 2P_(0) V_(0) In2` `DeltaU = nC_(v)DeltaT = (5)/(2) (2P_(0)V_(0) - P_(0)V_(0)) rArr DeltaQ_(ABC) = (5)/(2) P_(0)V_(0) +2P_(0)V_(0) In1`. |
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| 9. |
An ideal gas is taken through the process `ABC` as shown in figure. If the internal energy of the substance decreases by `10000J` and a heat of `7159` cal is released by the system, calculate the value of meachanical equivalent of heat `(J)`. |
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Answer» Correct Answer - `(30000)/(7159) = 4.19 J//cal` `DeltaW_(ABC) = DeltaW_(AB) +DeltaW_(BC)` `DeltaW_(ABC) = 0 +500 xx 10^(3) xx 0.04 = 20000J` `DeltaQ_(ABC) = DeltaW_(ABC) +DeltaU_(ABC)` `DeltaQ_(ABC) = 20000 +10000 = 30000J` `J = (DeltaQ(J))/(Delta Q(cal)) = (30000J)/(7159cal) = 4.19J//cal`. |
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| 10. |
Why can a ship not use the internal energy of sea water to operate its engine? |
| Answer» Correct Answer - For using the internal energy of sea water to operate the engine of a ship, the internal energy of the sea water has to be converted into meachanical energy. Since whole of the internal energy cannot be converted into mechanical energy, a part has to be rejected to a colder body (sink). since no such body is available, the internal energy of sea water cannot be used to operate the engine of teh ship. | |
| 11. |
Pick the correct statement (s) :A. The rms translational speed for all ideal-gas molecules at the same temperature is not the same but it depends on the mass.B. Each particle in a gas has average translational kinetic energy and the equation `(1)/(2)mv_(max)^(2) = (3)/(2)kT` establishes the relatiship between the average translational kinetic energy per particle and temperature of an ideal gas. It can be concluded that single particle has a temperatureC. Temperature of an ideal gas is doubled from `100^(@)C to 200^(@)C`. The average kinetic energy of each particle is also doubled.D. It is possible for both the pressure and volume of a monoatomic ideal gas to change simultaneously without causing the internal energy of the gas to change. |
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Answer» Correct Answer - A::D `V_(r.m.s.) = sqrt((3kT)/(m))` Since `PV = nRt` therefore `P` and `V` both can change simultaneously keeping temperature constant. |
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| 12. |
A system undergoes a cyclic process in which it absorbs `Q_(1)` heat and gives out `Q_(2)` heat. The efficiency of the process is `eta` and work done is `W.` Select correct statement:A. `W = Q_(1) - Q_(2)`B. `eta =(W)/(Q_(1))`C. `eta = (Q_(2))/(Q_(1))`D. `eta =1 -(Q_(2))/(Q_(1))` |
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Answer» Correct Answer - A::B::D `eta = (0//P)/(1//P) = (W)/(Q_(1)) = (Q_(1)-Q_(2))/(Q_(1)) eta = 1 -(Q_(2))/(Q_(1))`. |
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| 13. |
Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume `u=U/V propT^4` and pressure `P=1/3(U/V)`. If the shell now undergoes an adiabatic expansion the relation between T and R is :A. `T prop e^(-R)`B. `T prop e^(-3R)`C. `T prop (1)/(R )`D. `T prop (1)/(R^(3))` |
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Answer» Correct Answer - C `p = (1)/(3)(U)/(V)` `(nRT)/(V) prop (1)/(3)T^(4)` `VT^(3) = const` `(4)/(3)piR^(3)T^(3) = const` `T R = const` `T prop (1)/(R )` |
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| 14. |
Two samples `1` and `2` are initially kept in the same state. The sample `1` is expanded through an isothermal process where as sample `2` through an adiabatic process upto the same final volume. The final temperature in process `1` and `2` are `T_(1)` and `T_(2)` respectively, thenA. `T_(1) gt T_(2)`B. `T_(1) = T_(2)`C. `T_(1) lt T_(2)`D. The relation between `T_(1)` and `T_(2)` cannot be deduced. |
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Answer» Correct Answer - A For adiabatic `TV^(gamma -1) (gamma gt 1) ……….(i)` For isothermla `T =` const ……..(ii) From (i) and (ii) (i), `T_(2) lt T_(1)` |
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| 15. |
Define an isothermal process and state two essential conditions for such a process to take place. |
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Answer» Correct Answer - Isothermal process: An isothermal process is one, in which the pressure and volume of the system change but temperature remains constant. Conditions: For an isothermal process to take place (i) the cylinder should have conducting walls and (ii) the gas should be compessed or allowed to expand very slowly. |
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| 16. |
What are the conditions for thermodynamic equilibrium? |
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Answer» Correct Answer - 1. Temperature of every part of the system must be the same. 2. There should be no net unbalanced force on a part or whole of the syetm. 3. There should be no changes due to chamical reactions. |
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| 17. |
Define and adiabatic process and state two essential conditios for such a process to take place. |
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Answer» Correct Answer - Adiabatic process: An adiabatic process is one in which pressure volume and temperature of the system change but there is no exchange of heat between the system and the surroundings. Conditions. For an adiabatic process to take place (i) the cylinder should have insulating walls and (ii) the gas should be compressed or alolowed to expand very quickly. |
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| 18. |
Given two examples of reversible processes. Discuss their reversibility. |
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Answer» Correct Answer - 1. Meting of a solid and vaporisation of a liquid are reversible processes. On reversing the conditions under which these processes occur, the liquid can be converted black into solid and vapour into liquid. 2. All isothermal and adiabatic processes, in which no loss of heat occurs due to conduction, convection or radiation can be retraced reversing the boundary conditions and hence these are reversibly processes. |
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| 19. |
A vessel of volume `V` is evacuated by means of a piston air pump. One piston stroke captures the volume `Delta V`. How many strokes are needed to reduce the pressure in the vessel `eta` times ? The process is assumed to be isothermal, and the gas ideal. |
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Answer» Correct Answer - `n = (In eta)/(In(1+DeltaV//V))` `PV = constant` `PV = P_(1) (V +DeltaV)` `P_(1) = (PV)/(V +DeltaV) ..(1)` `P_(1)V = P_(2)(V +DeltaV)` `P_(2) = (P_(1)V)/(V +DeltaV) ..(2)` from (1) and (2) `P_(2) = P [(V)/(V +DeltaV)]^(2)` similarly `P_(n) = P [(V)/(V + DeltaV)]^(n)` according to problem `(P_(n))/(P) = (1)/(eta)` `(1)/(eta) = ((V)/(V + DeltaV))^(n)` `- ln(eta) = nln ((V)/(V +DeltaV))` `(ln eta)/(n = ln(1+(DeltaV)/(V)))` |
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| 20. |
Find the pressure of air in a vessel being evacuated as a function of evacuation time `t`. The vessel volume is `V`, the initial pressure is `p_0`. The process is assumed to be isothermal, and the evacuation rate equal to `C` and independent of pressure. The evatuation rate is the gas volume being evacuated per unit time, with that volume being measured under the gas pressure attained by that moment. |
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Answer» Correct Answer - `p = p_(0)e^(-Ct//V)` `m = V_(rho)` in small interval dt the increase in volume `dV = C dt` `m = V_(rho) = (V +C dt) (rho +d rho)` `V_(rho) = V_(rho) +Vd_(rho) +rhoC dt` `rho C dt =- V d rho ………..(i)` but `P alpha rho` `P = k rho` `(dP)/(P) = (d rho)/(rho).........(2)` `C dt = - (V d rho)/(rho) =- (VdP)/(P)` [from (1) and (2)] `C int_(0)^(t) dt =- V int_(P_(0))^(P) (dP)/(P)` `ln ((P)/(P_(0))) = (-C)/(V)t` `(P)/(P_(0)) = e^(-(Ct)/(V))` `P = P_(0) e^(-(Ct)/(V))` |
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| 21. |
A Carnot engine works as a refrigerator in between `250K` and `300K`. If it acquires `750` calories form heat source at low temperature, then what is the heat generated at higher temperature (in calories)? |
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Answer» Correct Answer - `900` Calories `eta = (Q_(2))/(Q_(1)-Q_(2)) = (T_(2))/(T_(1)-T_(2)) rArr (750)/(Q_(1) - 750) = (250)/(300-250)` `Q_(1) = 900` Calories |
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| 22. |
The coefficient of performance of a Carnot refrigerator working between `30^(@)C & 0^(@)C` isA. `10`B. `1`C. `9`D. `0` |
| Answer» Correct Answer - C | |
| 23. |
A Carnot engine work as refrigerator in between `0^(@)C` and `27^(@)C`. How much energy is needed to freeze `10 kg` ice at `0^(@)C`/ |
| Answer» Correct Answer - `879kcal` | |
| 24. |
Keeping the number of moles, volume and pressure the same, which of the following are the same for all ideal gas?A. rms speed of a moliculeB. densityC. temperatureD. average of magnitude of momentum. |
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Answer» Correct Answer - C `PV = nRT` `:.` temperature remains same for all ideal gas |
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| 25. |
Suppose a container is evacuated to leave just one molecule of a gas in it. Let `v_(mp)` and `v_(av)` represent the most probable speed and the average speed of the gas, thenA. `v_(mp) gt v_(av)`B. `v_(mp) lt v_(av)`C. `v_(mp) = v_(av)`D. none of these |
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Answer» Correct Answer - C One molecules has some single value of speed which is equal most probabla speed and average speed of the gas `:. V_(mp) = V_(av)`. |
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| 26. |
A given quantity of a ideal gas is at pressure P and absolute temperature T. The isothermal bulk modulus of the gas isA. `(2)/(3)P`B. `P`C. `(3)/(2)P`D. `2P` |
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Answer» Correct Answer - B `B = (VdP)/(dV) = - ((-PdV))/(dV)` (for isothermal process) `B = P` |
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| 27. |
The molecule of a given mas of gas have r.m.s. speed `200 ms^(-1)` at `27^(@)C` and `10^(5)Nm^(-2)` pressure. When the absolute temperature is doubled and the pressure is halved, then find rms speed of the molecules of the same gas. |
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Answer» Correct Answer - `200sqrt(2)ms^(-1)` `V_(rms) = sqrt((3RT)/(m)), (V_(rms_(2)))/(V_(rms_(1))) = sqrt((T_(2))/(T_(1))) = sqrt((2xx300)/(300))` `:. V_(rms2) = V_(rms1) sqrt(2) = 200 sqrt(2)` Note: There is no effect of change in pressure on r.m.s. velocity. This is because density also varies in this case. |
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| 28. |
Four containers are filled with monoatomic ideal gases. For each container , the number of moles , the mass of an individual atom and the rms speed of the atoms are expressed in terms of n , m and `v_("rms")` respectively . If `T_(A) , T_(B) , T_(C)` and `T_(D)` are their temperatures respectively then which one of the options correctly represents the order ? A. `T_(B) = T_(C) gt T_(A) gt T_(D)`B. `T_(D) gt T_(A) gt T_(C) gt T_(B)`C. `T_(D) gt T_(A) =T_(B) gt T_(C)`D. `T_(B) gt T_(C) gt T_(A) gt T_(D)` |
| Answer» Correct Answer - C | |
| 29. |
The avergae speed of nitrogen molecules in a gas is `v`. If the temperature is doubled and the `N_(2)` molecule dissociate into nitrogen atoms, then the average speed will beA. `v`B. `vsqrt(2)`C. `2v`D. `4v` |
| Answer» Correct Answer - C | |
| 30. |
An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are `Q_1=5960J, Q_2=-5585J, Q_3=-2980J and Q_4=3645J`, respectively. The corresponding quantities of work involved are `W_1=2200J, W_2=-825J, W_3=-1100J and W_4` respectively. (1) Find the value of `W_4`. (2) What is the efficiency of the cycle |
| Answer» Correct Answer - (i) `765 J` (ii) `(208)/(1921)` | |
| 31. |
Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram The net work done on the gas in the cycle ABCDA is :A. ZeroB. `276R`C. `1076R`D. `1904R` |
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Answer» Correct Answer - B `W_(ABCDA) = W_(AB) +W_(BC) +W_(CD) +W_(DA)` `= nR (DeltaT)_(AB) +nR(T_(B)) In (P_(B))/(P_(C))` `+nR(DeltaT)_(CD) +nR(T_(D)) In (P_(D))/(P_(A))` `= nR (200) +500 nR In 2 +nR (-200)` `+300 nR In (1)/(2)` `= 2 In 2[500R - 300R]` `= (400R)(In2) - (400R) (0.693) = 276R` `W_(ABCDA) = 276R` |
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| 32. |
Ideal monoatomic gas is taken through a process `dQ = 2dU`. Find the molar heat capacity (in terms of `R)` for the process? (where `dQ` is heat supplied and `dU` is change in internla energy) |
| Answer» Correct Answer - `3R` | |
| 33. |
In a certain gas, the ratio of the velocity of sound and root mean square velocity is `sqrt(5//9)`. The molar heat capacity of the gas in a process given by `PT = constant` is. (Take `R = 2 cal//mol K`). Treat the gas as ideal.A. `(R )/(2)`B. `(3R)/(2)`C. `(5R)/(2)`D. `(7R)/(2)` |
| Answer» Correct Answer - D | |
| 34. |
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. `1800J//"cycle"`B. `1000J//"cycle"`C. `2000 J//"cycle"`D. `1600 J//"cycle"` |
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Answer» Correct Answer - D `eta = (T_(1)-T_(2))/(T_(1)) - (W)/(Q) rArr Q = ((T_(1))/(T_(1)-T_(2)))W` `= (600)/((600-300)) xx 800 = 1600 J` |
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| 35. |
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» (i) For process `AC, P alpha V` `(2P_(0))/(V_(c)) = (P_(0))/(V_(0)) rArr V_(c) = 2V_(0)` (ii) Since process `AB` is isochoric hence `(P_(A))/(T_(A)) = (P_(B))/(T_(B)) rArr T_(B) = 2T_(0)` since process `BC` is isobaric therefore `(T_(B))/(V_(B)) = (T_(C))/(V_(C))` `rArr T_(C) = 2T_(B) = 4T_(0)` (iii) Since process is cyclic therefore `DeltaQ = W =` area under the cycle `= (1)/(2) P_(0)V_(0)`. (iv) Since `DeltaU` and `DeltaW` both are negative in process `CA` `:. DeltaQ` is negative in process `CA` and heat is rejected in process `CA` `DeltaQ_(CA) = W_(CA) +DeltaU_(CA)` `= -(1)/(2) [P_(0) +2P_(0)] V_(0) - (5)/(2) nR (T_(C)-T_(A))` `= - (1)/(2) [P_(0) +2P_(0)] V_(0) - (5)/(2) nR ((4P_(0)V_(0))/(nR)-(P_(0)V_(0))/(nR))` `=- 9 P_(0)V_(0)` (Heat rejected) (v) `eta =` efficiency of the cycle `= ("work done by the gas")/("heat injected") rArr eta = (P_(0)V_(0)//2)/(Q_("injected")) xx 100` `DeltaQ_(inj) = DeltaQ_(AB) +DeltaQ_(BC)` `= [(5)/(2)nR (2T_(0)-T_(0))] +[(5)/(2)nR(2T_(0))+2P_(0)(2V_(0)-V_(0))] = (19)/(2)P_(0)V_(0)`. `eta = (100)/(19) %` |
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| 36. |
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. |
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Answer» Since path `AB` and `Cd` are isochoric therefore work done during `AB` and `Cd` is zero. Path `BC` and `DA` are isobaric. Hence `W_(BC) = nRT DeltaT = nRT (T_(3)-T_(2))` `W_(DA) = nRT (T_(1)-T_(4))` Total work done `= W_(BC) +W_(DA) = nRT (T_(1) +T_(3)-T_(4)-T_(2))` |
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| 37. |
Which of the following is incorrect regarding the first law of thermodynamics?A. It is not applicable to any cycle processB. It is a restatement of the principle of conservation of energyC. It introduces the concept of the internal energyD. It introduces the concept of the entropy |
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Answer» Correct Answer - A::D Statement (a) and (d) ae wrong . Concept of entropy is associated with second law of thermodynamics. |
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| 38. |
A refrigerator transfers heat from the cold coling coils to the warm surroundings. Is it against the second law of thermodynamics? Justify your answer? |
| Answer» Correct Answer - According to the second law of thermodynamics, the heat energy by itself flow form a body at low temperature to a body at high temperature. However, it can flow, if some external agent performs work to do so. In a refrigerator, the external work performed by its compressor transfers heat from the cold cooling coils to the warm surroundings and therefore it is not aganist the second law of thermocynamics. | |
| 39. |
The given curve represents the variation of temperatue as a function of volume for one mole of an ideal gas. Which of the following curves best represents the variation of pressure as a function of volume ? .A. B. C. D. |
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Answer» Correct Answer - A Here, `T = V tan 45^(@) +T_(0) rArr T = V +T_(0)` And, `P = (nRT)/(V)`, `:. (PV)/(R ) = V +T_(0)` (Since, `n = 1)` `:. (P -R)V = RT_(0)` Therefore, Graph will be rectangular hyperbola. |
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| 40. |
A piece of leads is hammered. Does its internal energy increase? Does the heat enter the lead from outside ? |
| Answer» Correct Answer - The work done during hammering gets converted into heat energy. Due to this its internal energy increases. | |
| 41. |
In Fig., a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. the lower compartment of the container is filled with 2 moles of an ideal monoatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. the heat capacities per mole of an ideal monoatomic gas are `C_(upsilon) = (3)/(2) R and C_(P) = (5)/(2) R`, and those for an ideal diatomic gas are `C_(upsilone) = (5)/(2) R and C_(P) = (7)/(2) R.` Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. the total work done by the gases till the time they achieve equilibrium will be A. `250R`B. `200R`C. `100R`D. `-100R` |
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Answer» Correct Answer - D `DeltaW_(1) +DeltaU_(1) = DeltaQ_(1)` `Deltaw_(2) +DeltaU_(2) = DeltaQ_(2)` `Deltaq_(1) +DeltaQ_(2) = 0` `(7)/(2) R (T - 400) = (5)/(2)R (700-T)` `rArr = (6300)/(12) = 525 K` So `DeltaW_(1) +DeltaW_(2) = 2.R (525 -400) +2R (525-700)` `=+ 250R - 350 R =- 100R` |
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| 42. |
In Fig., a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. the lower compartment of the container is filled with 2 moles of an ideal monoatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. the heat capacities per mole of an ideal monoatomic gas are `C_(upsilon) = (3)/(2) R and C_(P) = (5)/(2) R`, and those for an ideal diatomic gas are `C_(upsilone) = (5)/(2) R and C_(P) = (7)/(2) R.` Consider the partition to be rigidly fixed so that it does not move. when equilibrium is achieved, the final temperature of the gases will be A. `550K`B. `525K`C. `513 K`D. `490K` |
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Answer» Correct Answer - D Let final temperature of gases is `T` Heat rejected by gas in lower compartment `(nC_(v)DeltaT) = 2 xx (3)/(2)R (700 -T)` Heat received by gas in above compartment `(nC_(p)DeltaT) = 2xx (7)/(2)R (T - 400)` Equating above `2100 - 3T - 7T - 2800` `rArr T = 490K` |
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| 43. |
To increase the efficiency of a carnot engine, will you prefer to (i) increase the temp. of source by`10K` or (ii) decrease the temp. of sink by `10K`? |
| Answer» Correct Answer - To increase the efficiency of the carnot engine the decreasing of temperature of the sink will be preferred. It is because, decreasing the temperature of sink by `10K` brings about greater increase in the efficiency of the heat engine then that is achieved by increasing the temperature of the source by `10K`. | |
| 44. |
One mole of a gas is subjected to two process AB and BC, one after the other as shown in the figure. BC is represented by `PV^(n) = constant`. We can conclude that (where T = temperature, W = work done by gas, V = volume and U = internal energy). A. `T_(A)= T_(B) = T_(C)`B. `V_(A) lt V_(B), P_(B)lt P_(C)`C. `W_(AB) lt W_(BC)`D. `U_(A) lt U_(B)` |
| Answer» Correct Answer - D | |
| 45. |
Is it possible to increase the temperature of a gas without giving it heat? OR Name the process, in which no heat is transfereed to or form a system but the temperature of the system changes. |
| Answer» Correct Answer - It happens during an adiabatic change. | |
| 46. |
A polytropic process for an ideal gas is represented by equation `PV^(n) = constant`. If g is ratio of specific heats `((C_(p))/(C_(v)))`, then value of n for which molar heat capacity of the process is negative is given asA. `gamma gt n`B. `gamma gt n gt 1`C. `n gt gamma`D. none, as it is not possible |
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Answer» Correct Answer - B `C = (R )/(gamma-1) - (R )/(n-1) = (R(n-gamma))/((n-1)(gamma-1))` `C` is negative if `gamma gt n gt 1`. |
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| 47. |
If `gamma` be the ratio of specific heats `(C_(p) & C_(v))` for a perfect gas. Find the number of degrees of freedom of a molecules of the gas? |
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Answer» Correct Answer - `(2)/(gamma-1)` `gamma = (C_(P))/(C_(V)) = (((f+2)/(2))R)/((f)/(2)R)` `= (f+2)/(f) =1+(2)/(f), f = (2)/(gamma-1)` |
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| 48. |
Graph shows a hypothetical speed distribution for a sample of `N` gas particle (for `V gt V_(0), (dN)/(dV) = 0)` A. The value of `V_(0)` is `2N`B. The ratio `V_(avg)//V_(0)` is equal to `2//3`C. The ratio `V_(rms)//V_(0)` is equal to `1//sqrt(2)`D. Three fourth of the total particle has a speed between `0.5 V_(0)` and `V_(0)`. |
| Answer» Correct Answer - A::B::C::D | |
| 49. |
An ideal gas has an adiabatic exponent `gamma`. In some process its molar heat capacity varies as `C = alpha//T`,where `alpha` is a constant Find : (a) the work performed by one mole of the gas during its heating from the temperature `T_0` to the temperature `eta` times higher , (b) the equation of the process in the variables `p, V`.A. `a ln eta`B. `(1)/(aln eta)`C. `aln eta -((eta-1)/(gamma-1))RT_(0)`D. `alneta -(gamma-1)RT_(0)` |
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Answer» Correct Answer - C `dW = dQ - dU , dW = nCdT - nC_(v)dT` `W = int CdT - int C_(v)dT , = int (a)/(T)dT - C_(v) DeltaT` ` = a In ((etaT_(0))/(T_(0))) - ((T_(2)-T_(1))R)/(gamma-1)` `W = a In eta (-(eta-1)T_(0)R)/(gamma-1)` |
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| 50. |
(a) Define two specific heats of a gas. Why is `C_(p) gt C_(v)`? (b) Shown that for an ideal gas, `C_(p) = C_(v) +(R )/(J)` |
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Answer» Correct Answer - (a) The specific heat capacity of a substance is defined as the heat supplied per unit mass of the substance per unit rise in the temperature. 1. Specific heat of gas at constant volume. The specific heat capacity of a gas is defined as the heat given per gram of the gas per unit rise in the temperature at constant volume. It is denoted by `C_(v)`. MOlar specific heat a constant volume. The molar heat capacity of a gas is defined as the heat given per mole of teh gas per unit rise in the temperature at constant volume. it is denoted by `C_(v)`. `C_(v) = M C_(v)` Here `M` is the molecular weight of the gas. 2. Specific heat of a gas at constant pressure. It is defined as the amount of heat required to raise the temperature of `1g` of a gas through `1^(@)C` at constant pressure. It is denoted by `C_(p)`. Molar specific heat of a gas at constant pressure. Molar specific heat of a gas at constant pressure. It is defined as the amount of heat required to raise the temperature of `1` mole of a gas through `1^(@)C` at constant pressure. It is denoted by `C_(p)`. Obviously, `C_(p) = M C_(p)` Here `M` is the molecular weight of the gas `C_(p)` is greater than `C_(v)` When a gas is heated both its volume and pressure change. Let us estimate the amount of heat required by the gas to heat it thorugh `1^(@)C`, when either its volume or its pressure is kept constant. (i) When volume is kept constant. The pressure on the gas is increased so that its volume remains constant. As the gas cannot perform work `(V =` constant), the heat supplied will increase only the temperature or the internal energy of the gas. Therefore, in case of `C_(v)`, heat is required only for increasing the temperature of the gas through `1^(@)C`. (ii) When pressure is kept constant. When the gas is heated at constant pressure, if expands alos. Therefore, the heat sullpied at constant pressure will partly increase its tempertaure (internal energy) and partly will be utillised in performing work aganist teh external pressure. Therefore in case of `C_(p)` more heat (as compared to the case of `C_(v))` will be required for increasing the temperature of the gas through `1^(@)C`. Hence the specific heat of a gas at constant pressure is greater then the specific heat at constant volume i.e., `C_(p) kt C_(v)`. The difference between the values of two specific heats is equal to the amount fo heat equivalent to the work performed by the gas during expansion at constant pressure. (b) The relation between two specific heats of a gas can be derived by applying first law of thermodynamics. Consider one mole of an ideal gas. Suppose that the gas is heated at constant volume, so that its temperature increases by `dT`. If `dQ` is the amount of heat supplied, then `dQ = 1 xx C_(v) xx dT` or `dQ = C_(v) dT ..........(i)` As the gas is heated at constant volume it will not performs any external work and in accordance with the first law of thermodynamics. `dQ = dU +0 = dU` Setting `dQ = dU` in the equation (i) we have `dU = C_(v)dT ..........(ii)` Now, heat the gas at constant pressure so as to again increase it temperature by `dT`. If `dQ` is the amount of heat supplied then `dQ = 1 xx C_(v) xx dT` or `dQ = C_(v) xx dT .........(iii)` the hea supplied at constant pressure increases the temperature of teh gas by `dT` i.e., increases its internal energy by `dU` and as well as enables the gas to perform work `dW`. if `dV` is the increase in volume of the gas, the work performed by the gas, `dW = P dV .........(iv)` Accoding to teh first law of thermodynamics, the total heat supplied to the gas to heat it at constant pressure, `dQ = dU +dW` Using the equations (i), (ii) and (iii) we get `C_(p) dT = C_(v) dT +P dV ......(v)` For one mole of a perfect gas, `PV = RT` The heat `dQ` is supplied to the gas at constant pressure `P`. Therefore, defferentiating both sides of the above equation w.r.t. by treating `P` as constant, `(d)/(dT) (PV) =(d)/(dT) (RT) or P(dV)/(dT) = R` or `PdV = R dT ..........(vi)` From the equations (v) and (vi) we have `C_(p)dT = C_(u)dT +R dT` or `C_(p) = C_(v) +R` |
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