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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.
| 151. |
`P - V` diagram of an ideal gas is as shown in figure. Work done by the gas in process `ABCD` is A. `4 P_(0) V_(0)`B. `2 P_(0) V_(0)`C. `3 P_(0) V_(0)`D. `P_(0) V_(0)` |
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Answer» Correct Answer - C c. `W_(AB) = - P_(0) V_(0)` `W_(BC) = 0` and `W_(CD) = 4 P_(0) V_(0)` `W_(ABCD) = - P_(0) V_(0) + 0 + 4 P_(0) V_(0)` `= 3 P_(0) V_(0)` |
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| 152. |
`P - V` diagram of an ideal gas is as shown in figure. Work done by the gas in process `ABCD` is A. `P_(0)V_(0)`B. `2P_(0)V_(0)`C. `3P_(0)V_(0)`D. `P_(0)V_(0)` |
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Answer» Correct Answer - C `W_(AB)=-P_(0)V_(0)` `W_(BC)=0` and `W_(CD)=4P_(0)V_(0)` `:. W_(ABCD)=-P_(0)V_(0)+4P_(0)V_(0)` `=3P_(0)V_(0)`. |
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| 153. |
An insulated `0.2 m^(3)` tank contains helium at `1200 kPa` and `47^(@)C`. A value is now opened, allowing some helium to escaped. The valve is closed when one-half of the initial mass has escped. The temperature of the gas is `(root(3)(4) =16)`A. `100 K`B. `200 K`C. `73^(@)C`D. `- 73^(@)C` |
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Answer» Correct Answer - B::D Initial mass of air, `m=(P_(1)V)/(RT_(1))` Final mass of air `=(m)/(2)=(P_(2)V)/(RT_(2))` `(P_(1)V)/(2RT_(1))=(P_(2)V)/(RT_(2))` `((T_(2))/(T_(1)))=2((P_(2))/(P_(1)))` As the tank is insulated, the process is adiabatic with `gamma=(5)/(3)` `T_(2)=T_(1)((P_(2))/(P_(1)))^((gamma-1)/(gamma))implies ((T_(2))/(T_(1)))^((5)/(2))=((P_(2))/(P_(1)))` `(T_(2))/(T_(1))=2((T_(2))/(T_(1)))^((5)/(2))impliesT_(2)=(T_(1))/(2)^((2)/(3))` `=(320)/(2^((2)/(3)))=200K=-73^(@)C[` Given `root(3)(4)=1.6]` |
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| 154. |
The emissive power of a sphere of area `0.07m^(2)` is 0.5 k cal/`m^(2)s`. The amount of heat radiated by the spherial surface in 20 second isA. 7 kcalB. 0.7 kcalC. 0.07 kcalD. 20 kcal |
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Answer» Correct Answer - B `E=(Q)/(At)` `therefore Q=EA t=0.5xx0.07xx20=0.7` kcal |
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| 155. |
The temperature of a gas consisting of rigid diatomic moleculoes is T = 300 K. Calculate the angular root-mean square velocity of a rotating molecules if its moment of inertia is ` I = 2.0 xx 10^(-40) kg m^(2)`. |
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Answer» The degrees of freedom of rotatioal motion = 2 `:. lt lt epsilon gt gt_("rot") = 2 xx (1)/(2) kT` (by equipartition principle) `= lt lt 2 xx (1)/(2) I omega^(2) gt gt` `kT = I lt lt omega^(2) gt gt implies omega_(rms) = sqrt((kT)/(I))` Therefore, `omega_(rms) = sqrt((1.38 xx 10^(-23) xx 300)/(2 xx 10^(-40))) = 4.5 xx 10^(9) "rad"//s` |
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| 156. |
One mole of a diatomic gas undergoes a process `P = P_(0)//[1 + (V//V_(0)^(3))]` where `P_(0)` and `V_(0)` are constant. The translational kinetic energy of the gas when `V = V_(0)` is given byA. `5 P_(0) V_(0)//4`B. `3 P_(0) V_(0)//4`C. `3 P_(0) V_(0)//2`D. `5 P_(0) V_(0)//2` |
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Answer» Correct Answer - B b. `P = (P_(0))/(1 + (V//V_(0))^(3))=(P_(0))/(2)` `T = (P_(0) V_(0))/(2R)` Therefore translational kinetic energy is equal to `(3)/(2) RT = (3R)/(2) (P_(0) V_(0))/(2R) = (3P(0) V_(0))/(4)` |
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| 157. |
As adiabatic vessel contains `n_(1) = 3` mole of diatomic gas. Moment of inertia of each molecule is `I = 2.76 xx 10^(-46) kg m^(2)` and root-mean-square angular velocity is `omega_(0) = 5 xx 10^(12) rad//s`. Another adiabatic vessel contains `n_(2) = 5` mole of a monatomic gas at a temperature `470 K`. Assume gases to be ideal, calculate root-mean-square angular velocity of diatomic molecules when the two vessels are connected by a thin tube of negligible volume. Boltzmann constant `k = 1.38 xx 10^(-23) J//:"molecule"`. |
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Answer» We know according to law of equipartition of energy, each gas molecule has `1//2 kT` energy associated with each of its degrees of freedom. As a diatomic gas molecule has two rotational degrees of freedom, final temperature the system is `T_(f) = (f_(1) n_(1) T_(1) + f_(2) n_(2) T_(2))/(f_(1) n_(1) + f_(2) n_(2))` Here for diatomic gas, `f_(1) = 5, n = 3` and `T_(1) = 250 K` For monatomic gas, `f_(2) = 3, n_(2) = 5` and `T_(2) = 470 K` Thus from Eq. (i), `T_(f) = (5 xx 3 xx 250 + 3 xx 5 xx 470)/(5 xx 3 + 3 xx 5) = 360 ` Thus final mixuture of the two gases is at temperature `360 `. If final rms angular velocity of diatomic gas molecules is `omega_(rms f)`, according to law of equipartition of energy, we have `(1)/(2) I omega_(rms f)^(2) = kT` or `omega_(rms f) = sqrt((2 kT)/(I))` `= sqrt((2 xx 1.38 xx 10^(-23) xx 360)/(2.76 xx 10^(-46)))` `omega_(rms f) = 6 xx 10^(12) rad//s` |
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| 158. |
The indicator diagram for two processes 1 and 2 carrying on an ideal gas is shown in Fig. If `m_(1)` and `m_(2)` be the slopes `(dP//dV)` for process 1 and process 2, respectively, then A. `m_(1) = m_(2)`B. `m_(1) gt m_(2)`C. `m_(1) lt m_(2)`D. `m_(2) C_(V) = m_(1) C_(P)` |
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Answer» Correct Answer - C::D During expansion, an isotherm lies above an adiabat Also `(` Slope of an adiabat `)` ` gamma(` Slope of an isotherm `)` `implies m_(2)=(C_(p))/(C_(v))(m_(1))` `implies m _(2) C_(v)= m _(1) C _(p)` Since `gamma -1`, ` implies m _(2) gt m_(1)` |
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| 159. |
Which one of the following is wrong statement.A. During free expansion, temperature of ideal gas does not change.B. During free expansion, temperature of real gas decrease.C. During free expansion of real gas temperature does no change.D. Free expansion is conducted in adiabatic manner. |
| Answer» Correct Answer - C | |
| 160. |
For the indicator diagram given below, select wrong statement? A. Cycle-II is heat engine cycleB. Net work is done on the gas in cycle-IC. Work done is positive for cycle-ID. Workdone is positive for cycle-II |
| Answer» Correct Answer - C | |
| 161. |
Which of the following will extinguish the fire quickly?A. Water at `100^(@)c`B. Steam at `100^(@)C`C. Water at `0^(@)C`D. Ice at `0^(@)C` |
| Answer» Correct Answer - A | |
| 162. |
The door of a running refrigerator inside a room is left open. The correct statement out of the following ones isA. you can cool the room to a certain degreeB. you can cool it to the temperature inside the refrigeratorC. you can ultimately warm the room slightlyD. you can neither cool nor warm the room |
| Answer» Correct Answer - C | |
| 163. |
A sample of a perfect gas occupies a volume V at a pressure P and obsolete temperature T. The mass of each molecules is m, which of the following expression given the number of molecules in the sample?A. `P/(kTV)`B. `mkT`C. `P/(kT)`D. `(Pm)/(kT)` |
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Answer» Correct Answer - D Ideal gas equation `PV = nRT=m/M RT` `(P)/(m//V) = (RT)/(M) implies P/(rho)=(RT)/(M)` `implies rho=(PM)/(rRT) = (Pxx(mN_(A)))/(kN_(A)T) = (Pm)/(kT)`. |
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| 164. |
Two gases A and B having the same temperature T, same pressure P and same volume V are mixed. If the mixture is at the same temperature and occupies a volume V. The pressure of the mixture isA. `P`B. `2 P`C. `4 P`D. `6 p` |
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Answer» Correct Answer - B Total volume is halved. Pressure is doubled. |
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| 165. |
A vessel of water is put in a dry, sealed room of volume `50m^(3)` at a temperature `27^(@)C`. The saturated vapour pressure of water at `27^(@)C` is `40mm` of mercury. How much water will evaporate before the water is in equilibrium with its vapour? (Relative density of mercury `=13.6, g=9.8ms^(-2)` and `=8.3Jmol^(-1)K^(-1)`) |
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Answer» Correct Answer - `1.927kg` |
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| 166. |
In a process, the pressure of an ideal gas is proportional to square of the volume of the gas. If the temperature of the gas increases in this process, then work done by this gasA. is positiveB. is negativeC. is zeroD. may be positive |
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Answer» Correct Answer - B `P alpha 1/(V^(2))rArr P=k/(V^(2)) rArr P V^(2)=k` `rArr PV.V=krArr nRTVrArr TV=k_(1)` Since temperature increases therefore volume decreases |
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| 167. |
Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300K. The piston of A is free to move, while that B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30K, then the rise in temperature of the gas in B isA. `30K`B. `18K`C. `50K`D. `42K` |
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Answer» Correct Answer - D In cylinder `A`, heat is supplied at constant pressure while in cylinder `B` heat is supplied at constant volume. `(DeltaQ)_(A)=nC_(p)(DeltaT)_(A)` and `(DeltaQ)_(B)=nC_(v)(DeltaT)_(B)` Given that `(DeltaQ)_(A)=(DeltaQ)_(B)` `:. (DeltaT)_(B) = (C_p)/(C_v)(DeltaT)_(A) = (1.4)(30)=42K`. |
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| 168. |
If an avarage person jogs, he produces `14.5xx10^(4) cal//min`. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming `1 kg` requites `580xx10^(3)` cal for evaporation) isA. 0.25kgB. 2.25kgC. 0.05kgD. 0.20kg |
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Answer» Correct Answer - A Amount of sweat evaprated/minute `("Sweat produced"//"minute")/("Number of calories required for evaporation"//kg)` `("Amount of heat produced per minute in jogging")/("Latenheat" ("in cal"//kg))` `(14.5xx10^(3))/(580xx10^(3))=(145)/(580)=0.25kg` |
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| 169. |
A gas undergoes change in its state from position `A` to position `B` via three different path as shown in Fig. Select the correct alternatives : A. Change in internal energy in all the three paths is equal.B. In all the three paths heat is absorbed by the gas .C. Heat absorbed `//` released by the gas is maximum in path `(1)`D. Temperature of the gas first increases and then decreases continuously in path `(1)` |
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Answer» Correct Answer - A::B::C Internal energy `(U)` depends only on the initial and final states. Hence. `Delta U` will be same in all the three paths. In all the three paths, work done by the gas is positive and the product `PV` or temperature `T` is increasing . Therefore, internal energy is also increasing. So, from the first law of thermodynamics, heat will be absorbed b the gas. Further, area under `P-V` graph is maximum in path 1 while `DeltaU` is same for all the three paths . Thereforr, heat absorbed by the gas is maximum in path 1. For temperature of the gas, we can see that product `PV` first increases in path 1 but whether it is decreasing or increasing later on we cannot say anything about it unless the exact values are known to us. |
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| 170. |
An ideal gas of mass `m` in a state `A` goes to another state `B` via three different processes as shown in Fig. If `Q_(1), Q_(2)` and `Q_(3)` denote the heat absorbed by the gas along the three paths, then A. `Q_(1) lt Q_(2) lt Q_(3)`B. `Q_(1) lt Q_(2) = Q_(3)`C. `Q_(1) = Q_(2) gt Q_(3)`D. `Q_(1) gt Q_(2) gt Q_(3)` |
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Answer» Correct Answer - A a. Area enclosed by curve 1 lt area enclosed b curve 2 lt Area enclosed by curve 3 `:. Q_(1) lt Q_(2) lt Q_(3)` (As `Delta U` is same for all the curves) |
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| 171. |
An ideal gas goes from State A to state B via three different process as indicate in the `P-V` diagram. If `Q_(2), Q_(3)` indicates the heat absorbed by the gas along the three processes and `DeltaU_(1), DeltaU_(2), DeltaU_(3)` indicates the change in internal energy along the three processes respectively, thenA. `Q_(1)gtQ_(2)gtQ_(3)` and `DeltaU_(1)=DeltaU_(2)=DeltaU_(3)`B. `Q_(3)gtQ_(2)gtQ_(1)` and `DeltaU_(1)=DeltaU_(2)=DeltaU_(3)`C. `Q_(1)=Q_(2)=Q_(3)` and `DeltaU_(1)=DeltaU_(2)=DeltaU_(3)`D. `Q_(3)gtQ_(2)gtQ_(1)` and `DeltaU_(1)gtDeltaU_(2)gtDeltaU_(3)` |
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Answer» Correct Answer - A For all process `1,2,and 3` `DeltaU=U_(B)-U_(A)` is the same `:. DeltaU_(1)=DeltaU_(2)=DeltaU_(3)` `Now, Q = DeltaU+W` Now `W` = work done by the gas `:. Q_(1) gt Q_(2) = Q_(3)`. |
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| 172. |
In a polytropic process an ideal gas (y= 1.40) was compressed from volume `V_1 = 10 litres to v_2 =5 litres`. The pressure increased from `p_1 = 10^(5) Pato p_(2) = 5xx10^(5)` Pa.Determine: (a) the polytropic expoment n, (b) the molar heat capacity of the gas for the process. |
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Answer» Correct Answer - (a)2.32 (b)1.7R In a polytropic process `p V^(n) = k(a constant)` `therefore p^(1)V_(1)^(n) = p_(2)V_(2)^(n) or (V_(1)/(V_(2)))^(n) = p_(2)/p_(1)` or `n = (Inp_(2)//p_(1))/(InV_(1)//V_(2))` Here `n =(In 5)/(In 2) = (1.6094)/(0.6931) = 2.32` In a polytropic process `C=(R)/(gamma-1) =(R)/(n-1) =(R)/(1.4-1)-(R)/(2.32-1) =1.74R` |
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| 173. |
The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process form an initial state `(P_1,V_1,T)` to the final state `(P_2,V_2 , T)` is equal toA. ZeroB. R In TC. `R" ln"(V_(1))/(V_(2))`D. `R" ln"(V_(2))/(V_(1))` |
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Answer» Correct Answer - D The Change in entropy of an ideal gas , `DeltaS=(DeltaQ)/(T)` ….(i) In isothermal process , there is no change in internal energy of gas , I .e `DeltaU=0` `therefore" " DeltaU=DeltaQ-W` `rArr" " 0=DeltaQ-WrArrDeltaQ=W` i.e `DeltaQ` = work done by gas in isothermal process which went through an isothermal process from `(p_(1),V_(1)T)` to `(p_(2),V_(2),T)` `"or" " " Delta Q=muRT"log"_(e)((V_(2))/(V_(1)))` (ii) For 1 mole of an ideal gas , `mu=1` So, from Eqs. (i) and (ii) ,we get `"or" " " DeltaS=R"log"_(e)((V_(2))/(V_(1)))=R "In"((V_(2))/(V_(1)))` |
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| 174. |
When an ideal gas undergoes an isothermal expansion, the pressure of the gas in the enclosure falls. This is due toA. decrease in the change of momentum per collisionB. decrease in the frequency of collisionC. decrease in both the frequency of collision and the change of momentum per collisionD. decrease in neither the frequency of collision nor the change of momentum per collision |
| Answer» Correct Answer - B | |
| 175. |
An ideal gas undergoes isothermal process from some initial state `i` to final state `f`. Choose the correct alternatives.A. dU=0B. dQ=0C. dQ=dUD. dQ=dW |
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Answer» Correct Answer - A::D For an isotehrmal process change in temperature of the system `dT=0rArrT=` constant . We know that for an ideal gas dU=change in internal energy =`nC_(v)dT=0` [where, n is number of moles and `C_(v)` is specific heat capacity at constant volume] From first law of thermodynamics, `dQ=dU+dW` `=0+dWrArrdQ=dQ` |
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| 176. |
A thermodynamic system undergoes cyclic process `ABCDA` as shown in figure. The work done by the system is A. `P_(0) V_(0)`B. `2P_(0) V_(0)`C. `P_(0) V_(0)/2`D. `zero` |
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Answer» Correct Answer - D `W_(BCOB) = -"Area of trian gle" BCO = -(P_(0)V_(0))/(2)` `W_(AODA)=+"Area of thra ngle"AOD =+(P_(0)V_(0))/(2)`. |
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| 177. |
Pressure versus temperature graph of an ideal gas is shown in figure. Density of the gas at point A is `P_(0)`. Density at B will be A. `(3)/(4)rho_(0)`B. `(3)/(2)rho_(0)`C. `(4)/(3)rho_(0)`D. `2rho_(0)` |
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Answer» Correct Answer - B `rho=(PM)/(RT)` or `rho prop(P)/(T)` `((P)/(T))_(A)=(P_(0))/(T_(0))` and `((P)/(T))_(B)=((3)/(2))(P_(0))/(T_(0))` `((P)/(T))_(B)=(3)/(2)((P)/(T))_(A) therefore rho_(B)=(3)/(2)rho_(A)=(3)/(2)rho_(0)`. |
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| 178. |
Suppose `0.5` moles of an ideal gas undergoes an isothermal expansion as energy is added to it as heat Q. graph shows the final volume `V_(f)` versus Q. the temperature of the gas is `(use In 9 = 2 and R = (25)/(3) J//mol-K)` A. `293 K`B. `360 K`C. `386 K`D. `412 K` |
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Answer» Correct Answer - B `Q = W = nRT 1n(V_f)/(V_i)` `=T = (Q)/(nR1nV_f//V_i) = (1500)/(0.5xx25/3xx1n3)` `=(1500)/(0.5xx25//3xx1) = 360K`. |
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| 179. |
A thermodynamic system undergoes cyclic process `ABCDA` as shown in figure. The work done by the system is |
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Answer» `W_(BCOB)=-"Area of triangle" BCO=(P_(0)V_(0))/(2)` `W_(AODA)=+"Area of triangle" AOD=+(P_(0)V_(0))/(2)` `therefore W_("net")=0` |
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| 180. |
Figure shows the volume versus temperature graph for the same mass of a gas (assumed ideal) corresponding to two different pressure `P_(1) and P_(2)`. Then A. `P_(1) gt P_(2)`B. `P_(1) lt P_(2)`C. `P_(1) = P_(2)`D. The information is insufficient |
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Answer» Correct Answer - A Since the graph is a straight line. so, `V = mT` where `m` is the slope, `=(nRT)//P` [from equation of state] or `m prop (1//P)` or, `m_(2) gt m_(1) :. P_(2) lt P_(1)`. |
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| 181. |
Pressure versus temperature graph of an ideal gas as shown in Fig. Corresponding density `(rho)` versus volume `(V)` graph will be A. B. C. D. |
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Answer» Along process AB,CD temperature is constant (isothermal process) i.e. `Pprop(1)/(V), rhoprop(1)/(V)` `rho-V` graph will be a rectangular hyperbola. Along BC and DA,V is constant `rArr rho` is constant |
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| 182. |
In the `P-V` diagram shown in figure `ABC` is a semicircle. The work done in the process `ABC` is |
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Answer» `W_(AB)` is negative (volume is decreasing) and `W_(BC)` is positive (volume is increasing) and since `|W_(BC)|gt|W_(AB)|` `therefore` Net work doen is positive `because` Workdone area between semicircle ltbrjgt `=pi` (pressure radius) (volume radius) `=pi(1 atm) ((1)/(2) "litre")=(pi)/(2)atm` (litre) |
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| 183. |
In the `P-V` diagram shown in figure `ABC` is a semicircle. The work done in the process `ABC` is A. zeroB. `(pi)/(2) atm-It`C. `-(pi)/(2) atm-It`D. `4 atm-It` |
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Answer» Correct Answer - B `W_(AB)` is negative (volume is decreasing ) and `W_(BC)` is positive (volume is increasing ) and since ,`|W_(BC)|gt|W_(AB)|` `:.` Net work done is positive and area between semicircle which is equal to `(pi)/(2)` atm-it. |
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| 184. |
Helium gas is subjected to a polytropic process in which the heat supplied to the gas is four times the work done by it. The molar heat capacity of the gas for the process is: (R is universal gas constant)A. `R//2`B. `R`C. `2//R`D. `3//R` |
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Answer» Correct Answer - C Given `Delta Q = 4 Delta W` But, `DeltaW = DeltaU = DeltaQ` `:. DeltaU = 3DeltaW = 3/4 DeltaQ` `implies nC_(v)DeltaT = 3/4 n CDeltaT or, C = 4/3 C_(v) = 4/3[(3R)/(2)] = 2R`. |
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| 185. |
For one complete cycle of a thermodynamic process gas as shown in the P-V diagram, which of following correct? A. `DeltaE_(int) = 0, Q lt 0`B. `DeltaE_(int) = 0, Q gt 0`C. `DeltaE_(int) gt 0, Q lt 0`D. `DeltaE_(int) lt 0, Q gt 0` |
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Answer» Correct Answer - A `DeltaE_(i nt) =0` for a complete cycle and for given cycle work done is negative, so from law of thermodynamics `Q` will be negative i.e., `Q lt 0`. |
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| 186. |
An isothermal process is aA. slow processB. quick processC. very quick processD. both 1 `&` 2 |
| Answer» Correct Answer - A | |
| 187. |
In which of the following processes all three thermodynamic variables, that is pressure volume and temperature can changeA. IsobaricB. Isothermal processC. IsochoricD. Adiabatic |
| Answer» Correct Answer - D | |
| 188. |
Two sample `A` and `B` of a gas initially at the same pressure and temperature are compressed from volume `V` to `V//2` (A isothermally and `B` adiabatically). The final pressure of `A` isA. A and B will be sameB. A will be more than in BC. A will be less than in BD. A will be double that in B |
| Answer» Correct Answer - C | |
| 189. |
A gas is compressed isothermally . The rms velocity of its moleculesA. increasesB. decreasesC. first increases and then decreasesD. remains the same |
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Answer» Correct Answer - D The root mean square velocity of a gas molecule is given by , `C"rms"=sqrt((3RT)/(M))` Gas is compressed isothermally , so T remains constant and hence, root mean square velocity will bremain same. |
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| 190. |
One mole of `O_(2)` gas having a volume equal to 22.4 litres at `0^(@)C` and 1 atmospheric pressure is compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process isA. 672.5 JB. 1728 JC. `-1728J`D. `-1572.5J` |
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Answer» Correct Answer - D `C_(P)-C_(V)=r` |
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| 191. |
An ideal gas `(C_p / C_v = gamma)` is taken through a process in which the pressure and volume vary as `(p = aV^(b))`. Find the value of b for which the specific heat capacity in the process is zero. |
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Answer» Given that `P = a V^(b)` or `PV^(-b) = a` Comparing with `PV^(r )` = constant, we have `r = -b` We know that `C = C_(v) - (R )/(r - 1)` Here `C = 0 = C_(V) - (R )/(gamma - 1)` `:. O = (R )/(gamma - 1) - (R )/(- b - 1)` or `b = - gamma` |
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| 192. |
In the above question, maximum pressure attainable isA. `(3)/(4) (a^(5//3) R^(2//3) T^(2//3)_(0))2^(1//3)`B. `(3)/(2) (a^(2//3) RT^(2//3)_(0))3^(1//2)`C. `(3)/(2) (a^(1//2) R^(2//3) T^(3//4)_(0))4^(1//3)`D. `(3)/(2) (a^(1//3) RT^(2//3)_(0))2^(1//3)` |
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Answer» Correct Answer - D `P=R[(T_(0))/(V)+aV^(2)] and V=((T_0)/(2a))^(1//3)` `implies P=3/2(a^(1)/(3)RT_(0)^(2/3))2^(1/3)`. |
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| 193. |
If the ratio of the specific heats of steam is 1.33 and R=8312 J/k mole k find the molar heat capacities of steam at constant pressure and constant volume.A. 33.5 kJ/k mole.B. 25.19 kJ/kg KC. 25.19 kJ/K mole.D. 24.12 kj/k mole 16.12kj/k mole |
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Answer» Correct Answer - A `(dU)/(dQ)=(1)/(gamma), (dW)/(dQ)=1-(1)/(gamma)` |
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| 194. |
The molar heat capacity of a perfect gas at constant volume is `C_v`. The gas is subjected to a process `T =T_0e^(AV)`, Where `T_0` and A are constant. Calculate the molar heat capacity of the gas a function of its volume |
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Answer» Correct Answer - `C_v+(R)/(AV)` `dW = PdV` But,` T=T_(0)e^(AV)` `therefore dW =(PdT)/(AT_(0)e^(AV))=(PdT)/(AT)=(RTdT)/(ATV)=(RdT)/(AV)` Also, `dQ =dU + dW `CdT = C_(v)dT+(RdT)/(AV)` therefore `C= C_(v)+(R)/(AV)` |
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| 195. |
One mole of an ideal gas undergoes a process in which `T = T_(0) + aV^(3)`, where `T_(0)` and `a` are positive constants and V is molar volume. The volume for which pressure with be minimum isA. `((T_(0))/(2a))^(1//3)`B. `((T_(0))/(3a))^(1//3)`C. `(a)/(2T_(0))^(2//3)`D. `(a)/(3T_(0))^(2//3)` |
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Answer» Correct Answer - A `T=T_(0)+aV^(3)implies (PV)/(nR) = T_(0)+aV^(3)` `P=nR[(T_0)/(V)+aV^(2)]` For minimum `P, (dP)/(dV)=0` `implies (-T_(0))/(V^2) + a2V=0implies V=((T_0)/(2a))^(1//3`. |
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| 196. |
A black body radiates heat at temperatures `T_(1)` and `T_(2) (T_(2) gt T_(1)` the frequency corresponding to maxium energy isA. less at `T_(1)`B. more at `T_(1)`C. equal in the two casesD. cannot say |
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Answer» Correct Answer - A Acoording to wien , Law `t_(m)prop(l)/(T^(-))rArr(C)/(v)prop(l)/(T)` i.e `vpropT^(-)` Since `T_(1)ltT_(2)^(-)` therefore `V_(1)ltV_(2)` |
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| 197. |
A black body radiates heat at temperatures `T_(1)` and `T_(2) (T_(2) gt T_(1)` the frequency corresponding to maxium energy isA. more at `T_(1)`B. more ar `T_(2)`C. equal for ` T_(1)andT_(2)`D. independent of `T_(1)andT_(2)` |
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Answer» Correct Answer - B Energy radiated by the bodies are `E_(1)=sigmaAT_(1)^(4)and E_(2)=sigmaAT_(2)^(4)` `rArr" " (E_(2))/(E_(1))=((T_(2))/(T_(1)))gt1" " (becauseT_(2)gtT_(1))` `rArr" " E_(2)gtE_(1)" or "hv_(2)gthv_(1)` `"or " v_(2)gtv_(2)` Thus,m frequency corresponding to `T_(2)` is more. |
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| 198. |
At upper atmosphere, an astronaut feelsA. exteremely hotB. slightly hotterC. extreely coolD. slightly cooler |
| Answer» Correct Answer - D | |
| 199. |
Assume that the temperature remains essentially constant in the upper parts of the atmosphere. The atmospheric pressure varies with height as. (the mean molecular weight of air is M, where `P_(0)=` atmospheric pressure at ground reference)A. `P_()e^((-3Mgh)/(2RT))`B. `P_(0)e^((-Mgh)/(2RT))`C. `P_(0)e^((-3Mgh)/(RT))`D. `P_(0)e^((Mgh)/(RT))` |
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Answer» Correct Answer - D Since pressure decreased with height `dp=pgdh` Consider a small volume `deltaV` of air of mass `Deltam` `P Delta V=((Deltam)/(M))RT` `P=(Deltam)/(Deltav)(RT)/(M)rArrP=(rhoRT)/(M)rArrrho=(PM)/(RT)` `therefore dp=(PM)/(RT)g dh` `int_(P_(0))^(P)(dP)/(P)=-(Mg)/(RT)int_(0)^(h)dh` In `((P)/(P_(0)))=-(-Mgh)/(RT)` `P=P_(0)e^((-Mgh)/(RT))` |
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| 200. |
Assume that the temperature remains essentially constant in the upper part of the atmosphere. Obrain an epression for the variation in pressure in the upper atmosphere with height, the mean molecular weight of air is M. |
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Answer» Suppose the pressure at height h is p and that at `h+dh is p+dp. Then ` `dp=-rhog dh. …(i)` Now considering any small volume `DeltaV` if air of mass `Deltam, ` `pDeltaV=nRt=(Deltam)/(M)RT` `p=(Deltam)/(DeltaV) (RT)/(M)=(rhoRT)/(M)` `or, =rho=(M)/(RT)p.` Putting in (i), `dp=-(M)/(RT)pg dh`ltbr.`or, `int_(p_(0))^(p)(dp)/(p)=int_(0)^(h)-(M)/(RT)g dh` `or, In(p)/(p_(0))=-(Mgh)/(RT)` where p_(0) is the pressure at h=0.` Thus, `p=p_(0) e^(-(Mgh)/(RT)).` |
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