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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.
| 51. |
When a gas enclosed in a closed vessel was heated so as to increase its temperature by `5^(@)C`, its pressure was seen to have increased by 1%. The initial temperature of the gas was nearlyA. `500^(@)C`B. `227^(@)C`C. `273^(@)C`D. `150^(@)C` |
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Answer» Correct Answer - B `(DeltaT)/(T)=(DeltaP)/(P)=(1)/(100)` `(5)/(T)=(1)/(100)" "therefore T=500K` |
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| 52. |
A small hole is made in a metallic hollow enclosure whose walls are maintained at a temperature of `10^(3)`K. then the amount of radiant energy emitted per metre square per second from the hole isA. 56.7 JB. 56.7 KJC. `10^(3)J`D. data is in sufficient |
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Answer» Correct Answer - B `E=sigmaT^(4)` `=5.67xx10^(-8)xx(10^(3))^(4)` `=5.67xx10^(-8)xx10^(12)=5.67xx10^(4)` `=56.7xx10^(3)J` `=56.7kJ`. |
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| 53. |
A sphere and a cube of equal volumes both are made of iron and have similar surface. If both are cool in identical surrounding, at a lower temperature, then the ratio of the initial rates of loss of heat isA. `((pi)/(6))^(1//3):1`B. `((3)/(4pi))^(-1//3):1`C. `((3)/(4pi))^(2//3):1`D. `1:1` |
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Answer» Correct Answer - A Volume of sphere=Volume of cube `(4pi)/(3)r^(3)=l^(3)` `(r)/(l)=((3)/(4pi))^(1//3)` . . . (i) `((dQ)/(dt))_(S)=4pir^(2)(T^(4)-T_(0)^(4))` `((dQ)/(dt))_(C)=6l^(2)(T^(4)-T_(O)^(4))` `((dQ//dt)_(S))/((dQ//dt)_(C))=(4pir^(2))/(6l^(2))` Substitute r/l from equation (i), we have, `=(4pi)/(6)xx((3)/(4pi))^(2//3)=((pi)/(6))^(1//3)` |
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| 54. |
Equal volumes of hydrogen and oxygen gasses of atomic weights 1 and 16 respectively are found to exert equal pressure on the walls of two separate containers the ratio of rms speed of hydrogen and oxygen gas isA. `1:4`B. `4:1`C. `1:32`D. `32:1` |
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Answer» Correct Answer - B `C_(1) prop (1)/(sqrt(M_(1))) and C_(2)=(1)/(sqrt(M_(2)))(C_(2))/(C_(1))=sqrt((M_(1))/(M_(2)))=sqrt((1)/(16))=(1)/(4)` `therefore (C_(1))/(C_(2))=(4)/(1)` |
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| 55. |
When two gases having volumes 1 litres and 2 liters are at the same temperature, then the ratio of the average kinetic energy of the molecules in the two gasesA. `1:1`B. `1:2`C. `2:1`D. `3:2` |
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Answer» Correct Answer - A `(KE_(1))/(KE_(2))=((3)/(2)RT)/((3)/(2)RT)=(1)/(1)` |
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| 56. |
A gaseous mixture consists of 16g of helium and 16 g of oxygen. The ratio `(C_p)/(C_v)` of the mixture isA. `1.4`B. `1.54`C. `1.59`D. `1.62` |
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Answer» Correct Answer - D For mixture of gases, `C_v = (n_1C_(v_1)+n_(2)C_(v_2))/(n_(1)+n_(2))` where `C = f/2 R , f` is degree of freedom and `C_p = (n_1C_(P_1)+n_(2)C_(P_2))/(n_(1)+n_(2))` where `C_(p) = (1+f/2)R` For helium, `n_(1) = 4, f=3` For oxygen , `n_(2) = 1/2 , f = 5` `:. (C_p)/(C_v) = (4xx(5R)/(2)+1/2xx(7R)/(2))/(4xx(3R)/2 + 1/2 xx (5R)/2) = (47)/(29)` `=1.62`. |
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| 57. |
The ratio of the densities of the two liquid is 2:3 and the ratio of their specific heats is 3:2. what will be the ratio of their thermal heat capacities, when same volume of both liquids are taken?A. `2:3`B. `3:2`C. `9:4`D. `1:1` |
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Answer» Correct Answer - D `(C_(1))/(C_(1))=(M_(1)C_(1))/(M_(2)C_(2))=(rho_(1))/(rho_2)xx(C_(1))/(C_(2))` `=(2)/(3)xx(3)/(2)=(1)/(1)` |
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| 58. |
The first law of thermodynamics is based on the law of conservation ofA. energyB. massC. momentumD. pressure |
| Answer» Correct Answer - A | |
| 59. |
A block returns to its initial position after dissipating mechanical energy in the from of heat through frictionn. Is this process reversibe or irreversible ? |
| Answer» This process is irreversible because the same amount of mechanical energy cannot be regained from heat. It follows form the second law of thermodynamics which states that all the heat can never be converted to useful work. | |
| 60. |
The average kinetic energy per molecule of all monatomic gases is the same at the same temperature. Is this ture or false ? |
| Answer» True, because average kinetic energy per molecule is `(3//2) kT`. | |
| 61. |
A certain mass of is taken from an initial thermodynamics state `A` to another state `B` by process I and II. In process I for the gas does `5 J` of work and absorbs `4 J` of heat energy. In process II, the gas absorbs `5 J` of heat. The work done by the gas in process II is |
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Answer» Correct Answer - 6 For process I : `dQ = dU + dW` `implies 4 = dU + 5 implies dU = -1` For process II : `5 = dU + dW` `implies = - 1 + dW implies dW + dW` |
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| 62. |
The molar specific heat capacity of all monatomic gases is the same. Is this ture of false ? (Assumed ideal nature.) |
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Answer» Ture. The molar specific heat capacity is given by `C = (R )/(gamma -1)` `R` is a universal constant and `gamma` is the same for all monoatomic gases, and hence the molar specific heat capacity of all monoatomic gases is the same. |
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| 63. |
Figure shows an insulated cylinder of volume `V` containing monatomic gas in both the compartments. The pistone is diathermic. Initially the piston is kept fixed and the system is allowed to acquire a state of thermal equilibrium. The initial pressures and temperatures are as shown in the figure. Calculate The heat that flows from `RHS` to `LHS`, Given `T_(2)gt T_(1)`. Now, the pin which was keeping the piston fixed is removed. and the piston is set free to move. The piston is allowed to slide slowly, such that of mechanical equilibrium is also achieved and `T_(1)P_(2)gt P_(1)T_(2)`. Find.A. `(4)/(3)P_(1)P_(2)V((T_(2)-T_(1)))/(P_(1)T_(2)+P_(2)T_(1))`B. `(3)/(4)P_(1)P_(2)V((T_(2)-T_(1)))/(P_(1)T_(2)+P_(2)T_(1))`C. `(3)/(4)P_(1)P_(2)V((T_(2)+T_(1)))/(P_(1)T_(2)+P_(2)T_(1))`D. `(3)/(4)P_(1)P_(2)V((T_(2)-T_(1)))/(P_(1)T_(2)-P_(2)T_(1))` |
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Answer» Correct Answer - B Heat flowing from right to left `Q=n_(1)C_(V)(T-T_(1))=(P_(1)V)/(2 R T_(1)2)(3)/(2)R{((P_(1)+P_(2))T_(1)T_(2))/(P_(1)T_(2)+P_(2)T_(1))-T_(1)}` `=(3)/(4)P_(1)P_(2)V((T_(2)-T_(1)))/(P_(1)T_(2)+P_(2)T_(1))` |
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| 64. |
A thermos flask contains coffee. It is vigorously sheken, considering the coffee as the system. (a) Does its temperature rise ? (b) Has heat been added to it ? (c ) Has work been done on it ?A. decreasesB. increasesC. remains sameD. decreases if temperature is below `4^(@)C` and increses if temperature is equal to or more than `4^(@)C` |
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Answer» Correct Answer - B We know that there is no loss of heat in an ideal flask. Therefore mechanical energy wasted in shaking is changed into heat and hence, temperature rises. |
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| 65. |
Can one distinguish between the internal energy of a body acquired by het transfer and that acquired by the performance of work on it by an etenal agent ? |
| Answer» No. because the only external manifestation is a rise in temperature in both cases. Always keep in mind that in ternall energy is a state function. Any change in internal energy depends only on initial and final state , not on the path followed. | |
| 66. |
The two conducting cyliner-piston systems shows below are linked. Cyliner 1 is filled with a certain molar quantity of a monatomic ideal gas, and cylinder 2 is filled with an equal molar quantity of a diatomic ideal gas. The entire apparatus is situated inside an oven whose temperature is `T_(a) = 27^(@)C`. The cylinder volumes have the same initial value `V_(0) = 100 cc`. When the oven temperature is slowly raised to `T_(b) = 127^(@)C`. What is the volume change `Delta V` (in cc) of cylinder 1 ? |
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Answer» `P_(0) V_(0) = nRT` `(P_(0) + Delta P) (V_(0) + Delta V_(0)) = nRT_(b)` `(P_(0) + Delta P) (V_(0) - Delta V) = nRT_(b)` `Delta V = 0` Pressure depends on number of moles and is independent of nature of gas. |
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| 67. |
A theoms flask contains coffee. It is vigorously sheken, considering the coffee as the system. (a) Does its temperature rise ? (b) Has heat been added to it ? (c ) Has work been done on it ? |
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Answer» a. Yes, the temperature of the coffee increases. b. No, heat has not been added to the coffee as it is thermally insulated. c. Yes, wor is done on it by the man who shakes the flask. |
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| 68. |
A flask is filled with `13 g` of an ideal gas at `27^(@)C` and its temperature is raised to `52^(@)C`. The mass of the gas that has to be released to maintain the temperature of the gas in the flask at `52^(@)C`, the pressure remaining the same isA. `2.5 g`B. `2.0 g`C. `1.5 g`D. `1.0 g` |
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Answer» Correct Answer - D d. `PV prop` Mass of gas `xx` Temperature In this problem pressure and volume remain constant, so `M_(1) T_(1) = M_(2) T_(2) =` constant `:. (M_(2))/(M_(1)) = (T_(1))/(T_(2)) = ((27 + 273))/((52 + 273)) = (300)/(325) = (12)/(13)` `implies M_(2) = M_(1) xx (12)/(13) = 13 xx (12)/(13) g = 12 g` i.e., the mass of gas released from the flask `= 13 g - 12 g = 1g`. |
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| 69. |
The temperature of 3 kg of nitrogen is raised form `10^(@)C` to `100^(@)C`, Compute the heat added, the work done, and the change in internal energy if (a) this is done at constant volume and (b) if the heating is at constant pressure. For nitrogen `C_(p) = 1400 J kg^(-1) K^(-1)` and `C_(v) = 740 J kg^(-1) K^(-1)`. |
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Answer» At constant volume `Delta W = 0`. Now `Delta U` is always given by `Delta U = mC_(v) Delta T` `= 3 xx 740 xx (100 - 10) = 199800 J` According to the first law of thermodynamics `Delta Q = Delta U + Delta W` `Delta Q = Delta U = 199800 J` b. At constant pressure `Delta Q = mC_(P) Delta T = 3 xx 1040 xx (100 - 10) = 280800 J` `Delta U` is always given by `Delta U = ,mC_(v) Delta T = 199800 J` According to the first law of thermodynamics `Delta Q = Delta U + Delta W` or `Delta W = Delta Q - Delta U = 280800 - 199800 = 81000 J` |
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| 70. |
Explain why the temperature of a gas drops in an adiabatic expansion, on basis of the kinetic theory of gases, |
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Answer» In an adiabaticf expansion `Delta Q = 0`. But `Delta Q = Delta U = + Delta W` by the first law of thermodynamics. Therefore, in an adiabatic expanison, `Delta W = - Delta U` Thus, work down by a gas in an adiabatic expanison is associated with a decrease of internal energy. So its temperature decreasese. |
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| 71. |
In what process is the heat added entirely converted into internal energy of the system ? |
| Answer» In an isochoric process where `Delta W = 0` and, therefore, `Delta Q = Delta U` by the first of thermodynamics. | |
| 72. |
If at same temperature and pressure, the densities for two diatomic gases are respectively `d_(1) and d_(2)` , then the ratio of velocities of sound in these gases will beA. `sqrt((rho_(2))/(rho_(1)))`B. `sqrt((rho_(1))/(rho_(2)))`C. `rho_(1)rho_(2)`D. `sqrt(rho_(1)rho_(2))` |
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Answer» Correct Answer - A `P_(1)=P_(2)` `(1)/(3)rho_(1)C_(1)^(2)=(1)/(3) rho_(2)C_(2)^(2)` `((C_(1))/(C_(2)))^(2)=(rho_(2))/(rho_(1))" "therefore (C_(1))/(C_(2))=sqrt((rho_(2))/(rho_(1)))` |
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| 73. |
The luminousity of a star is 10000 times that of the sun. if the surface temperature of the sun is 6000 K, then the surface temperature of the star isA. 8446 KB. 84860 KC. 848600 KD. 60,000 K |
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Answer» Correct Answer - D `P_(1)=I_(1)A_(1) and P_(2)=I_(2)A_(2)` But `P_(1)=sigmaA_(1)T_(1)^(4) and P_(2)=sigmaA_(1)T_(1)^(4)` `therefore sigmaT_(1)^(4)A_(1)=I_(1)A_(1)` `sigmaT_(2)^(4)A_(2)=I_(2)A_(2)` `((T_(2))/(T_(1)))^(4)=(I_(2))/(I_(1))` `(T_(2))/(T_(1))=((I_(2))/(I_(1)))^(1//4)=(10^(4))^(1//4)=10` `T_(2)=T_(1)xx10=6000K` |
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| 74. |
Two stars emit maximum radiation at wavelength 3600 Å and 4800 Å respectively. The ratio of their temperatures isA. `1:2`B. `3:4`C. `4:3`D. `2:1` |
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Answer» Correct Answer - C `(T_(1))/(T_(2))=(lamda_(m_(2)))/(lamda_(m_(1)))=(4800)/(3600)=(48)/(36)=(4)/(3)` |
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| 75. |
The filament of an evacuated light bulb has a length 10 cm, diameter 0.2 mm and emissivity 0.2. then the power it radiates at `1727^(@)C` is [`sigma=5.67xx10^(-8)` SI units]A. 11.4 WB. 1140 WC. 114 WD. `1.4xx10^(5)W` |
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Answer» Correct Answer - A `P=sigmaArT^(4)` |
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| 76. |
Vapour is injected at a uniform rate in a closed vessel which was initially evacuated. The pressure in the vesselA. (a)increases continuouslyB. (b)decreases continuouslyC. (c)first increases and then decreasesD. (d)first increases and then becomes constant. |
| Answer» Correct Answer - D | |
| 77. |
A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300K. The ratio of the average rorational kinetic energy per `O_2` molecules to that per `N_2` molecules isA. `1 : 1`B. `1 : 2`C. `2 : 1`D. depends on the momets of inertia of the two molecules |
| Answer» Correct Answer - A | |
| 78. |
A lamp of voume 50 cc was sealed off during manufacture at a pressure `0.1` newton per square metre at `27^(@)C`. Calculate the mass of the gas enclosed in the lamp. Molecular weight of the gas `=10` and `R=8.3Jmol^(-1)K^(-1)` |
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Answer» Correct Answer - `2xx10^(-11)kg` |
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| 79. |
Consider a sample fo oxygen at 300K. Find the average time taken by a molecule to travel a distance equal to the diameter of the earth. |
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Answer» `V_avg = (sqrt (8RT))/(sqrt(piM)) = (sqrt(8 xx 8.83 xx 300))/(sqrt(3.14 xx 0.032))` ` = 445.25 m/s` ` T = (distance/ speed) =(6400000 xx 2 sec)/(445.25)` `=28747.83 hrs/3600` ` =7.985 hrs =8 hrs.` |
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| 80. |
A vessel contains 1.60g of oxygen and 2.80g of nitrogen. The temperature is maintained at 300K and the voume of the vessel is `0.166m^(3)`. Find the pressure of the mixture. |
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Answer» (P_(O_2)) = (n_(O_2))RT/ V` ` (P_(N_2)) = ((n_(N_2)RT)/V)` ` (n_(O_2)) = (m/(M_(O_2))) = 1.60/32 ` ` = 0.05` ` (n_(N_2)) = (m/M_(N_2)) = 2.80/ 28 = 0.1` ` Now, (P_mix)= ((N_(O_2)) + (n(N_2))/V) RT` ` P_mix = ((0.05 xx 0.1) xx 8.3 xx 300 / 0.168)` `= 2250 N/(m^2)` |
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| 81. |
A vertical cylinder of height 100cm contains air at a constant temperature. The top is closed by a friction less light piston. The atmospheric pressure is equal to 75 cm of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston. |
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Answer» Given, `P_1 = Atomospheric pressure` ` = 75 xx rho g, ` ` V_1= 100 xx A` ` (P_2) = Atmospheric pressure` + Mercury pressure `= 75 rho g + h rho g ` ` (if h = height of mercury )` ` V_2 = (100 - h)` ` ((P_1)(V_1))= ((P_2)(V_2))` ` rArr 75 rho g (100 A)` ` = ((75 +h) rho g (100 - h) A)` ` rArr (75 xx 100) = ((75+h)(100-h))` ` rArr 7500= 7500 - 75h +100h - (h^2)` ` rArr (h^2) - 100 h + 75h = 0 ` ` h^2 - 25 h = 0 ` ` rArr (h^2) = 25 h ` ` h = 25 cm ` Height of Mercury that can be poured ` = 25 cm ` . |
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| 82. |
The average translational kinetic energy of air molecules is `0.040 eV (1eV=1.6xx10^(-19)J).` Calculate the temperature of the air. Blozmann constant `K=1.38xx10^(-23) J K^(-1).` |
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Answer» Average K.E.= 3/2 `(3/2 KT) = (0.04 xx 1.6 xx 10^-19 )` ` rArr (3/2 xx 1.38 xx 10^-23 xx T)` =0.04 xx 1.6 xx 10^-19` `rArr T= (2 xx 0.04 xx 1.6 xx 10^-19)/(3 xx 1.38 xx 10^-23)` ` =0.0309178 xx 10^4` ` =309.178 =310K` |
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| 83. |
A vessel of volume `V_(0)` contains an ideal gas at pressure `p_(0)`and temperature T. Gas is continuously pumped out of this vessel at a constant volume-rate `dV/dt=r` keeping the temperature constant, The pressure inside the vessel. Find (a) the pressure of the gas as a function of time,(b) the time taken before half the original gas is pumped out. |
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Answer» We have, ` dV/dt = r ` ` rArr dV = rdt ` Let the pressure pumped out gas = dp ` ` Volume of container = `V_0` At a pump dV amount of gas has been pumped out ` PdV = -(V_0) dP ` ` rArr Prdt = - (V_0) dP` ` rArr dp/P = (- rdt / V_0) ` On integration , we get ` P = (e^ (-rt/V_0))` ` Half of the gas been pumped out , pressure will be half ` i.e. 1 = (1/2 e ^ (-rt/ V_0))` ` rArr ` 1 n 2 = rt/ V_0 ` ` rArr t = In 2 xx (V_0/r )` |
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| 84. |
A container of volume 50cc contains air (mean molecular weight =28.8g) and is open to atmosphere where the pressure is 100kPa. The container is kept in a bath containing melting ice`(0^(@)C.(a) Find the mass of the air in the container when thermal equlibrium is reached.(b) the container is mow placed in another bath containing boling water`(100^(@)C)`. Find the mass of air in the container. (C) The container is mow closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached . |
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Answer» Given, `(V= 50 cc. = (50 xx (10^-6)(m^3)))` ` P= 100KPa = (10^5)Pa,` ` M = 28.8 g ` ` (a) PV = n(RT_1) ` ` rArr PV = (m/M)(RT_1)` ` rArr m = (PMV/ RT_1)` ` = ((10^5) xx 28.8 xx 50 xx (10^-6)/ 8.3 xx 273)` ` = 50 xx 28.8 xx (10^1)/ 8.3 xx373)` ` 0.0465 gm ` (c) When the vessel is closed ` P xx 50 xx (10^-6) = (0.0465/ 28.8) xx 8.3 xx 273` rArr P=(0.0465 xx 8.3 xx 273 / 28.8 xx 50 xx (10^-6))` ` = 0.07316 xx (10 ^6)Pa` ` = 73.16 KPa` ` = 73 KPa ` |
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| 85. |
Hydrogen gas is contained ina closed vessel at 1atm(100kPa) and 300K.(a) Calculate the mean speed pf the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of `45^(@)`with it. How many molecules strike each squre metre ifj the wall per second? |
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Answer» `P= 1 atm =10^5 pascal` `T = 300K` ` M= 2g= 2 xx 10^-3kg ` (a) V_avg = (sqrt (8 RT)/ (sqrt(pi M)))` ` = (sqrt (8 xx 8.3 xx 300)/(sqrt(3.14 xx 2 xx (10^-3))))` ` = (sqrt (8 xx 8.3 xx 300 xx (10^3))/ (sqrt(3.14 xx 2)))` `= 1781.004 =1780 m//s` (b) When the molecules strike at an angle `45^@,` force exerted `=mV(cos 45^@)-(-mV(cos 45^@))` ` = 2 m V (cos 45^@)` = `(2mV/(sqrt 2)) = (sqrt 2)mV` Number of molecules striking per unit area `= force/((sqrt 2)mV xx Area)` ` = Pressure/((sqrt 2)m V)` ` (10^5)/(((sqrt 2)xx 2 xx (10^-3) xx 1780)/(6 xx (10^23))` ` =(3 xx (10^31)/2516.92)` ` =1.19 xx (10^-3) xx (10^31)` ` = 1.19 xx (10^28) xx =1.2 xx (10^28)` |
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| 86. |
Oxygen is filled in a closed metal jar of volume `1.0xx10^(-3)m^(3)` at a pressure of `1.5xx10^(5)Pa.` and temperature 400K.The jar has a small leak in it. The atmospheric pressure is `1.0xx10^(5)`Pa and the atmospheric temperature is 300K. Find the mass of the gas that leaks out by time the pressure and the temperature inside the jar equlise with the surrounding. |
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Answer» Given ` (V_1)=(1 xx (10^-3)(m^3))` ` P_1= (1.5 xx (10^5)Pa)` ` T_1= 400K` We know ` (P_1)(V_1)= (n_1)(RT_1)` ` rArr n_1 = ((P_1)(V_1)/RT_1)` ` =(1.5 xx (10^5)xx 1 xx (10^-3)/(8.3 xx 400))` ` rArr (m_1) = (1.5 xx M/8.3 xx 4) = (1.5 xx 32/ 4 xx 8.3)` ` = 1.457 ~~1.446` ` Again, (P_2)= 1 xx (10^5)Pa,` ` V_2 = 1 xx (10^-3)(m^3)` ` T_2 = 300K` We know, ` (P_2)(V_2)= (n_2)RT_2` ` rArr (n_2)= ((P_2)(V_2)/(RT_2)) = (10^5 xx (10^-3)/ (8.3 xx 300))` ` = (1/3 xx 8.3)= 0.040` ` rArr m_2= 0.04 xx 32 = 1.285` ` Delta m = (m_1-(m_2))=1.446 -1.285` ` = 0.1608 g ~~ 0.16g.` |
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| 87. |
At what temperature will be the oxygen molecules have the same root mean square root mean square speed as hydrogen molecules at 300 K ?A. 1600 KB. 2400 KC. 3200 KD. 4800 K |
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Answer» Correct Answer - D Root mean square speed `V(rms)=sqrt((3RT)/(M))rArrTpropM` `(thereforev_("rms"),Rrarr"constant")` `or" " (T_(O_(2)))/(T_(H_(2)))=(M_(O_(2)))/(M_(O_(2)))` `rArr " " (T_(O_(2)))/(300)=(32)/(2)` `rArr "temperature"" "T_(O_(2))=4800K` |
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| 88. |
The temperature at which the mean KE of the molecules of gas is one - third of the mean KE of its molecules at `180^(@)C` isA. `-122^(@)C`B. `-90^(@)C`C. `60^(@)C`D. `-151^(@)C` |
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Answer» Correct Answer - D Kinetic energy of a gas is directly proportional to its temperature. `therefore " " (K_(1))/(K_(2))=(T_(1))/(T_(2))` `rArr " " (K)/(K//3)=(273+180)/(T_(2))` `rArr " " 3=(453)/(T_(2))` `T_(2)=151^(@)C` |
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| 89. |
The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root mean square velocity of the gas molecules is v, at 480 K it becomesA. 4 vB. 2 VC. `(v)/(2)`D. `(v)/(4)` |
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Answer» Correct Answer - B The root mean square volocity is given by `v"rms"=sqrt((3RT)/(M))` `So " "(v_(1))/(v_(2))=sqrt((T_(1))/(T_(2)))` Now , `T_(1)=120K,T_(2)=480K,v_(1)=v` `so, " " (v_(1))/(v_(2))=sqrt((120)/(480))=sqrt((1)/(4))=(1)/(2)` `rArr" " (v)/(v_(2))=(1)/(2)` `rArr" "v_(2)=2v` |
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| 90. |
Suppose ideal gas equation follows `VP^(3) = contant`. Initial temperature and volume of the gas are T and V respectively. If gas expand to `27V` temperature will becomeA. `T`B. `9T`C. `27T`D. `T//9` |
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Answer» Correct Answer - B `VP^(3)` = constant = `k=(k)/(V^(1//3))` Also, `PV = mu RT implies k/(V^(1//3))*V = mu RT implies V^(2//3) = (muRT)/(k)` Hence, `((V_1)/(V_2))^(2//3) = (T_1)/(T_2) implies ((V)/(27V))^(2//3) = T/(T_2) implies T_(2) = 9T`. |
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| 91. |
Suppose ideal gas equation follows `VP^(3) = contant`. Initial temperature and volume of the gas are T and V respectively. If gas expand to `27V` temperature will becomeA. TB. 9TC. 27TD. T/9 |
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Answer» Correct Answer - B |
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| 92. |
At a given volume and temperature the pressure of a gasA. varies inversely as its massB. varies inversely as the square of its massC. varies linearly as its massD. is independent of its mass |
| Answer» Correct Answer - C | |
| 93. |
The root mean square speed of a group of gas moecules, having speeds `v_(1),v_(2),....v_(N)` isA. `(1)/(N)sqrt((V_(1)+V_(2)+................+V_(N))^(2))`B. `(1)/(N)sqrt((V_(1)^(2)+V_(2)^(2)+...+V_(N)^(2)))`C. `sqrt( (1)/(N)(V_(1)^(2)+V_(2)^(2)+..+V_(N)^(2)))`D. `sqrt((V_(1)+V_(2)+.....+V_(N))^(2))` |
| Answer» Correct Answer - C | |
| 94. |
Calculate `gamma` of a gaseous mixture consisting of 3 moles of nitrogen and 2 moles of carbon dioxide. |
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Answer» The `gamma` of a mixture is given by `(n_(1) + n_(2))/(gamma -1) = (n_(1))/(gamma_(1) - 1) + (n_(2))/(gamma_(2) - 1)` Here `n_(1) = 3, n_(2) = 3, gamma_(1) = 7//5` (as nitrogen is diatomic and `gamma_(2) = 4//3` (as carbon dioxide is triatomic) Therefore, `(5)/(gamma -1) = (3)/((7)/(5) - 1) + (2)/((4)/(3) - 1) implies gamma = 1.37` |
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| 95. |
A helium ballon occupies 8.0 L at `20^(@)C` and 1.0- atm pressure. The ballon rises to an altitude where the air pressure is 0.65 atm and the temperature is `-10^(@)C`. What is its volume when it reaches equilibrium at the new altitude? Note: Neglect tension forces in te material of the balloon. |
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Answer» Correct Answer - 11.05L `(P_(1)V_(1))/(T_1) = (P_(2)V_(2))/(T_2)` `(P_1) =1atm` `(V_1) = 8L` `T_(1) = 20^(@) = 293 K` `(P_2) =0.65 atm` `(V_2) = ?` `T_(2) = - 10^(@)C = 263 K` `P_(2) = 0.65 atm` `V_(2) =?` `T_(2) =-10^(@)`C = 263 K` `(1xx8)/(293) = (0.65V_(2))/(263)` `v_(2) =(8xx263)/(293xx0.65)` `V_(2) = 11.05 L` |
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| 96. |
Calculate the value of `gamma` for a gaseous mixture consisting of `n_(1)` moles of oxygen and `n_(2)` moles of carbon of carbon dioxide. The values of `gamma` for oxygen and carbon dioxide are `gamma_(1) and gamma_(2)` respectively. Assume the gases to be ideal. [Hint: `U_(m) = U_(1) + U_(2) and U = (nRT)/(gamma-1)`] |
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Answer» `U_(m) = U_(1) + U_(2)` `U_(1) = (n_(1)RT)/(r_(1) - 1) U_(2) = (n_(2)RT)/(r_(2)) - 1` `U_(m) = (n_(1) +n_(2)RT)/(r_(mix) - 1)` `(n_(1) + n_(2)RT)/(r_(mix) - 1) = (n_(1)RT)/(r_(1) -1) + (n_(2)RT)/(r_(2) - 1)` `(n_(1) + n_(2))/(r_(mix) - 1) = (n_(1)(r_(2) - 1))/(r_(1)-1) + (n_(2)(r_(1)-1)/(r_(2)-1)` `(n_(1)+n_(2))(r_(1)-1)(r_(2)-1)/(n_(1)(r_(2)-1) + (n_(2)r_(2)-1) + r_(mix)` `r_(mix) = (r_(1)r_(2)n_(1) +n_(2)) -n_(1)r_(1) +n_(2)r_(2))/(n_(1)r_(2)-1)+n_(2)r_(2)-1)` |
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| 97. |
An ideal gas under goes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by `PV^n=constant`, then n is given by (Here `C_P and C_V` are molar specific heat at constant pressure and constant volume, respectively):A. `n=(C_(P))/(C_(V))`B. `n=(C-C_(P))/(C-C_(V))`C. `n=(C_(P)-C)/(C-C_(V))`D. `n=(C-C_(V))/(C-C_(P))` |
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Answer» Correct Answer - B For polytropic process `C=C_(v)+(R)/(1-n)` `R=C_(P)-C_(v)` on simplyfincation |
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| 98. |
Find the minimum radius of a planet of mean density `5500kgm^(-3)` and temperature `400^(@)C` which has retained oxygen in its atmosphere. Density of oxygen at STP `=1.424kgm^(-3). G=6.6xx10^(-11)Nm^(2)kg^(-2)` |
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Answer» Correct Answer - `4.15xx10^(5)m` |
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| 99. |
(a) Assume the atmosphere as an ideal gas in static equilibrium at constant temperature T0. The pressure on the ground surface is P0. The molar mass of the atmosphere is M. Calculate the atmospheric pressure at height h above the ground. (b) To be more realistic, let as assume that the temperature in the troposphere (lower part of the atmosphere) decreases with height as shown in the figure. Now calculate the atmospheric pressure at a height h0 above the ground. |
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Answer» Correct Answer - (a) `P=P_(0)`e^((-Mgh)/(RT)) " "(b) P=P_(0)((5)/(4))^(-(5Mgh_(0))/(RT_(0)))` |
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| 100. |
Atmosphere of a planet contains only an ideal gas of molar mass `M_(0)`. The temperature of the atmosphere varies with height such that the density of atmosphere remains same throughout. The planet is a uniform sphere of mass M and radius ‘a’. Thickness of atmosphere is small compared to ‘a’ so that acceleration due to gravity can be assumed to be uniform throughout the atmosphere. (a) Find the temperature difference between the surface of the planet and a point at height H in its atmosphere (b) If the rms speed of gas molecules near the surface of the planet is half the escape speed, calculate the temperature of the atmosphere at a height H above the surface. Assume `(C_(p))/(C_(V)) = gamma` for the atmospheric gas |
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Answer» Correct Answer - (a) `DeltaT=(GMM_(0)H)/(Ra^(0)) " " (b) T_(H)=(GMM_(0))/(R_(a))[(1)/(2 gamma)-(H)/(a)]` |
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