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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.
| 601. |
calculate the number of moleculesin each cubic metre of a gas at 1 atm and `27^(@)C`. |
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Answer» We have `pV=NkappaT` Or, `N=(pV)/(kappaT)` `=((1.0xx10^(5)Nm^(-2))(1 m^3))/((1.38xx10^(-23)JK^(-1))(300K)) ~~ 2.4xx10^(25).` |
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| 602. |
A glass contains some water at room temperature `20^(@)C`.Refrigerated water is added to it slowly. When the tempareture of the glass reaches `10^(@)C,`small droplets condense on the outer surface. Calculate the relative humidity in the room. The boling point of water at a jprjjessure of 17.5mm of mercury is `20^(@)C`and at 8.9mm of mercury is `10^(@)C`. |
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Answer» Given : SVF at the dew point = 8.9 mm SVP at room temperature = 17.5 mm Dew point = `10^@C` as at this temperature the condensation starts Room temperature = ` 20^@C` ` (R_H) = SVP at dew point / SVP at room temp. ` ` = 8.9 / 17.5 = 51 %` . |
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| 603. |
Water falls from a height `500m`, what is the rise in temperature of water at bottom if whole energy remains in the water ? `(J=4.2)`A. `0.96^(@)C`B. `1.02^(@)C`C. `1.16^(@)C`D. `0.23^(@)C` |
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Answer» Correct Answer - C By the law of conservation of energy JmC dT=mgh `dT=(gh)/(Jc)=(9.8xx500)/(4.2xx10^(3))=1.17^(@)C` |
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| 604. |
A glass bulb of volume `400 cm^(3)` is connected to another bulb of volume `200 cm^(3)` by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of `20^(@) C` and 1.000 atm, respectively. The larger bulb is immersed in steam at `100^(@) C` and the smallar in melting ice at `0^(@)`. Find the final common pressure. |
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Answer» Let `n_(1)` and `n_(2)` donote the number of moles of gas in the large and small bulbs, in the final configuration, respectively. Denoting the final temperature by `T_(1)` and `T_(2)` and the final pressure by `P_(f)` the ideal gas law implies that `p_(f) V_(1) = n_(1) RT_(1)` and `p_(1) V_(2) = n_(2) RT_(2)` Where `p_(0), V_(0) = V_(1) + V_(2)` and `T_(0)` are the initial pressure, volume, and temperature, respectively. Using Eqs. (i) and (ii) in the equation `n_(1) + n_(2) = n`, we get `(p_(f) V_(1))/(RT_(1)) + (p_(f) V_(2))/(RT_(2)) = (p_(0) V_(0))/(RT_(0))` Solving for `P_(f)`, we obtain `p_(f) = (p_(0) V_(0))/(T_(0) ((V_(1))/(T_(1)) + (V_(2))/(T_(2))))` Inserting the numerical value, `p_(0) = 1.00 atm, T_(0) = 293.15` `K (t_(0) = 20^(@) C)`, `T_(1) = 373.15 K (t_(1) = 100^(@) C)`, and `T_(2) = 237.15 K` `(t_(2) = 0^(@) C)`, we find `p_(f) = ((1.00)(600))/((293.15) [((400)/(373.15)) + ((200)/(273.15))]) = 1.13` atm |
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| 605. |
A container of volume `1m^(3)` is divided into two equal compartments, one of which contains an ideal gas at 300K. The oterh compartment is vaccum. The whole system is thermally isolated from its surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would beA. 300KB. 250KC. 200KD. 100K |
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Answer» Correct Answer - A `P_(2)//P_(1)=(d_(2)//d_(1))^(gamma)` |
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| 606. |
Caculate the mass of `1cm^(3)`of oxygen kept at STP. |
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Answer» `n=(PV)/(RT)= (1 xx 1 xx 10^-3)/( 0.082 xx 273 )= 10^-3/22.4` ` mass= ((10^-3 xx 32)g)/224 ` ` = 1.428 xx 10^-3 g ` ` =1.428 mg ` |
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| 607. |
An electric bulb of volume 250cc was sealed during manufacturing at a pressure of `10^(-3)`mm of mercury at `27^(@)`C. Compute the number of air molecules contained in the bulb.Avogadro constant `=6xx10^(23)mol^(-1),` density of mercury`=13600kgm^(-3)and g=10ms^(-2)`. |
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Answer» `V=250cc =250 xx 10^-3. l` ` P=(10^-3mm)= (10^-3 xx 10^-3 m)` ` =(10^-6 xx 13600 xx 10)` ` 136 xx 10^-3 pascal ` ` T= 27^@C = 300 K ` ` n=(PV)/(RT)` ` = (136 xx 10^-3 xx 250 xx 10^-3)/ (8.3 xx 300) ` ` = (136 xx 250 xx 10^-6)/ (8.3 xx 300)` No. of molecules `=((136 xx 250)/(8.3 xx 300)) xx 10^-6 xx 6 xx 10^23` ` = 81 xx 10^17 = 0.81 xx 10^16` |
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| 608. |
Show that the internal energy of the air (treated as an ideal gas) contained in a room remains constant as the temperature changes between day and gight. Assume that the atmospheric pressure around remains constant and the air in the room maintains this prssure by communicating with the surrounding thourgh the widows, doors, etc. |
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Answer» Internal energy = nRT ` Now, PV = nRT` ` nT = PV` ` here P and V constant So, nT is constant` ` :. Internal energy = R xx constant ` ` = constant `. |
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| 609. |
At ordinary temperatures, the molecules of an ideal gas have only translational and rotational kinetic energies. At high temperatures they may also have vibrational energy. As a result of this, at higher temperatureA. `C_(V) = (3 R)/(2)` for a monatomic gasB. `C_(V) gt (3 R)/(2)` for a monatomic gasC. `C_(V) lt (3 R)/(2)` for a diatomic gasD. `C_(V) gt (3 R)/(2)` for a diatomic gas |
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Answer» Correct Answer - A::D Vibrational kinetic energy of a monatomic gas `=0` at all temperature. So, `C_(v)=3 R//2` for a monoatomic gas at high temperatures also. In case of a diatomic gas `C_(v)=5 R//2` at low temperatures while , `C_(v)=5R//2` at high temperatures due to vibrational `KE`. |
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| 610. |
A gas has volume `V` and pressure `p`. The total translational kinetic energy of all the molecules of the gas isA. 3/2 PV only if the gas is monoatomicB. 3/2PV only if the gas is diatomicC. 3/2 PV if the gas is diatomicD. 3/2 PV in all cases |
| Answer» Correct Answer - D | |
| 611. |
Statement-1: The total translational kinetic energy of fall the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume because. Statement-2: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.A. Statement-1 is True, Statement-2 is True,Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True,Statement-2 is NOT a correct expalnation for Statement-1C. Statement-1 is True, statement-2 is FalseD. Statement-1 is false, statement-2 is True |
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Answer» Correct Answer - B |
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| 612. |
If the ratio of specific heat of a gas of constant pressure to that at constant volume is `gamma`, the change in internal energy of the mass of gas, when the volume changes from `V` to `2V` at constant pressure `p` isA. `(PV)/(gamma-1)`B. `PV`C. `gamma-1`D. `(PV)/(gamma)` |
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Answer» Correct Answer - A `DeltaU=nC_(V)DeltaT` But `C_(V)=(R)/(gamma-1)` and PdV=RdT |
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| 613. |
If the ratio of specific heat of a gas of constant pressure to that at constant volume is `gamma`, the change in internal energy of the mass of gas, when the volume changes from `V` to `2V` at constant pressure `p` isA. `R/((gamma-1))`B. PVC. `(PV)/((gamma-1))`D. `(gammaPV)/((gamma-1))` |
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Answer» Correct Answer - C `DeltaU=f/2 n R DeltaT=(1/(gamma-1))PdV=((PV)/(gamma-1))` |
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| 614. |
Water is used as a collent becauseA. lower densityB. easily availabilityC. high specific heatD. low specific heat |
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Answer» Correct Answer - C For a particular change in temperature in would draw maximum heat. |
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| 615. |
In previous, Find the value of `gamma = C_(P) // C_(v) :`A. 1.6B. 1.5C. 1.6D. 1.66 |
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Answer» Correct Answer - D d. Straight line represents the constant pressure `W_(12) = 0` `dQ = n_(P). dT implies dU = nC_(v) dT` `dW = n (C_(P) - C_(v)) dT` `(dQ)/(dW) = (5)/(2) = (nC_(P) . dT)/(n (C_(P) - C_(v)) dT` `= (1)/(1 - C_(v)//C_(P)) = (gamma)/(gamma - 1)` `gamma = (5)/(3) = 1.66` |
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| 616. |
The ratio of heat absorbed and work done by the gas in the process, as shown is `5//2`. Find the parameters representing `x` and `y` axes : A. Temperature and pressureB. Temperature and volumeC. Volume and pressureD. Volume and density |
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Answer» Correct Answer - B b. If parameters are temperature and pressure straight line will represent constant volume for which `dW = 0`. Hence, option `(a)` is wrong. If parameters are volume and pressure, `(P prop (1)/(V))` for constant temperature. `Delta U = 0` for constant temperature and `Delta Q = Delta W`. Hence, `dQ//dW != 5//2`. So `(c )` is wrong. Obviously, option `(d)` is not possible Thus, option `(b)` is correct. Analyes it. |
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| 617. |
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. (a) Find the value of `W_4`. (b) What are the efficiency of the cycle?A. ZeroB. 275 JC. 765 JD. 1040 J |
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Answer» Correct Answer - C `triangleW = triangleQ` for cylicli process |
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| 618. |
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 cycleA. `1315 J`B. `275 J`C. `765 J`D. `675 J` |
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Answer» Correct Answer - C c. `Delta Q = Q_(1) + Q_(2) + Q_(3) + Q_(4)` `= 5960 - 5585 - 2980 + 3645 = 1040 J` `Delta W = W_(1) + W_(2) + W_(3) + W_(4)` `= 220 + 825 - 1100 - W_(4) = 275 + W_(4)` For a cyclic process, `U_(f) = U_(f)` `Delta U = Uf - U_(i) = 0` Form the first law of thermodynamics, `Delta Q = Delta U + Delta W` `1040 = 0 = 275 + W_(4)` or `W_(4) = 765 J` |
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| 619. |
A black body emits radiations of maximum intensity at a wavelength of Å 5000 , when the temperature of the body is `1227^(@)C`. If the temperature of the body is increased by `1000^(@)C` , the maximum intensity of emitted radiation would be observed atA. `2754.8`AB. 3000 ÅC. 3500 ÅD. 4000 Å |
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Answer» Correct Answer - B `lamda_(m_(2))=(T_(1))/(T_(2))xxlamda_(m_(1))=(1500)/(2500)xx5000=3000A^(@)` |
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| 620. |
Speed of sound in a gas is `v` and `rms` velocity of the gas molecules is `c`. The ratio of `v` to `c` isA. `(3)/(gamma)`B. `(gamma)/(3)`C. `sqrt((3)/(gamma))`D. `sqrt((gamma)/(3))` |
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Answer» Correct Answer - D Speed of sound `v_(s) = v = sqrt((gamma RT)/(M_(0)))` `rms` speed `v_(rms) = c = sqrt((3 RT)/(M_(0)))` `(v)/( c) = sqrt((gamma)/(3))` |
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| 621. |
A gas is compressed at a constant pressure of `50N//m^(2)` from a volume `10m^(3)` to a volume of `4m^(3)`. 100J of heat is added to the gas then its internal energy isA. Increases by 400JB. Increases by 200JC. Decreases by 400JD. Decreases by 200J |
| Answer» Correct Answer - A | |
| 622. |
A gas undergoes a change of state during which 100J of heat is supplied to it and it does 20J of work. The system is brough back to its original state through a process during which 20 J of heat is released by the gas. What is the work done by the gas in the second process? |
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Answer» `dQ_(1)=dU_(1)+dW_(1)` `100=dU_(1)+20rArrdU_(1)=80J` `dQ_(2)=dU_(2)+dW_(2) (therefore dU_(1)=-dU_(2))` `-20=-80+dW_(2)rArrdW_(2)=60J` |
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| 623. |
When 20J of work was done on a gas, 40J of heat energy was released. If the initial internal enrgy of the gas was 70J, what is the final internal energy?A. 50JB. 60JC. 90JD. 110J |
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Answer» Correct Answer - A `dQ-U_(f)-U_(i)+dW` |
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| 624. |
We consider a thermodynamic system. If `DeltaU` represents the increase in its internal energy and W the work done by the system, which of the following statements is true?A. `DeltaU=-W` is an adiabatic processB. `DeltaU=W` in an isothermal processC. `DeltaU=-W` in an isothermal processD. `DeltaU=W` in an adiabatic process |
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Answer» Correct Answer - A An isothermal process is a constant temperature process. In this process T= constant or `DeltaT=0` `therefore" " DeltaU=nC_(v)DeltaT=0` An adiabatic process is defined as one with no heat trasfer into or out of a system . therefore , Q = 0 . From the first law of thermodynamics. ` W=-DeltaU " or " DeltaU=-W` |
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| 625. |
For V versus T curves at constant pressure `P_1 and P_2` for and ideal gas shown in figure-A. `P_1 gt P_2`B. `P_1 lt P_2`C. `P_1 = P_2`D. `P_1 P_2` |
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Answer» Correct Answer - A Slope `= (nR)/(P)` P increases, slop decreases |
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| 626. |
Four moles of an ideal gas is initially in state `A` having pressure ` 2 xx 10^(5) N//m^(2)` and temperature ` 200 K` . Keeping the pressure constant the gas is taken to state `B` at temperature of `400K`. The gas is then taken to a state `C` in such a way that its temperature increases and volume decreases. Also from `B` to `C` , the magnitude of `dT//dV` increases. The volume of gas at state `C` is eqaul to its volume at statem `A`. Now gas is taken to initial state `A` keeping volume constant. A total `1000 J` of heat is withdrawn from the sample of the cyclic process . Take `R=8.3 J// K// mol`. The volume of gas at state `C` isA. `0.0332 m ^(3)`B. `0.22 m^(3)`C. `0.332m ^(3)`D. `3.32 m ^(3)` |
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Answer» Correct Answer - A The volume of gas at state `C` is `V_(C)=V_(A)=(nRT_(A))/(P_(A))` `=0.0332m^(3)` |
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| 627. |
A cylindrical container is divided into three parts by two tight fitting pistons. The pistons are connected by a spring. The region between the pistons is vacuum and the other two parts have same number of moles of an ideal gas. Initially, both the gas chambers are at temperature `T_(0)` and the spring is compressed by 1 m. Length of both gas chambers is 1 m in this position. Now the temperature of the left and right chambers are raised to `(4T_(0))/(3)` and `(5T_(0))/(3)` respectively. Find the final compression in the spring in equilibrium. Assume that the pistons slide without friction |
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Answer» Correct Answer - `(sqrt(13)-1)/(2)m` |
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| 628. |
The air tight and smooth piston of a cylindrical vessel are connected with a string as shown. Initially pressure and temperature of the gas are `P_(0)` and `T_(0)`. The atmospheric pressure is also `P_(0)`. At a later time, tension in the string is `(3)/(8) P_(0)A` where A is cthe cross-sectional are of the cylinder. at this time, the temperature of the gas has become. A. `(3)/(8) T_(0)`B. `(3)/(4) T_(0)`C. `(11)/(8) T_(0)`D. `(13)/(8) T_(0)` |
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Answer» Correct Answer - C |
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| 629. |
P-T graph for same number of moles of two ideal gases are shown . Find the path along which volume decreases . A. A to BB. B to AC. C to DD. D to C |
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Answer» Correct Answer - C |
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| 630. |
What will be the rms speed f argon at `40^(@)C`? If the rms speed of oxygen molecule at 1092 K is 920 /s. (molecular weight of oxygen is 32 and that of argon is 40).A. 460 m/sB. 404.5 m/sC. 44 m/sD. 4405.0 m/s |
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Answer» Correct Answer - A `(C_(2))/(C_(1))=sqrt((T_(2))/(T_(1))xx(M_(1))/(M_(2)))=sqrt((313xx32)/(1092xx40))` `=sqrt((313)/(1092)xx(4)/(5))=sqrt((313)/(273)xx(1)/(5))` `C_(2)=(920)/(2)=460m//s`. |
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| 631. |
For nitrogen `C_(p)-C_(V)=x` and for argon `C_(P)-C_(V)`=Y.The relation between x and y is given byA. a=16 bB. a=4 bC. 16 b=aD. a=b |
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Answer» Correct Answer - D For hydrogen gas, `C_(p)-C_(V)=R=a` . .. (i) for oxygen gas, `C_(p)-C_(V)=R=b` . . (ii) `therefore a=b`. |
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| 632. |
Calculate the heat absorbed by the system is going through the process shown in figure A. `31.4J`B. 3.14JC. `3.14xx10^(4)J`D. none |
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Answer» Heat absorbed`=pir^(2)` `=pi(P_(r))(V_(r))` `=3.14(100xx10^(3))(100xx10^(-6))` `=3.14J` |
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| 633. |
The molar heat capacity in a process of a diatomic gas if it does a work of `Q/4` when a heat of `Q` is supplied to it is |
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Answer» `dU=C_(v)dT=((5)/(2)R)dT ("or")dT=(2)(dU)/(5R)` From first law of thermodynamics `dU=dQ-dW=Q-(Q)/(4)=(3Q)/(4)` Now molar heat capacity `C=(dQ)/(dT)=(Qxx5R)/(2(dU))=(5RQ)/(2((3Q)/(4)))=(10)/(3)R` |
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| 634. |
The molar heat capacity in a process of a diatomic gas if it does a work of `Q/4` when a heat of `Q` is supplied to it isA. `(2)/(5) R`B. `(5)/(2) R`C. `(10)/(3) R`D. `(6)/(7) R` |
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Answer» Correct Answer - C `dU = C_(v)dT = (5/2 R) dT or dT = (2(dU))/(5R)` ..(i) From first law of thermodynamics `dU = dQ - dW = Q -Q/4 = (3Q)/(4)`. Now molar heat capacity `C = (dQ)/(dT) = (Q)/((2(dU))/(5R)) = (5RQ)/(2((3Q)/(4))) = 10/3 R`. |
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| 635. |
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` |
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Answer» Correct Answer - D `(V_(sound))/(V_(rms)) =(sqrt((gammaP)/(rho)))/(sqrt((P3)/(rho))) = sqrt(5/9)x` `PT =` constnat `P^(2)V = const implies PV^(1//2)=` constant ltbr gt`implies x = 1/2` `C = C_(v)+(R)/(1-x) = 3/2 R +2R = (7R)/(2)`. |
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| 636. |
Argon gas is adiabatically compressed to half its volume. If `P, V` and `T` represent the pressure, volume and temperature of the gasous, respectively, at any stage, then the correct equation representing the process isA. `TV^(2//5) = constant`B. `VP^(5//3) = constant`C. `TP^(2//5) = constant`D. `PT^(2//5) = constant` |
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Answer» Correct Answer - C For an adiabatic process `PV^(gamma)` = constant `TV^(-1)` = constant and `T^(gamma)P^(1//gamma)` = constant Putting `gamma = 5//3` (argon being a monoatomic gas check the options. |
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| 637. |
An ideal gas is initially at temperature T and volume V. Its volume is increased by `DeltaV` due to an increase in temperature `DeltaT,` pressure remaining constant. The quantity `delta=(DeltaV)/(VDeltaT)` varies with temperature asA. B. C. D. |
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Answer» Correct Answer - C c. From ideal gas equation `PV = RT` `P Delta T = R Delta T` Dividing Eq. (ii) by Eq. (i), we get `(Delta T)/(V) = (Delta T)/(T) implies (Delta V)/(V Delta T) = (1)/(T) = delta` (given) `:. Delta = (1)/(T)` So the graph between `delta` and `T` will be a rectangular hyperbola. |
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| 638. |
As a result of the isobaric heating by `DeltaT=72K`, one mole of a certain ideal gas obtain an amount of heat `Q=1.6kJ`. Find the work performed by the gas, the increment of its internal energy and `gamma`. |
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Answer» By the first law of thermodynamics `Delta Q = Delta U + Delta W` In an isobaric process `Delta Q = C_(P)Delta T` and `Delta U = C_(V)Delta T` `:. C_(P) Delta T = C_(V) Delta T + Delta W` or `Delta W = (C_(P) - C_(V)) Delta T` or `Delta W = R Delta T` `( :. C_(P) - C_(V) = R)` `:. Delta W = 8.3 xx 72 = 597.6 J` `Delta Q = C_(P) Delta T` `Delta Q = C_(P) Delta T` `:. 1.6 xx 1000 = C_(P) xx 72` `implies C_(P) = (1.6 xx 1000)/(72) = 22.2 J mol^(-1) K^(-1)` `Delta U = Delta Q = Delta W = 1.6 xx 1000 - 597.6 = 1002.4 J` But `Delta U = C_(V) Delta T` `:. C_(V) = (1002.4)/(72) = 13.9 J mol^(-1) K^(-1)` `:. gamma = (C_(P))/(C_(V)) = (22.2)/(13.9) = 1.60` |
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| 639. |
When heat in given to a gas in an isobaric process, thenA. The work is done by the gasB. Internal energy of the gas increasesC. Both (a) and (b)D. None from (a) and (b) |
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Answer» Correct Answer - C When heat is supplied at constant pressure, a part of it goes in the expansion of gas and remaining part is used to increase the temperature of the gas which in turn increases the internal energy . |
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| 640. |
A spherical body of 5 cm radius is maintained at a temperature of `327^(@)C`. The wavelength at which maximum energy radiated will be nearly `(b=2.898xx10^(-3)m" "K)`A. 482 ÅB. 4.82 ÅC. 482 `mu` mD. 4.82 `mu` m |
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Answer» Correct Answer - D `lamdaT=b` `lamda=(b)/(T)=(2.898xx10^(-3))/(600)=4.83mum` |
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| 641. |
A girl weighing 42 kg eats bananas whose energy is 980 calories. If this energy is used to height h find the value of h. (J=4.2J calorie) |
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Answer» Energy gained by the girl in eating bananas =980 calories =980xx4.2J. W=H(in S.I.) `980xx4.2=mgh rArr 980xx4.2=42xx9.8xxh` `rArrh=(980xx4.2)/(42xx9.8)=10m` |
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| 642. |
An engine is supposed to operate between two reservoirs at temperature `727^(@)C` and `227^(@)C`. The maximum possible efficiency of such an engine isA. `1//2`B. `1//4`C. `3//4`D. `1` |
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Answer» Correct Answer - A `eta = (T_1-T_2)/(T_1) = ((273+727)-(273+227))/(273+727)` `=(100-500)/(1000) = 1/2`. |
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| 643. |
Consider the vaporization of 1g of water at `100^(@)C` at one atmosphere pressure. Compute the work done by the water system in the vaporization and change internal energy of the system. |
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Answer» To change a system of mass m of liquid to vapour heat required is `Q=ml_("vapour")` The process takes place at constant pressure so the work done by the system is the work Where `DeltaV=(V_("vapour")-V_("liquid"))` From first law of thermodynamics `DeltaU=Q-W=mL_(v)-P(V_("vapour"-V_("liquid")))` Latent heat of vaporization of water `L_("vapour")=22.57xx10^(5)J//kg` `Q=(1.00xx10^(-3))(22.57xx10^(5))=2.26xx10^(3)` `because` No. of moles=weight /gram molecular weight `therefore` Moles of water in `1g=(1)/(18)=0.0556` mole `V_("vapour")=(nRT)/(P)=((0.0556)(8.315)(373))/(1.013xx10^(5))=1.70xx10^(-3)m` The density of water is `1.00xx10^(3) kg//m^(3)=1.00 g//cm^(3)` `V_("liquid")=1.00xx10^(-6) m^(3)` Thus the work done by the water system Vaporization is `W=P(V_("vapour")-V_("liquid"))` `=(1.013xx10^(5))(1.70xx10^(-3)-1.00xx10^(-6))=172J` The work done by the system is positive since the volume of the system has increased. From first law, `DeltaU=Q-QrArrDeltaU=2.26xx10^(3)-172=2.09xx10^(3)J` |
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| 644. |
Figure shown a cylindrical tube of radius `r` and lengh `l` fitted with a cork. The friction coefficient between the cork and the tube is mu. The tube contains an ideal gas at temperature `T`, and atmospheric pressure `P_(0)` The tube is slowly heated , the cork pipe out when temperature is boubled. What is normal force per unit length exerted by the periphery of tube? Assume uniform temperature throughout gas any instant. |
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Answer» Since volume of the gas is constant, `(P_(i))/(T_i)=(P_f)/(T_(f)` `P_(f)=P_(i)((T_(F))/T_(i))=2P_(i)=2P_(0)` The Forces actihg on the cork are shown in the figure in equilibrium. `P_(0)xxA+muN=2P_(0)A` `N=(P_(0)A)/mu` `N` is the total normal force exerted by the tube on the cork, hence contact force per unil length is `(dN)/(dl)=(N)/(2pir)`=`(P_(0)A)/(2pimur)` |
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| 645. |
Which of the process described below are irrevesible?A. The increase in temperatre of an iron rod by hammering itB. gas in a small container at a temperature `T_(1)` is brough in contanct with a big reservoir at a higher temperature `T_(2)` which increase the temperature of the gasC. A quasi-static isothermal expansion of a n ideal gas in cylinder fitted with a frictionless pistonD. An ideal gas is enclosed in apiston cylinder arrangement with adiabatic walls. A weight w is added to the piston, resulting in compression of gas. |
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Answer» Correct Answer - A::B::D (a) When the rod is hammered, the external work is done on the rod which increases its temperature. The process cannot be retracted itself (b) In this process energy in the form of heat is transferred to the gas in the small container by big reservoir at temperature `T_(2)`. (d) As the widht is added to the cylinder arangement in the form of external pressure hence, it cannot be reversed back itself. |
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| 646. |
Assertion : The rms velocity of gas molecules is doubled, when temperature of gas becomes four times. Reason : `c oo sqrt(T)`A. If both assertion and reason are true and the reason is correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true, but the reason is false.D. If assertion is false but the reason is true |
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Answer» Correct Answer - A The rms velocity of gas molecules is given by `v_(rms)= c = sqrt((3kT)/(m))` so, `c prop sqrt(T)` Hence, it is clear that when temperature becomes four times rms velocity will be two times. |
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| 647. |
Assertion : The number of degrees of freedom of triatomic molecules is `6` Reason : Triatomic molecules have three translational degrees of freedom and three rotational degrees of freedom.A. If both assertion and reason are true and the reason is correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true, but the reason is false.D. If assertion is false but the reason is true |
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Answer» Correct Answer - A We know that for a diatomic gas `C_(v)=3R, C_(P)=4R` `:. Gamma = (C_P)/(C_v) = 4/3` But `gamma=1+2/f` so , `1+2/f=4/3implies 2/f = 4/3-1=1/3` `f=6` Tus, option (a) is correct. |
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| 648. |
During the adiabatic change of ideal gas, the realation between the pressure and the density will be:A. `(P_1)/(P_2)= ((d_2)/(d_1))^(1/gamma)`B. `P_1d^gamma1= P_2d_2^gamma`C. `P_1d_1^(-gamma) = P_2d_2^(-gamma)`D. `(P_1)/(P_2)=((d_1)/d_2)^(1/gamma)` |
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Answer» Correct Answer - C `(i) PV^(r) = constant` `(ii) d =(m)/(v)` `rArr v = (m)/(d)` `rArr P((m^1)/(d))^(r) = constant` `rArr (p)/(dr) = constant` `rArr P_(1) d_(1)^-r = P_(2) d_(2)^-r` |
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| 649. |
During an experiment, an ideal gas is found to obey a condition `Vp^2 =` constant. The gas is initially at a temperature (T), pressure (p) and volume (V). The gas expands to volume (4V).A. The pressure of gas changes to `P/2`B. The temperature of gs changes to 4TC. The graph of above process on the P - T diagram is parabolaD. The graph of above lprocess on the P -T diagram is hyperbola. |
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Answer» Correct Answer - A::D `UP^2 = constant` `PV = nRT` replace `P = (nRT)/(V)` |
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| 650. |
The average translational kinetic energy of the molecules of a gas will be doubled if at constantA. Volume, its pressure is doubledB. temperature, its pressure is doubledC. pressure, its volume is halvedD. temperature, its volume is doubled |
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Answer» Correct Answer - A `P` doubled, average translational kinetic energy doubled. |
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