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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.
| 851. |
One mole of an ideal monatomic gas undergoes the process `p=alphaT^(1//2)`, where `alpha` is a constant. (a) Find the work done by the gas if its temperature increases by 50K. (b) Also, find the molar specific heat of the gas. |
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Answer» `P = alpha T^(1//2)` where `alpha` is constant `n = 1 mol`, monatomic `W = int PdV = int (alpha T^(1//2)) dV` From Eq. (i), `T = (alpha T^(1//2))/(nR) implies T^(1//2) = (alpha V)/(nR)` Differntiating both sides, `(1)/(2 sqrtT) dT = (alpha dV)/(nR) implies dV = (nR)/(alpha 2 sqrtT) dT` `W = int alpha T^(1//2) (nR)/(2T^(1//2)) dT` `= (n alpha R)/(2) int_(T_(1))^(T_(2)) dT = (R )/(2) xx 50 = 25R = 207.7 J` b. `Q = Delta U + W` `C (T_(2) - T_(1)) = (R )/((gamma -1)) (T_(2) - T_(1)) + (R )/(2) (T_(2) - T_(1))` `C = (R )/((gamma - 1)) + (R )/(2) = ((gamma + 1)/(gamma - 1)) (R )/(2)` |
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| 852. |
One mole of an ideal monatomic gas undergoes a process described by the equation `PV^(3)`= constant. The heat capacity of the gas during this process isA. `2R`B. `R`C. `3/2R`D. `5/2 R` |
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Answer» Correct Answer - B `PV^(x)` = constant (polytropic process) Heat capacity in polytropic process is given by `[C=C_(v)+(R)/(1-x)]` Given that `PV^(3) = consta nt implies x=3` ..(1) Also gas is monoatomic `f=3` ..(2) BY formula `C = (fR)/(2)+R/(1-x)` `C = 3/2 R -R/2 = R`. |
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| 853. |
If one mole of a monatomic gas `(gamma=5/3)` is mixed with one mole of a diatomic gas `(gamma=7/5),` the value of gamma for mixture isA. 1.5B. 1.54C. 1.4D. 1.45 |
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Answer» Correct Answer - A `gamma=(n_(1)gamma_(1)+n_(2)gamma_(2))/(n_(1)+n_(2))` |
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| 854. |
The volume of one mode of an ideal gas with adiabatic exponent `gamma` is varied according to the law `V = a//T`, where a is constant . Find the amount of heat obtained by the gas in this process, if the temperature is increased by `Delta T`. |
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Answer» Correct Answer - `(2-gamma)RtriangleT/(gamma-1) We have `triangleW = intpdv and triangle U = intC_(v)dT ,for an ideal gas pV = RT`, `therefore " " triangle W = overset(T+triangleT)underset(T)int(RT)/(V)dV = overset(T+triangleT)underset(T)int(RT^2)/(a)(-a/T^(2)dT)=-RtriangleT` `triangleU = overset(T+triangleT)underset(T)int(R)/(gamma-1)dT = (RtriangleT)/(gamma-1)` `therefore " " triangleQ = triangleU + triangleW =(RtriangleT)/(gamma-1)+(-RtriangleT) = ((2-gamma)RtriangleT)/(gamma-1)` |
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| 855. |
If one mole of a monatomic gas `(gamma=5/3)` is mixed with one mole of a diatomic gas `(gamma=7/5),` the value of gamma for mixture isA. `1.40`B. `1.50`C. 1.53D. 3.07 |
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Answer» Correct Answer - B `gamma_("mix")=((mu_(1)gamma_(1))/(gamma_(1)-1)+(mu_(2)gamma_(2))/(gamma_(2)-1))/((mu_(1))/(gamma_(1)-1)+(mu_(2))/(gamma_(2)-1))=((1xx(5)/(5))/([(5)/(3)-1])+(1xx(7)/(5))/([(7)/(5)-1]))/([(1)/((5)/(3)-1)]+[(1)/((7)/(5)-1)])=(3)/(2)=1.5` |
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| 856. |
When `1` mole of a monatomic gas is mixed with `3` moles of a diatomic gas, the value of adiabatic exponent `gamma` for the mixture isA. `5/3`B. `1.5`C. `1.4`D. `13/9` |
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Answer» Correct Answer - D `(n_1+n_2)/(gamma-1) = (n_1)/(gamma_(1)-1) + (n_2)/(gamma_(2)-1)` `(1+3)/(gamma-1) = (1)/(5/3-1) + 3/(7/5 -1)` `4/(gamma-1) = 3/2+15/2 = 9` or `gamma -1 =4/9 implies gamma = 1 + 4/9 = 13/9`. |
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| 857. |
An ideal refrigerator has a freezer at a temperature of `-13^(@)C`. The coefficient of performance of the engine is `5`. The temperature of the air (to which heat is rejected) will beA. `325^(@)C`B. `325 K`C. `39^(@)C`D. `320^(@)C` |
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Answer» Correct Answer - C Coefficient of performance `K = (T_2)/(T_(1)-T_(2)) implies 5=((273-13))/(T_(1)-(273-13)) = (260)/(T_(1)-260)` `= 5T_(1)-1300=260implies 5T_(1) = 1560` `implies T_(1)=312K rarr 39^@C`. |
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| 858. |
During an adibatic compression ,830 J of work is done on 2 moles of a diatomic ideal gas to reduce its volume by 50%. The change in its temperature is nearly:A. 40KB. 33KC. 20kD. 14K |
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Answer» Correct Answer - C `W_("adiabatic")=(nR)/(gamma-1)DeltaT`. |
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| 859. |
using euipartion of energy, the specific heat `("in" jkg^(-1)K^(-1)`of aluminium at room temperature can be estimated to be (atomic weigh of aluminium=27)A. 410B. 25C. 1850D. 925 |
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Answer» Correct Answer - D using law of equipartition of energy ,the heat developed in aluminium is `2xx(3)/(2)K_BT=msDeltaT=ms(T-0)` `rArr" " 3K_(B)=ms=(M)/(N_(A))s` `rArr" " s=(3K_(B)N_(A))/(M)` (i) `"Here", " " M=27,K_(B)=1.38xx10^(-23)JK^(-1)` `"and " N_(A)=6.02xx10^(23)//"mole" ` from Eq (i), specific heat of aluminium at room temperature, `s=(3xx1.38xx10^(-23)xx6.02xx10^(23))/(27xx10^(-3))" " [because1g=10^(-3)kg]` `s=(138xx60.2)/(9)=923.06~~JKg^(-1)K^(-1)` |
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| 860. |
Two spheres A and B with masses in the ratio 2:3 and specific heat 2:3 fall freely from rest. If the rise in their temperature on reaching the ground are in the ratio 1:2 the ratio of their heights of fall isA. `3:1`B. `1:3`C. `4:3`D. `3:4` |
| Answer» Correct Answer - B | |
| 861. |
Hailstones fall from a certain height. If they melt completely on reaching the ground, find the height from which they fall. (`g=10ms^(-2),L=80` calorie/g and J=4.2 J/calorie.) |
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Answer» On reaching the ground, a hailstone of mass M losses potential energy which is converted into he energy required to melt it. In Sl, potential energy required for melting the hailstons `Mgh=MLrArrgh=L` `h=(L)/(g), h=(80xx4.2xx1000)/(10)` `=33.6xx1000m=33.6km`. |
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| 862. |
A gas expand with temperature according to the relation `V = KT^(2//3)`. What is the work done when the temperature changes by `30^(@)C`?A. `10 R`B. `20 R`C. `30 R`D. `40 R` |
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Answer» Correct Answer - B `W = int PdV = int (RT)/(V) dV` Since `V = kT^(2//3) KT^(-1//3) dT` Eliminating `K` we find `(dV)/(V) = 2/3(dT)/(T)` Hence `W = int_(T_1)^(T_2) 2/3 (RT)/(T) dT = 2/3 R(T_(2)-T_(1)) = 2/3 R(30) = 20R`. |
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| 863. |
One mole of a gas expands with temperature T such thaht its volume, V=`KT^(2)`, where K is a constant. If the temperature of the gas changes by `60^(@)C` then the work done by the gas is `120RA. R ln 60B. kR In 60C. 60 kRD. 120R |
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Answer» Correct Answer - D `V=kT^(2) therefore 2kT dT=dV` and `P=(RT)/(V)=(RT)/(kT^(2))=(R)/(kT)` `dW=PdV=2RdT, W=int PdV` |
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| 864. |
A gas expands with temperature according to the relation `V = kT^(2//3)`. What is the work done when the temperature changes by `30^(@)C`?A. `166.2 J`B. `136.2 J`C. `126.2 J`D. none of these |
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Answer» Correct Answer - A a. `PV = RT` for 1 mol `W = int PdV = int (RT)/(V) dV` `V = CT^(2//3)` `dV = (2)/(3) CT^(1//3) dT` or `(dV)/(V) = (2)/(3) (dT)/(T)` `W = int_(T_(1))^(T_(2)) RT ((2)/(3)) (dT)/(T) = (2)/(3) R (T_(2) - T_(1)) = 166.2 J` |
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| 865. |
Diatomic molecules like hydrogen haven energy due to both translational as well as rotational motion. From the equation in kinetic theory `PV=(2)/(3)E,E` isA. The total energy per unit volumeB. Only the translational part of energy because rotational energy is very small compared to translational part of energy because rotational energy is very small compared to translational energyC. Only the translational part of the energy because during collisions with the wall pressure relates changes in linear momentumD. The trnaslational part of the energy because roatational enegies of molecules can be of either sign and its average over all the molecules is zero |
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Answer» Correct Answer - C Sol, (c) According to Kinetic theory equation, `PV=(2)/(3)E ["where P = pressure V=volume"]` E is representing only translational part of energy. Internal energy contains all types of energies like translation, rotational, vibrational etc. But the molecules of an ideal gas is treated as point masses in kinetic theory, so its kinetic energy is only due to translational motion. point mass does not have rotational or vibrational motion. Here we assumed that the walls only exert perpendiucalr forces on molecules. They do not exert any parallel force, hence there will not be any tuype of rotation present. The wall produces only charges in translational motion. |
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| 866. |
The cofficient of performance of a refrigerator is 5. If the temperature inside freezer is `-20^(@)C`, the temperature of the surroundings to which it rejects heat is :A. `21^(@)C`B. `31^(@)C`C. `41^(@)C`D. `11^(@)C` |
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Answer» Correct Answer - B Coefficient of performance of refregetor `COP = (T_L)/(T_(H)-T_(L))` Where `T_(L) rarr` lower temperature and `T_(H) rarr` higher temperature so, `5=(T_L)/((T_(H)-T_(L))` `implies T_(H) = 6/5T_(L)=6/5(253)=303.6K`. |
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| 867. |
The heat supplied to one mole of an ideal monoatomic gas in increasing temperature from `T_(0)` to `2T_(0)` is `2RT_(0)`. Find the process to which the gas followsA. `P^(2)V^(-1) = constant`B. `PV^(-1) = constant`C. `P^(-1)V = constant`D. `PV^(-1) = constant` |
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Answer» Correct Answer - D The supplied to the gas, `Q = nC(DeltaT)` `n` is the number of moles in our case `n=1`, and `C` is the molar heat capacity of the gas For polytropic molar heat capacity is given by `C = R/((gamma-1))+(R)/((1-N))` `=R/((5//3)-1)+(R)/((1-N))= 3/2 R+(R)/((1-N))` `Q = n[(3R)/(2)+(R)/(1-N)] (2T_(0)-T)` `(1)RT_(0)[(5-3N)/(2(1-N))] = 2RT_(0)` (given) or `5-3N=4(1-N)` `or N=-1` Hence equation of process , `PV^(-1)= consta nt`. |
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| 868. |
A cylindrical container having non-conducting walls is partitioned in two equal parts such that the volume of the each parts is eqaul to `V_(0)` A movable non-conducting piston is kept between the two parts. Gas on left is slowly heated so that the gas on right is compressed upto volume `V_(0)//8`. Find pressure and temperature on both sides if initial pressure and temperature, were `P_(0)` and `T_(0)` respectively. Also find heat given by the heater to the gas (number of moles in each parts is n) |
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Answer» Since the process on the right is adiabatic therefore `PV^(gamma)` = constant `implies P_(0) V_(0)^(gamma) = P_("final ") ((V_(0))/(8))^(gamma) implies P_("final") = 32 P_(0)` `T_(0) V_(0)^(gamma - 1) = T_("final") ((V_(0))/(8))^(gamma -1) implies T_("final" = 4 T_(0)` Let volume of the left part is `V_(1)` `implies 2V_(0) = V_(1) + (V_(0))/(8) implies V_(1) = (15 V_(0))/(8)` Since number of moles on the left parts remains constant therefore the left part `(PV)/(T)` = constant. Final pressure on both sides will be same `implies (P_(0) V_(0))/(T_(0)) = (P_("final") V_(1))/(T_("final")) implies T_("final") = 60 T_(0)` `Delta Q = Delta u + Delta W` `Delta Q = (5 R)/(2) (60 t_(0) - T_(0)) + n (3R)/(2) (4T_(0) - T_(0))` `Delta Q = (5 nR)/(2) xx 59 T_(0) + (3 nR)/(2) xx 3T_(0)` |
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| 869. |
The figure shows the graph of logarithmic reading of pressure and volume for two ideal gases A and B undergoing adiabatic process. From figure, it can be concluded that A. gas B is diatomicB. gas A and B both are diatomicC. gas A is monoatomicD. gas B is monoatomic & gas A is diatomic |
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Answer» Correct Answer - A Slope of adiabatic Process` = -rp` |
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| 870. |
A gas is undergoing an adiabatic process. At a certain stage A, the values of volume and temperature `(V_(0), T_(0))`. From the details given in the graph, find the value of adiabatic constant `gamma` A. `(V_(0))/(T_(0) tan theta) + 1`B. `(V_(0) tan theta)/(T_(0) + 1`C. `(V_(0) tan^(2) theta)/(T_(0)) + 1`D. `(V_(0))/(T_(0)) + tan theta` |
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Answer» Correct Answer - A We have, for an adiabatic process, in relation wiht temperature and volume `TV^(gamma-1)` = constant Differentiating with respect to `T` `V^(gamma-1).1+T.(gamma-1)V^(gamma-1)=(V_0)/(T_(0)tan theta)` `gamma = (V_0)/(T_0 tan theta) + 1`. |
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| 871. |
When thermodynamic system returns to its original state, which of the following is NOT possible?A. The work done is ZeroB. The work done is positiveC. The work done is negativeD. The work done is independent of the path followed |
| Answer» Correct Answer - D | |