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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.
| 451. |
On the basis of information available from the reaction `(4)/(3)Al+O_(2) rarr (2)/(3)Al_(2)O_(3),DeltaG = -827kJ mol^(-1)` of `O_(2)`, the minimum emf required to carry out of the electrolysis of `Al_(2)O_(3)` is `(F=96,500 C mol^(-1))`A. 6.42 VB. 8.56 VC. 2.14 VD. 4.28 V |
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Answer» Correct Answer - C `DeltaG= -nFE_(cell)^(@)` |
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| 452. |
1 mole of an ideal gas `A(C_(v.m)=3R)` and 2 mole of an ideal gas B are `((C_(v,m)=(3)/(2)R)` taken in a container and expanded reversible and adiabatically from 1 litre of 4 letre starting from initial temperature of 320 K. `DeltaE ` or `DeltaU` for the process is :A. `-240 R`B. 240RC. 480 RD. `-960 R` |
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Answer» Correct Answer - D Aev. `C_(v, m) = (n_(1)C_(V, m_(1))+n_(2)C_(v, m_(2)))/(n_(1)+n_(2)) = 2R` For adiabatic process dU = dW `(dT)/(T) = -(R)/(C_(v, m))((dV)/(V))` `n_(1)C_(V), m_(1)dT + n_(2)C_(V)m_(2) dT = -(n_(1)RT + n_(2)RT) xx (dV)/(V)` `ln.(T_(2))/(T_(1)) = -(1)/(2)ln((V_(2))/(V_(1))) rArr T_(2) = 320 xx ((1)/(4))^(1//2) = 160 K` `Delta U = (n_(1)C_(v), m_(1) + n_(2)C_(V), m_(2))Delta T = -960 R` |
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| 453. |
1 mole of an ideal gas `A(C_(v.m)=3R)` and 2 mole of an ideal gas B are `((C_(v,m)=(3)/(2)R)` taken in a container and expanded reversible and adiabatically from 1 litre of 4 letre starting from initial temperature of 320 K. `DeltaE ` or `DeltaU` for the process is :A. `240` RB. 240 RC. 480 RD. `-960` R |
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Answer» Correct Answer - D `"Av. C"_(V,m)=(n_(1)C_(V,m_(1))+n_(2)C_(V,m_(2)))/(n_(1)+n_(2))=2R` for adiabatic process dU = dW `(dT)/(T)=-(R)/(C_(V,m))((dV)/(V))` `n_(1)C_(V,m_(1))dT+n_(2)C_(V,m_(2))dT` `=-(n_(1)RT+n_(2)RT)xx(dV)/(V)` `"ln"(T_(2))/(T_(1))=-(1)/(2)"ln"((V_(2))/(V_(1)))` `rArr" "T_(2)=320xx((1)/(4))^(1//2)` = 160 K `DeltaU=(n_(1)C_(V,m_(1))+n_(2)C_(V,m_(2)))DeltaT=-960R` |
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| 454. |
Anhydrous `AlCl_(3)` is covalent. From the date given below, predict whether it would remain covalent or become ionic in aqueous solution. (Ionisation energy for `Al` is `1537 kJ mol^(-1)`) `Delta_("hydration") for Al^(3+) = - 4665 kJ mol^(-1)` `Delta_("hydration") for Cl^(Ө) = - 381 kJ mol^-1`. |
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Answer» Correct Answer - IONIC Total hydration energy of `Al^(3+)` and `3 Cl^(-)` ions of `AlCl_(3). (Delta H_("Hydration")) = ` Hydration energy of `Al^(3+) + 3 xx` Hydration energy of `Cl^(-)` `=[-4665 + 3 xx (-381) ] kJ mol^(-1) = - 5880 kJ mol^(-1)` This amount of energy exceeds the energy needed for the isonisation of `Al` to `Al^(3+)` (i.e., `5808 gt 5137`). Beacuse of this `AlCL_(3)` becomes less ionic in aqueous solution. In aqueous solution `AlCl_(3)` exists in ionic form as `[Al (H_(2)O)_(6)]^(3+)` and `3 Cl^(-)` `AlCl_(3) + 6H_(2)O rarr [Al (H_(2)O)_(6)]^(3+) + 3Cl^(-)` `AlCl_(3) + aq rarr AlCl_(3) (aq) , Delta H = ?` `Delta H =` (Energy released during hydration) - (Energy used during hydration) `(-4665) - (3 xx 381) + 5137 = - 671 kJ mol^(-1)` Thus, formation of ions will take places. |
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| 455. |
Classify the following as open, closed, or isolated system. a. A beaker containing as opne, boiling water. b. A chemical recation taking place in an enclosed flask. c. A cup of tea placed on a table. d. Hot water placed in perfectly insulated closed container. e. A thermos flask containing hot coffee. |
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Answer» Types of system: a. Open system b. Closed system c. Open system d. Isolated system e. Isolated system |
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| 456. |
`N_(2(g))+2O_(2(g)) rarr 2NO_(2) +X kJ` `2NO_((g))+O_(2(g)) rarr 2NO_(2(g))+Y kJ` The enthalpy of formation of `NO` isA. `(2X-2Y)`B. `X-Y`C. `1//2(Y_X)`D. `1//2(X-Y)` |
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Answer» `N_(2)(g)+2O_(2)(g)rarr 2NO_(2)(g),DeltaH =+X kJ …(i)` `2NO(g) +O_(2)(g) rarr 2NO_(2)(g),DeltaH =+YkJ …(ii)` Substracting equation (ii) from equation (i), we get `N_(2)(g) +O_(2)(g) rarr 2NO(g),DeltaH = X-Y` or `Delta_(f)H^(Theta) (NO) = ((X-Y))/(2) kJ mol^(-1)` |
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| 457. |
Equal volumes of `1M HCI` and `1M H_(2)SO_(4)` are neutralised by `M NaOH` solution and `x` and `y kJ//` equivalent of heat are liberated, respectively. Which of the following relations is correct?A. `x = 2y`B. `x = 3y`C. `x = 4y`D. `x = (1)/(2)y` |
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Answer» Since, `H_(2)SO_(4)` gives `2 "moles" H^(o+), HCI` gives `1"mole" H^(o+)` from 1mole after ionisation. Hence, `H_(2)So_(4)` will release double amount of heat as compared to `HCI`. i.e., `y = 2x or x = (y)/(2)` |
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| 458. |
White phosphorous is less stable than red phosphorus. Mention whether the process of conversion of white phosphorous to red phosphorous is exothermic or endothermic reaction. |
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Answer» Exothermic reaction. |
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| 459. |
A volume of `10m^(3)` of a liquid is supplied with `100 kal` of heat and expands at a constant pressure of `10 atm` to a final volume of `10.2m^(3)`. Calculate the work done and change in internal energy. |
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Answer» Correct Answer - `48 Kcal. 52 Kcal`. Here, `V_(1)= 10m^(3), dQ=100 Kcal = 10^(5)cal.` `P= 10 atm = 10^(6)N//m^(2)` `V_(2)= 10.2 m^(3), U=?, dW=?` `dW=pdV=(10xx10^(5))(10.2-10)= 2xx10^(5)J` `= (2xx10^(5))/(4186)Kcal= 48 Kcal` From first law of thermodynamics, `dU=dQ-dW= 100-48= 52 Kcal` |
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| 460. |
Assertion : If both `DeltaH^(@) and DeltaS^(@)` are positive then reaction will be spontaneous at high temperature. Reason : All processes with positive entropy chang are spontaneous.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C Spontaneity of a process depends upon `DeltaG^(@)` which combines both factors `DeltaH^(@)` and `DetlaS^(@)`. `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` For the spontaneity, `DeltaG^(@)` should be negative. If `DeltaH^(@)` is `+ve , DeltaS^(@)` is +ve then, `DeltaG^(@)` to be `-ve` only when `TDeltaS gt DeltaS` which is possible at high temperature. |
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| 461. |
For the real gas reaction, `2A(g)+B(g)to2C(g) deltaH=-440kJ//mol` If the reaction is carried out in 10 litre rigid vessol, the initial pressure is `50"bar"` bar which decreses to `20"bar"` in the course of reactioin. The change in internal energy for the reaction is:A. `-434kJ`B. `-140kJ`C. `-443kJ`D. `-470kJ` |
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Answer» Correct Answer - B |
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| 462. |
Which of the following is trueA. `C_(P)-C_(v)=R` is applicable for all gasesB. For an ideal gas undergoing adiabatic process, temperature will not always change under normal conditions.C. If `deltaT=0`, process must be isothermalD. If isothermal reversible and irreversible process are started from same initial state to same final pressure then both process ends at same date. |
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Answer» Correct Answer - D |
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| 463. |
1 mole of argon is expanded isothermiocally and irreversably ( not against vaccum) from 10L to 100L. Which of the following is incorrect the process ?A. `DeltaU=0`B. `DeltaH=0`C. Heat supplied(q)=0D. `DeltaT=0` |
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Answer» Correct Answer - C |
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| 464. |
Calculate entropy change when `10mol` of an ideal gas expands reversible and isothermally from an initial volume of `10L` to `100L` at `300K`. |
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Answer» `DeltaS = 2.3030 nR log_(10) ((V_(2))/(V_(1)))` `= 2.303 xx 10 xx 8.314 log_(10) ((100)/(10))` `= 191.24 J K^(-1)` |
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| 465. |
Which of the following are intensive properties?A. VolumeB. EnthalpyC. Refractive indexD. Internal energy |
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Answer» Correct Answer - C Those properties whose value do not depend on the quanitity of size of matter present in the system are known as intensive properties. Volume, ethalpy, and internal energy are extensive proterties. An extensive property is a property whose value depends on the quantity of size of amatter. As a memory aid, remember "i" for intensive and independent. |
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| 466. |
The difference between `Delta H` and `Delta U` for the combustion of methane at `27^(@)C` will be (in `J mol^(-1)`)A. `8.314 xx 27 xx - 3`B. `8.314 xx 300 xx (-3)`C. `8.314 xx 300 xx (-2)`D. `8.314 xx 300 xx1` |
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Answer» Correct Answer - C Thermochemical equaiton for the combustion of methane is `CH_(4) (g) + 2 O_(g) rarr CO_(2) (g) + 2H_(2) O (1)` According to thermodynamics, `Delta H = Delta U + Delta n_(g) RT` or `Delta H - Delta U = Delta n_(g) RT` Where `Delta n_(g) = sum n_(P) (g) - sum n_(R ) (g)` `= (1) - (1 + 2)` `= - 2` `:. Delta - Delta U = (-2) (8.314) (300)` |
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| 467. |
Which of the following is the best description of heat capacity?A. `C = (q)/(Delta T)`B. `C = underset(Delta T rarr 0)(lim) (q)/(Delta T)`C. `C = underset(qrarr 0)(lim) (q)/(Delta T)`D. `C = (Delta q)/(Delta T)` |
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Answer» Correct Answer - B Since that heat capacity varies with temperature, the value of `C` must be considered over a very narrow temperature range. |
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| 468. |
Which of the following is an extensive property?A. Heat capacityB. Molar heat capacityC. Specific heat capacityD. both (2) and (3) |
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Answer» Correct Answer - A The magnitude of heat capacity depends upon the amount of substance being heated. In other cases, the amount of substance is fixed. |
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| 469. |
Calculate enthalpy of ionisation of `OH^(-)` ion. Given: `H_(2)O_((l)) rarr H_((aq))^(+)+OH_((aq))^(-), ? H^(0) =57.32 kJ` `H_(2_((g)))+1/2 O_(2_((g))) rarr H_(2)O_((l)), ? H^(0) =-285.83 kJ` |
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Answer» On adding given equations, `H_(2_((g)))+1/2 O_(2_((g))) rarr H_((aq))^(+)+OH_((aq))^(-)` `DeltaH^(@)=-228.51 kJ` but for `H_((aq))^(+), DeltaH^(@)=0` therefore heat of form ation of `OH^(-)` is `-228.51 kJ`. |
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| 470. |
Which of these reaction has `DeltaS^(@)gt0`?A. `S_(8)(l)toS_(8)`(s,monoclinic)B. `H_(2)(g)+O_(2)(g)toH_(2)O_(2)(aq)`C. `H_(2)(g)+2Ag^(+)(aq)to2H^(+)(aq)+2Ag(s)`D. `PCl_(5)(g)toPCl_(3)(g)+Cl_(2)(g)` |
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Answer» Correct Answer - D |
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| 471. |
A substance has latent heat of vaporisation (at its boiling point 300K) =`3kJ//g`. If molar mass of substance is 40, the molar entropy change for condensation process will be:A. `10J//K`B. `400J//K`C. `-400J//K`D. `-10J//K` |
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Answer» Correct Answer - C |
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| 472. |
Standard emtropy of `X_(2),Y_(2)" and " X_(2)Y_(3)` are 60, 40 and 150 in `J//K` mol `X_(2)+(3)/(2)Y_(2)toX_(2)Y_(3),DeltaH=+30kJ` The temperature at which reaction will attain equilibrium is :A. 250 KB. 1000 KC. 750 KD. 200 K |
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Answer» Correct Answer - B |
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| 473. |
Find the emtropy increment of one mole of carbon dioxide when its absolute temperature increases `n = 2.0` times if the process of heating is (a) isochoric , (b) isobaric. The gas is to be regarded a ideal. |
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Answer» (a) In an isochoric process the entropy change will be `Delta S = int_(T_i)^(T_f) (C_V dT)/(T) = C_V 1n (T_f)/(T_i) = C_V 1n n = (R 1n n)/(gamma -1)` For carbon dioxide `gamma = 1.30` so, `Delta S = 19.2 "Joule"//.^@K - "mole"` (b) For an isobaric process, `Delta S = C_p 1n(T_f)/(T_i) = C_p 1n n = (gamma R 1n n)/(gamma -1)` =`25 "Joule"//.^@K - "mole"`. |
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| 474. |
In which case will the efficiency of a carnot cycle be higher : When the hot body temperature is increased by `Delta T`, or when the cold body temperature is decreased by the same magnitude ? |
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Answer» The efficiency is given is `eta = (T_1 - T_2)/(T_1), T_1 gt T_2` Now in the two cases the efficiencies are `eta_h = (T_1 + Delta T - T_2)/(T_1 + Delta T), T_1` encreased `eta_l = (T_1 - T_2 + Delta T)/(T_1), T_2` decreased Thus `eta_h lt n_l`. |
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| 475. |
Find the force exerted on a particle by a uniform field if the concentrations of these particles at two levels separated by the distance `Delta h = 30 cm` (along the field) differ by `eta = 2.0` times. The temperature of the system is equal to `T = 280 K`. |
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Answer» If the force exerted is `F` then the law of variation of concentration with height reads `n(Ƶ) = n_0 e^(- F Ƶ//kT)` So, `eta = e^(F Delta h//kT)` or `F = (kT 1n eta)/(Delta h) = 9 xx 10^-20 N`. |
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| 476. |
From the conditions of the foregoing problem find : (a) the number of molecules whose potential energy lies within the internal from `U` to `U + dU` , (b) the most probable value of the potential energy of a molecule , compare this value with the potential energy of a molecule located at its most probable distance from the centre of the field. |
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Answer» Write `U = ar^2`or `r = sqrt((U)/(a))`, so `dr = sqrt((1)/(a)) (dU)/(2 sqrt(U)) = (dU)/(2 sqrt(a U))` so `dN n_0 4 pi(U)/(a) (dU)/(2 sqrt(a U)) exp ((U)/(kT))` =`2 pi n_0 a^(-3//2) U^(1//2) exp ((-U)/(kT)) dU` The most probable value of `U` is given by `(d)/(dU) ((dN)/(dU)) = 0 ((1)/(2 sqrt(u)) -U^(1//2)/(kT)) exp ((-U)/(kT))`or `U_(pr) = (1)/(2) kT` From `2.111 (b)`, the potential energy at the most probable distance is `kT`. |
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| 477. |
From the conditions of the foregoing problem find the number of molecules reaching a unit area of a wall with the velocities in the interval from `v` to `v + dv` per unit time. |
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Answer» Simiarly the number of molecules reaching the wall (per unit area of the wall with velocities in the interval `v` to `v + dv` per unit time is `dv = int_(theta = 0)^(theta = pi//2) dn (v) (d Omega)/(4 pi) v cos 0` =`int_(theta = 0)^(theta = pi//2) n((m)/(2 pi kT))^(3//2) e^(-mv^2//kT) v^3 dv sin theta cos theta d theta xx 2pi` =`n pi ((m)/(2 pi kT))^(3//2) e^(-mv^2//2kT) v^3 dv`. |
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| 478. |
What fraction of molecules in a gas at a temperature `T` has the kinetic energy of translational motion exceeding `epsilon_0` if `epsilon_0 gt gt k T` ? |
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Answer» `(Delta N)/(N) = (2)/(sqrt(pi)) (kT)^(-3//2) int_0^oo sqrt(epsilon) e^(-epsilon//kT) d epsilon` `~~ (2)/(sqrt(pi)) (kT)^(-3//2) sqrt(epsilon_0) int_(epsilon_0)^oo e^(-epsilon//kT) d epsilon(epsilon_0 gt gt kT)` =`(2)/(sqrt(pi)) (kT)^(-3//2) sqrt(epsilon_0) kT e^(-epsilon_0//kT) = 2 sqrt((epsilon_0)/(pi kT)) e^(-epsilon_0//kT)` (In evaluating the integral, we have taken out `sqrt(epsilon)` as `sqrt(epsilon_0)` since the integral is dominated by the lower limit). |
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| 479. |
Calculate the most probable, the mean, and the root mean square velocities of a molecule of a gas whose density under standard atmospheric pressure is equal to `rho = 1.00 g//1`. |
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Answer» `v_p = sqrt((2 kT)/(m)) = sqrt((2 RT)/(M)) = sqrt((2 p)/(rho)) = 0.45 km//s`, `v_(av) = sqrt((8)/(pi) (p)/(rho)) = .51 km//s` and `v_(rms) = sqrt((3 p)/(rho)) = 0.55 km//s`. |
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| 480. |
Determine the gas temperature at which (a) the root mean square velocity of hydrogen molecules exceeds their most probable velocity by `Delta v = 400 m//s` , (b) the velocity distribution function `F (v)` for the oxygen molecules will have the maximum value at the velocity `v = 420 m//s`. |
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Answer» (a) `v_(rms) = v_p = (sqrt(3) - sqrt(2)) sqrt((k T)/(m)) = Delta v`, `T = (m)/(k) ((Delta v)/((sqrt(3) - sqrt(2))))^2//K = 384 K` (b) Clearly `v` is the most probable speed at this temperature. So `sqrt((2 kT)/(m)) = v` or `T = (mv^2)/(2 k) = 342 K`. |
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| 481. |
In the case of gaseous nitrogen find : (a) the temperature at which the velocities of the molecules `v_1 = 300 m//s` and `v_2 = 600 m//s` are associated with equal values of the Maxwell distribution friction `F (v)` , (b) the velocity of the molecules `v` at which the value of the Maxwell distribution function `F (v)` for the temperature `T_0` will be the same as that for the temperature `eta` times higher. |
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Answer» (a) We have, `(v_1^2)/(v_p^2) e^(-v_1^2//v_p^2) = (v_2^2)/(v_p^2) e^(-v_2^2//v_p^2)` or `((v_1)/(v_2))^2 = e^(v_1^2 - v_2^2//v_p^2)` or `v_p^2 = (2 kT)/(m) = (v_1^2 - v_2^2)/((1 n v_1^2//v_2^2))` So `T = (m(v_1^2 - v_2^2))/(2 k 1n (v_1^2)/v_2^2) = 330 K` (b) `F (v)=(4)/(sqrt(pi)) (v^2)/(v_p^2) e^(-v^2//v_p^2) xx (1)/(v_p) ((1)/(v_p)` comes from `F(v) dv = df (u), du = (dv)/(v_p))` Thus `(v^2)/(v^3) e^(-v^2//v^2) p_1 = v^2//v_(p_2) e^(-v_2//v_(2_2)) v_(p_1)^2 = (2 kT_0)/(m), v_(p_2)^2 = (2 kT_0)/(m) eta ` now `e -(mv^2)/(2 kT_0)(1 - (1)/(eta)) = (1)/(eta^(3//2))` or `(mv^2)/(2 kT_0)(1 - (1)/(eta)) = (3)/(2) 1n eta` Thus `v = sqrt((3 kT_0)/(m)) sqrt((1 n eta)/(1 - 1//eta))`. |
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| 482. |
Find the fraction of gas molecules whose velocities differ by less than `delta eta = 1.00 %` from the value of (a) the most probable velocity , (b) the root mean square velocity. |
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Answer» (a) The formula is `df (u) = (4)/(sqrt(pi)) u^2 e^(-u^2) du`, where `u = (v)/(v_p)` Now prob `(|v - v_p|/(v_p) lt delta eta) = int_(1 - delta eta)^(1 + delta eta) df (u)` =`(4)/(sqrt(pi)) e^(-1) xx 2 delta eta = (8)/(sqrt(pi) e) delta eta = 0.0166` (b) `Prob (|(v - v_rms)/(v_(rms))| lt delta eta) = Prob(|(v)/(v_p) - v_(rms)/(v_p)|lt delta eta (v_(rms))/(v_p))` =`Prob (|u - sqrt((3)/(2))| lt sqrt((3)/(2)) delta eta)` `sqrt((3)/(2)) + sqrt((3)/(2)) delta eta` =`int (4)/(sqrt(4)) u^2 e^(- u^2) du` `sqrt((3)/(2)) - sqrt((3)/(2)) delta eta` =`(4)/(sqrt(pi)) xx (3)/(2) e^(-3//2) xx 2 sqrt((3)/(2)) delta eta = (12 sqrt(3))/(sqrt(2 pi)) e^(-3//2) delta eta = 0.0185`. |
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| 483. |
What fraction of monatomic molecules of a gas a in a thermal equilibrium possesses kinetic energies differenting from the mean value by `delta eta = 1.0 %` and less ? |
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Answer» The mean kinetic energy is `lt epsilon gt = int_0^oo epsilon^(3//2) e^(- epsilon//kT)// int_0^oo epsilon^(1//2) e^(-epsilon//kT) d epsilon = kT (Gamma(5//2))/(Gamma(3//2)) = (3)/(2) kT` Thus `(delta N)/(N) = int_((3)/(2)kT(1 - delta eta))^((3)/(2)(1 + delta eta)kT) (2)/(sqrt(pi)) (kT)^(-3//2) e^(-epsilon//kT) epsilon^(1//2) epsilon` =`(2)/(sqrt(pi)) e^(-3//2) ((3)/(2))^(3//2) 2 delta eta = 3 sqrt((6)/(pi)) e^(3//2) delta eta` If `delta eta = 1 %` this gives `0.9 %`. |
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| 484. |
At what temperature of a nitrogen and oxygen mixture do the most probable velocities of nitrogen and oxygen molecules differ by `Delta v = 30 m//s` ? |
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Answer» `v_(pN) = sqrt((2 kT)/(m N)) = sqrt((2 RT)/(M_N)), v_(p_0) = sqrt((2 RT)/(M_0))` `v_(P_N) - v_(p_0) = Delta v = sqrt((2 RT)/(M_N))(1 - sqrt((M_N)/(M_0)))` `T = (M_N (Delta v)^2)/(2 R(1 - sqrt((M_N)/(M_0)))^2) = (m_N(Delta v)^2)/(2k(1 - sqrt((m_N)/(M_0)))^2) = 363 K`. |
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| 485. |
At what temperature of a gas will the number of molecules, whose velocities fall within the given interval from `v` to `v + dv`, be the greatest ? The mass of each molecule is equal to `m`. |
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Answer» `dN (v) = (N 4)/(sqrt(pi)) (v^2 dv)/(v_p^3) e^(-v^2//v_p^2)` For a given `v` to `v + dv` (i.e., given `v` and `dv`) this is maximum when `(delta)/(delta v_p) (dN(v))/(N v^2 dv) = 0 = (-3 v_p^-4 + (2 v^2)/(v_p^6)) e ^(-v^2//v_p^2)` or, `v^2 = (3)/(2) v_p^2 = (3 kT)/(m)`. Thus `T = (1)/(3) (mv^2)/(k)`. |
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| 486. |
A gas consists of molecules of mass `m` and is at a temperature `T`. Making use of the Maxwell velocity distribution function, find the corresponding distribution of the molecules over the kinetic energies `epsilon`. Determine the most probable value of a gas the kinetic energy `epsilon_p`. Does `epsilon_p` correspond to the most probable velocity ? |
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Answer» `dN (v) = N((m)/(2 pi kT))^(3//2) e^(-mv^2//2kT) 4 pi v^2 dv = dN(epsilon) = (dN(epsilon))/(d epsilon) d epsilon` or, `(dN(epsilon))/(d epsilon) = N((m)/(2 pi kT))^(3//2) e^(-mv^2//2RT) 4 pi v^2 (dv)/(d epsilon)` Now, `epsilon = (1)/(2) mv^2` so `(dv)/(d epsilon) = (1)/(mv)` or, `(dN(epsilon))/(d epsilon) = N((m)/(2 pi kT))^(3//2) e^(- epsilon//kT) 4 pi sqrt((2 epsilon)/(m))(1)/(m)` =`N (2)/(sqrt(pi))(kT)^(-3//2) e^(-epsilon//kT) epsilon^(1//2)` i.e., `dN (epsilon) = N(2)/(sqrt(pi)) (kT)^(-3//2) e^(- epsilon//kT) d epsilon` The most probable kinetic energy is given from `(d)/(d epsilon) (dN(epsilon))/(d epsilon) = 0` or, `(1)/(2) epsilon^(-1//2) e^(-epsilon//kT) - (epsilon^(1//2))/(kT) e^(epsilon//kT) = 0` or `epsilon = (1)/(2) kT = epsilon_(p r)` The corresponding velocity is `v = sqrt((kT)/(m)) = v_(pr)`. |
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| 487. |
The temperature of a hydrogen and helium mixture is `T = 300 K`. At what value of the molecules velocity `v` will the Maxwell distribution function `F (v)` yield the same magnitude for both gases ? |
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Answer» `(v^2)/(v_(P_H)^3) e^(-v^2//v_(P_H)^2) = (v^2)/(v_(P_(H_e))^3) e^(-v^2//v_(P_(H_e))^2)` or `e^(v^2((.^m He)/(2 kT) -(.^m H)/(2 kT)))=((.^mHe)/(.^m H))^(3//2)` `v^2 = 3kT (1 n^m He//^m H)/(m_(H_e) -m_H)`, Putting the values we get `v = 1.60 km//s`. |
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| 488. |
Which correctly represents the entropy (s) of an isolated system during a process.A. B. C. D. |
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Answer» Correct Answer - C `Delta S gt 0` for spontaneity and `Delta S = 0` at equilibrium |
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| 489. |
In isothermal process if heat is evolved from the system thenA. Internal energy remains constantB. Change in internal energy is zeroC. Change in entropy is zeroD. Change in free energy is zero |
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Answer» Correct Answer - A::B `U = f(T)` for ideal gas |
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| 490. |
The change in free energy accompainied by the isothermal reversible expansion of 1 mol of an ideal gas when it doubles its volume is `Delta G_(1)`. The change in free energy accompanied by sudden isothermal irreversible doubling volume of 1 mole of the same gas is `Delta G_(2)`. Ratio of `Delta G_(1)` and `Delta G_(2)` isA. 1B. `(1)/(2)`C. `-1`D. `-(1)/(2)` |
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Answer» Correct Answer - A free energy is a state function .i.e. indpendent on the path followed |
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| 491. |
Statement-1:A liquid crystallises into a solid and is accompainied by decrease in entropy. Statement -2: In crystals, moles cules organise in an ordered manner.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true and R is not the correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - A | |
| 492. |
A change in the free energy of a system at constant temperature and pressure will be: `Delta_(sys)G = Delta_(sys)H -T Delta_(sys)S` At constant temperature and pressure `Delta_(sys) G lt 0` (spontaneous) `Delta_(sys)G = 0` (equilibrium) `Delta_(sys)G gt 0` (non-spontaneous) If both `DeltaH` and `Deltas` are negative, the reaction will be spontaneousA. At high temperatureB. At all temperaturesC. At low temperatureD. At high pressure |
| Answer» `DeltaC = DeltaH - T Deltas` | |
| 493. |
One mole of ice is converted into water at `273K`. The entropies of `H_2O(s)` and `H_2O(l)` are `38.20` and `60.01Jmol^(-1)K^(-1)` respectively. The enthalpy change for the conversion is:A. `59.54 J mol^(-1)`B. `5954 J mol^(-1)`C. `595.4 J mol^(-1)`D. `320.6 J mol^(-1)` |
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Answer» Correct Answer - B `DeltaG=DeltaH-TDeltaS`, at equilibrium `DeltaG=0 :. DeltaH=TDeltaS` Or `DeltaH=273xx(60.01-38.20)` `=5954.13 J mol^(-1)` |
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| 494. |
One mole of ice is converted into water at 273 K. The entropies of `H_(2)O(s)` and `H_(2)O(l)` are 38.20 and 60.01 J `mol^(-1)K^(-1)` respectively. Calculate the enthalpy change for this conversion a ? |
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Answer» At 273 K ice and water are in equilibrium with each other and at equilibrium `Delta G = 0` `because Delta G = Delta H-T Delta S " " Delta H=T Delta S =273xx(60.01-38.20)=5954.13 J mol^(-1)` |
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| 495. |
One mole of a gas is heated at constant pressure to raise its temperature by `1^(@)C`. The work done in joules isA. `-4.3`B. `-8.314`C. `-16.62`D. Unpredicatable |
| Answer» `w =- n R DetaT or w =- p DeltaV =- 1 xx 8.314 xx1 =- 8.314J` | |
| 496. |
Assertion (A). A liquid crystallises into a solid and is accompanied by decrease in entropy. Reason (R). In crystals, molecules organise in an ordered manner.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not the correct explanation of AC. A is true but R is falseD. A is false but R is true |
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Answer» Correct Answer - A Both assertion and reason are true and reason is the correct explanation of assertion when a liquid crytallises, entropy decrease because in crystalline form the molecules are more ordered as compared to the liquid |
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| 497. |
Which of these coversios has a poitive `DeltaS^(@)`? (p)combustion of charcoal (Q) condensation of `Br_(2)(g)` (R ) precipitation ofAgCl(s)A. P onlyB. QonlyC. R onlyD. Qand R only |
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Answer» Correct Answer - a |
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| 498. |
One mole of ice is melted at `0^(@)C` and then is heated to `100^(@)C`. What is the difference in entropies of the steam and ice? The heats of vaporisation and fusion are `540 cal g^(-1) and 80 cal g^(-1)` respectively. Use the average heat capacity of liquid water as 1cal `g^(-1) degree^(-1)`A. `18((80)/(373)+(540)/(273)+In (373)/(273))`B. `18((80)/(373)+(540)/(373)+In (373)/(273))`C. `((80)/(273)+(540)/(373)+In (373)/(373))`D. `((80)/(273)+(540)/(373)+100)` |
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Answer» Correct Answer - B |
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| 499. |
3.0 moles of ideal gas is heated at constant pressure from `27^(@)C` to `127^(@)C`. Then the work expansion of gas isA. `-2.494 kJ`B. `+2.494 kJ`C. `-10.5 kJ`D. `+10.5 kJ` |
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Answer» Correct Answer - A `W= - nR DeltaT` |
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| 500. |
Choose the correct statement(s).A. For phase transformation of liquid water at 1 atm , 373 K to water vapour at 1 atm, 373 K , `DeltaG` will be zero .B. Two solid blocks of same material and same mass having different temperature are kept in an isolated system then entropy of system must increase.C. Intensive properties are not additive in nature while extensive properties are additive in nature.D. No cyclic process is possible in which the sole result is absorption of heat from a hot reservoir and its complete conversion into work |
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Answer» Correct Answer - a,b,c,d |
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