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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.
| 1851. |
These question consists of two statements each, printed as Statement-I and Statement-II . While answering these Question you are required to choose any one of the following four responses. Statement-I : There is no change in enthalpy of an ideal gas during compression at constant temperature . Statement -II : Enthalpy of an ideal gas is a function of temperature and pressure.A. If both Statement -I & Statement are True & the Statement-II is a correct explanation of the Statement -IB. If both Statement -I & Statement are True & the Statement-II is not a correct explanation of the Statement -IC. If Statement-I is True but the Statement-II is False.D. If Statement-I is True but the Statement-II is True . |
| Answer» Correct Answer - C | |
| 1852. |
An ideal gas `(C_(P)//C_(V)=gamma)` is expanded so that the amount of heat transferred to the gas the is equal to the decrease in its internal energy . What is the magnitude of work performed by one mole of the gas when its volume increase eight times if the initinal temperature of the gas is ` 300 K `? `C_(V)` for the gas is `1.5R` .(`R=2 cal//mol//K`)A. 900 calB. 450 calC. 1247.7 calD. 623.8 cal |
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Answer» Correct Answer - A `T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1//2)=300 xx((1)/(8))^((5)/(3)-1//2)=150 K` Now, `w= DeltaU - q = 2DeltaU = 2 n C_(v) DeltaT` `= 2 xx 1 xx 1.5 R xx(150 - 300) = - 900 al.` |
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| 1853. |
What is temperature? Explain. |
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Answer» The temperature of a body is a quantitative measure of the degree of hotness or coolness of the body. According to the kinetic theory of gases, it is a measure of the average kinetic energy per molecule of the gas. Temperature difference determines the direction of flow of heat from one body to another or from one part of the body to the other. Its SI unit is the kelvin (K). |
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| 1854. |
What is heat? Explain. |
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Answer» When two bodies are in thermal contact with each other, there is a transfer of energy from the body at higher temperatures to the body at lower temperatures. The energy in the transfer is called heat. Also when two parts of a body are at different temperatures, there is a transfer of energy from the part at a higher temperature to the other part. The SI unit of heat is the joule. [Note: Count Rumford [Benjamin Thompson] (1753-1814) Anglo-American adventurer, social reformer, inventor, and physicist, measured the relationship between work and heat. When he visited Arsenal in Munich, he found that a tremendous amount of heat was produced in a short time when a brass cannon was being bored. He found that even with a blunt borer a lot of heat can be produced from a piece of metal. At that time it was thought that heat consists of a fluid called caloric. Rumford’s experiments showed that caloric did not exist and heat is the motion of the particles of a body. He measured the relation between work done and corresponding heat produced. The result was not accurate, but important in the development of thermodynamics.] |
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| 1855. |
What is thermodynamics ? |
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Answer» Thermodynamics is the branch of physics that deals with the conversion of energy (including heat) from one form into another, the direction of energy transfer between a system and its environment with the resulting variation in temperature, in general, or changes of state, and the availability of energy to do mechanical work. |
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| 1856. |
What is meant by thermal equilibrium ? What is meant by the expression “two systems are in thermal equilibrium” ? |
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Answer» A system is in a state of thermal equilibrium if there is no transfer of heat (energy) between the various parts of the system or between the system and its surroundings. Two systems are said to be in thermal equilibrium when they are in thermodynamic states such that, if they are separated by a diathermic (heat conducting) wall, the combined system would be in thermal equilibrium, i.e., there would be no net transfer of heat (energy) between them. [Note : It is the energy in transfer that is called the heat.] |
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| 1857. |
State the zeroth law of thermodynamics |
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Answer» Zeroth law of thermodynamics : If two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other. [Note :The zeroth law is fundamental to the other laws of thermodynamics. That this law is assumed by the other laws of thermodynamics was realized much later. This law has no single discoverer. It was given the status of a law, following the suggestion by R. H. Fowler (in 1931), only after the first, second and third laws were named.] |
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| 1858. |
In an isothermal process for an ideal gasA. `DeltaQ=0`B. `DeltaW=0`C. `DeltaU=0`D. `DeltaV=0` |
| Answer» Correct Answer - C | |
| 1859. |
What do you mean by adiabatic process ? |
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Answer» A process, in which pressure, volume and temperature of the system change, but there is no exchange of heat between the system and its surroundings, is called an adiabatic process. |
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| 1860. |
Define the term isothermal process. |
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Answer» A process in which pressure and volume of the system change at constant temperature is called an isothermal process. |
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| 1861. |
Define reversible process. |
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Answer» A process is said to be reversible when the various stages of an operation to which it is subjected, can be traversed back in the opposite direction in such a way that the substance passes through exactly the same conditions at every step in the reverse process as in the direct process as in the direct process. |
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| 1862. |
Entropy is a measure of randomess of system. When a liquid is converted to the vapour state entropy of the system increases. Entropy in the phase transformation is calculated using `Delta S = (Delta H)/(T)` but in reversible adiabatic process `Delta`S will be zero. The rise in temperature in isobaric or isochoric process increases the randomness of system, which is given by `Delta S = "2.303 n C log"((T_(2))/(T_(1)))` `C = C_(P) or C_(V)` The change in entropy when 1 mole `O_(2)` gas expands isothermally and reversibly from an initial volume 1 litre to a final volume 100 litre at `27^(@)C`A. `20.5JK^(-1(mol^(-1)`B. `38.29 JK^(-1)mol^(-1)`C. `42.50 JK^(-1)mol^(-1)`D. `50.65 JK^(-1)mol^(-1)` |
| Answer» Correct Answer - B | |
| 1863. |
Entropy is a measure of randomess of system. When a liquid is converted to the vapour state entropy of the system increases. Entropy in the phase transformation is calculated using `Delta S = (Delta H)/(T)` but in reversible adiabatic process `Delta`S will be zero. The rise in temperature in isobaric or isochoric process increases the randomness of system, which is given by `Delta S = "2.303 n C log"((T_(2))/(T_(1)))` `C = C_(P) or C_(V)` Entropy change in a reversible adiabatic process isA. ZeroB. Always positiveC. Always negativeD. Sometimes positive and sometimes negative |
| Answer» Correct Answer - A | |
| 1864. |
Heat of reaction is defined as the amount of heat absorbed or evolved at a given temperaturewhen the reactants have combined to form the products is represented by a balanced chemcial equation. If the heat is denofed by q then the numerical value of q depends on the manner in which the reaction is performed for the two methods of conducting chemical reactions in calorimeters. Constant volume W = 0 and `q_(v) = Delta E` Bomb calorimeter Constant pressure W = - V `Delta`P, therefore `q_(P) = Delta E + P Delta V rar (V. Delta P)` When maltose `C_(12)H_(22)O_(11)(s)` burns in a calorimetric bomb at 298 K yielding carbon dioxide and water, the heat of combustion is -1350 kcal/mol, the heat of combustion of maltose at constant pressure will beA. `-2650` kcal/molB. `-675` kcal/molC. `-1350` kcal/molD. `-1100` kcal/mol |
| Answer» Correct Answer - C | |
| 1865. |
Heat of reaction is defined as the amount of heat absorbed or evolved at a given temperaturewhen the reactants have combined to form the products is represented by a balanced chemcial equation. If the heat is denofed by q then the numerical value of q depends on the manner in which the reaction is performed for the two methods of conducting chemical reactions in calorimeters. Constant volume W = 0 and `q_(v) = Delta E` Bomb calorimeter Constant pressure W = - V `Delta`P, therefore `q_(P) = Delta E + P Delta V rar (V. Delta P)` The heat capacity of a bomb calorimeter is 300 JK When 0.16 gm of methane was burnt in this calorimeter the temperature rose by `3^(@)`C. The value of `Delta`U per mole will beA. 100 kJB. 90 kJC. 900 kJD. 48 kJ |
| Answer» Correct Answer - B | |
| 1866. |
For the isothermal reversible expansion of an ideal gas from volume `V_(1)` to volume `V_(2)`, the work done is given by `w_(rev)="…................."` whereas for the irreversible expansion , it is given by `w_("irrev") = "…........."`. |
| Answer» `w_("rev") = -2.303 n RT log. (V_(2))/( V_(1)), w_("irrev") =-P( V_(2) -V_(1)) = - P DeltaV` | |
| 1867. |
A reversible reaction has `DeltaG^(@)` negative for forward reaction ? What will be sign of `DeltaG^(@)` for backwork reaction ? |
| Answer» Correct Answer - Negative. | |
| 1868. |
A thermodynamic system consists of a cylinder-piston attangement with ideal gas in it. It goes from the state `i` to the state `f` as shwon in the figure. The work done by gas during the process is A. ZeroB. NegativeC. PositibeD. Nothing can be predicted |
| Answer» Correct Answer - Zero | |
| 1869. |
Match the following `{:(,"Column-I",,"Column-II"),(,("Enthalpy change in kcal"),,("Neutralisation")),((A),lt "13.7 kcal",(p),underset(("1 mol"))(HCI)+underset(("1 mol"))(NaOH)),((B),= "13.7 kcal",(q),underset(("1 mol"))(HF)+underset(("1 mol"))(NaOH)),((C),"gt 13.7 kcal",(r),underset(("1 mol"))(NH_(4)OH)+underset(("1 mol"))(HCI)),((D),"= 27.4 kcal",(s),underset(("2 mol"))(NaOH)+underset(("1 mol"))(H_(2)SO_(4))),(,,(t),underset(("1 mol"))(NaOH)+underset(("1 mol"))(CH_(3)COOH)):}` Assume heat of neutralisation of strong acid with strong base is 13.7 kcal. |
| Answer» Correct Answer - A(r, t), B(p), C(q, s), D(s) | |
| 1870. |
If a system absorbs heat and expands through a volume `DeltaV` against external pressure, P accompanied by increas in internal energy, `Delta U `, then `q = "…..............."`. |
| Answer» `q = DeltaU +PDeltaV` | |
| 1871. |
The heat of neutralisation of aqueous hydrochloric acid by `NaOH` si `x kcal mol^(-1)of HCI`. Calculate the heat of neutralisation per mol of aqueous acetic acid.A. `0.5 xkcal`B. `xkcal`C. `2x kcal`D. Cannot be calculated from the given data |
| Answer» It cannot be determined. | |
| 1872. |
Under the same conditions, how many `mL` of `M KOH` and `0.5M H_(2)SO_(4)`solutions, respectively, when mixed to form a total volume of `100mL`, produces the highest rise in temperature?A. `67,33`B. `33,67`C. `40,60`D. `50,50` |
| Answer» `50:50`, because `H_(2)SO_(4)` is biprotic acid. It liberates `2H^(o+)` per mole of acid. | |
| 1873. |
Internal energy change of a system is the heat absorbed or evolved at constant `"…............."` whereas enthalpy change is that at constant `"…..............."`. |
| Answer» Correct Answer - volume, pressure | |
| 1874. |
Why for predicting the spontaneity of a reaction , free energy criteria is better thanthe entropy criteria ? |
| Answer» Criteria of free energy change is better because it requires free energy change of the system only whereas the entropy change requires the total entropy change of the system and the surroundings. | |
| 1875. |
A gas expands by `0.5L` against a constant pressure of `1 atm`. Calculate the work done in joule and calorie. |
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Answer» Work `=- P_(ex) xx` Volume change `= - 1xx 0.5 =- 0.5L atm` `=- 0.5 xx 101.328J =- 50.664J` `0.5L tam =- 0.5 xx 24.20 cal =- 12.10cal` |
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| 1876. |
A vessel contains `N` molecules of an ideal gas. Dividing mentally the vessel into two halves `A` and `B`, find the probability that the half `A` contains `n` molecules. Consider the cases when `N = 5` and `n = 0,1,2,3,4,5`. |
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Answer» The probabilities that the half `A` contains `n` molecule is `N_(C_n) xx 2^-N = (N!)/(n!(N - n)!) 2^-N`. |
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| 1877. |
A vessel of volume `V_0` contains `N` molecule of an ideal gas. Find the probability of `n` molecules getting into a certain separated part of the vessel of volume `V`. Examine, in particle, the case `V = V_0//2`. |
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Answer» The probability of one molecule being confined to the marked volume is `p = (V)/(V_0)` We can choose this molecule in many `(N_(C_(1)))` ways. The probability that `n` molecules get confined to the marked volume is cearly `N_(C_n) p^n (1 - p)^(N - n) = (N !)/(n!(N - n)!) p^n (1 - p)^(N - n)`. |
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| 1878. |
One mole of an ideak gas consisting of monatomic molecules is enclosed in a vessel at a temperature `T_0 = 300 K`. How many times and in what way will the statistical weight of this system (gas) vary if it heated isochorically by `Delta T = 1.0 K` ? |
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Answer» For a monoatomic gas `C_V = (3)/(2) R` per mole The entropy change in the process is `Delta S = S - S_0 = int_(T_0)^(T_0 + Delta T) C_V (dT)/(T) = (3)/(2) R 1n (1 + (Delta T)/(T_0))` N ow from the Boltzmann equation `S = k 1n Omega` `(Omega)/(Omega_0) = e^((S - S_0)//k) = (1 + (Delta T)/(T_0))^((3 N_A)/(2)) = (1 + (1)/(300))^((3 xx 6)/(2) xx 10^23) = 10^23 xx 10^21` Thus the statistical weight increases by this factor. |
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| 1879. |
A vertical capillary is brought in contact with the water surface. What amount of heat is liberated while the water rises along the capillary ? The wetting is assumed to be complete, the surface tension equals `alpha`. |
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Answer» If the liquid rises to a height `h`, the energy of the liquid column becomes `E = rho g pi r^2 h. (h)/(2) - 2pi rh alpha = (1)/(2) rho g pi (r h - 2(alpha)/(rho g))^2 - (2 pi alpha^2)/(rho g)` This is minimum when `rh = (2 alpha)/(rho g)` and that is relevant height to which water must rise. At this point, `E_(min) = - (2 pi alpha^2)/(rho g)` Since `E = 0` in the absence of surface tension a heat `Q = (2 pi alpha^2)/(rho g)` must have been liberated. |
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| 1880. |
Find the capillary pressure (a) in mercury droplets of diameter `d = 1.5 mu m` , (b) inside a soap bubble diameter `d = 3.0 mm` if the surface tension of the soap water solution is `alpha = 45 mN//m`. |
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Answer» (a) `Delta p = alpha ((1)/(d//2) + (1)/(d//2)) = (4 alpha)/(d)` =`(4 xx 490 xx 10^-3)/(1.5 xx 10^-6) (N)/(m^2) = 1.307 xx 10^6 (N)/(m^2) = 12` atmosphere (b) The soap bubble has two surfaces so `Delta p = 2 alpha ((1)/(d//2) + (1)/(d//2)) = (8 alpha)/(d)` =`(8 xx 45)/(3 xx 10^-3) xx 10^-3 = 1.2 xx 10^-3` atmosphere. |
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| 1881. |
Find the free energy of the surface layer of a mercury droplet of diameter `d = 1.4 mm` , (b) a soap bubble of diameter `d = 6.0 mm` if the surface tension of the soap water solution is equal to `alpha = 45 mN//m`. |
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Answer» (a) The free energy per unit area being `alpha`, `F = pi alpha d^2 = 3 mu J` (b) `F = 2 pi alpha d^2` because the soap bubble has two surfaces. Substitution gives `F = 10 mu J`. |
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| 1882. |
Find the increment of the free energy of the surface layer when two identical mercury droplets, each of diameter `d = 1.5 mm`, merge isothermally. |
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Answer» When two mercury drops each of diameter `d` merge, the resulting drop has diameter `d_1` where `(pi)/(6) d_1^3 = (pi)/(6) d^3 xx 2` or, `d_1 = 2^(1//3) d` The increase in free energy is `Delta F = pi 2^(2//3) d^2 alpha-2 pi d^2 alpha = 2pid^2 alpha(2^(-1//3) -1) = - 1.43 mu J`. |
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| 1883. |
A soap bubble of radius `r` is inflated with an ideal gas. The atmospheric pressure is `p_0`, the surface tension of the soap water solution is `alpha`. Find the difference between the molar heat capacity of the gas during its heating inside the bubble and the molar heat capacity of the gas under constant pressure, `C - C_p`. |
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Answer» When heat is given to a soap bubble the temperature of the air inside rises and the bubble expands but unless the bubble bursts, the amount of air inside does not change. Further we shall neglect the variation of the surface tension with temperature. Then from the gas equations `(p_0 + (4 alpha)/(r))(4 pi)/(3) r^3 = v R T, v = Constant` Differentiating `(p_0 + (8 alpha)/(3 r))4 pi r^2 dr = v R dT` or `dV = 4 pi r^2 dr = (vRdT)/(p_0 + (8 alpha)/(3 r))` Now from the first law `₫ Q = v CdT = v C_V dT + (v R dT)/(p_0 + (8 alpha)/(3 r)).(p_0 + (4 alpha)/( r))` or `C = C_V + R (p_0 + (4 alpha)/(r))/(p_0 (8 alpha)/(3 r))` using `C_p = C_V + R, C = C_p + ((1)/(2) R)/(1 + (3 p_0 r)/(8 alpha))`. |
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| 1884. |
A saturated water vapour is contained in a cylindrical vessel under a weightless piston at a temperature `t = 100 ^@C`. As a result of a slow introduction of the piston a small fraction of the vapour `Delta m = 0.70 g` gets condensed. What amount of work was perfomed over the gas ? The vapour is assumed to be ideal, the volume of the liquid is to be neglected. |
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Answer» The condensation takes place at constant pressure and temperature and the work done is `p Delta V` where `Delta V` is the volume of the condensed vapour in the vapour phase. It is `p Delta V = (Delta m)/(M) RT = 120.6 J` where `M = 18 gm` is the molecular weight of water. |
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| 1885. |
Assertion: Reversible systems are difficult to find in real world. Reason: Most processes are dissipative in nature.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of t he assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 1886. |
Under which of the following conditions is the relation, `Delta H = Delta E + P Delta V` valid for a system :-A. Constant pressureB. Constant temperatureC. Constant temperature and pressureD. Constant temperature, pressure and composition |
| Answer» Correct Answer - A | |
| 1887. |
For the case of nitrogen under standard conditions find : (a) the mean number of collisions experienced by each molecule per second , (b) the total number of collisions occurring between the molecules within `1 cm^3` of nitrogen per second. |
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Answer» (a) `v = (1)/(tau) = (1)/(lamda//lt v gt) = (lt v gt)/(lamda)` =`sqrt(2) pi d^2 n lt v gt = .74 xx 10^10 s^-1 (sec 2.223)` (b) Total number of collisions is `(1)/(2) nv ~~ 1.0 xx 10^29 s cm^-3` Note, the factor `(1)/(2)`. When two molecules collide we must not count it twice. |
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| 1888. |
How many times does the mean free path of nitrogen molecules exceed the mean distance between the molecules under standard conditions ? |
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Answer» The mean distance between molecules is of the order `((22.4 xx 10^-3)/(6.0 xx 10^23))^(1//3) = ((224)/(6))^(1//3) xx 10^-9 metres ~~ 3.34 xx 10^-9 metres` This is about `18.5` times smaller than the mean free path calculated in `2.223(a)` above. |
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| 1889. |
Oxygen is enclosed at the temperature `0 ^@C` in a vessel with the characteristic dimension `l = 10 mm` (this is the linear dimension determining the character of a physical process in question). Find : (a) the gas pressure below which the mean free path of the molecules `lamda gt l` , (b) the corresponding molecular concentration and the mean distanace between the molecules. |
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Answer» (a) `lamda gt l` if `p lt (kT)/(sqrt(2) pi d^2 l)` Now `(kT)/(sqrt(2) pi d^2 l)` for `O_2 of O is 0.7 Pa` (b) The corresponding `n` is obtained by dividing by `kT` and is `1.84 xx 10^20` per `m^3 = 1.84^14 per c.c` and the correspending mean distance is `(l)/(n^(1//3))`. =`(10^-2)/((0.184)^(1//3) xx 10^5) = 1.8 xx 10^- m ~~ 0.18 mu m`. |
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| 1890. |
An acoustic wave alphaagates through nitrogen under standard conditions. At what frequency will the wavelength be equal to the mean free path of the gas molecules ? |
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Answer» The velocity of sound in `N_2` is `sqrt((gamma p)/(rho)) = sqrt((gamma RT)/(M))` so, `(1)/(v) = sqrt((gamma RT_0)/(M)) = (RT_0)/(sqrt(2) pi d^2 p_0 N_A)` or, `v = pi d^2 p_0 N_A sqrt((2 gamma)/(MRT_0))`. |
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| 1891. |
What will be the change in internal energy when 12 kJ of work is done on the system and 2 kJ of heat is given by the system?A. `+10kJ`B. `-10kJ`C. `+5kJ`D. `-5kJ` |
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Answer» Correct Answer - A Heat evolved = -2 kJ Work done on the system `=+12kJ` `DeltaU=q+w=-2+12=+10k` |
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| 1892. |
The following is not a combustion reactionA. `CO+1/2 O_(2) rarr CO_(2)`B. `C+O_(2) rarr CO_(2)`C. `C+1/2 O_(2) rarr CO`D. `CH_(4)+2O_(2) rarr CO_(2)+2H_(2)O` |
| Answer» Correct Answer - C | |
| 1893. |
`NH_(4)Cl(s) +H_(2)O rarr NH_(4) Cl (aq)` `DeltaH=16.3 kJ` `DeltaH` in the above reaction representsA. Heat of solutionB. Integral heat of solutionC. Heat of dilutionD. Heat of ionization |
| Answer» Correct Answer - A | |
| 1894. |
Entropy is a measure ofA. DisorderB. Internal energyC. EfficiencyD. Useful work done by the system |
| Answer» Correct Answer - A | |
| 1895. |
When ammonium chloride is dissolved in water the solution becomes cold becauseA. Heat of solution of ammonium chloride is positiveB. Heat of solution of ammonium chloride is negativeC. Heat of dilution of ammonium chloride is positiveD. Heat of formation of ammonium chloride is positive |
| Answer» Correct Answer - A | |
| 1896. |
Predict the enthalpy change, free energy change and entropy change when ammonium chloride is dissolved in water and the solution becomes colder. |
| Answer» Correct Answer - `DeltaH = + ive, DeltaS = + ve , DeltaG = - ive`. | |
| 1897. |
In which of the following entropy increases?A. Rusting of ironB. Melting of iceC. Crystallization of sugar from solutionD. Vaporisation of camphor |
| Answer» Correct Answer - a,b,d | |
| 1898. |
The criteria for the spontaneity of a process areA. `(dG)_(T,P) lt 0`B. `( dE)_(S,V) lt 0`C. `( dH ) _(S,P ) lt0`D. ` ( dS) _( E,V) lt 0` |
| Answer» Correct Answer - a,b,c,d | |
| 1899. |
At 100 K from tata, `N_(2)(g)+3H_(2)(g) rarr2NH_(3)(g)DeltaH^(@)=-123.77"kJmole"^(-1)` `{:("Substance",N_(2),H_(2),NH_(2)),(Cp//R,3.5,3.5,4):}` Calculate the heat of formation of `NH_(3)` to `300 K.` writer your answer, excluding the decimal places. |
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Answer» Correct Answer - 44 |
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| 1900. |
A substance has a normal boiling point of 400K. Which of the following options correct set of thermodynamic parameters for the reaction: `(l)(1atm.400K)toA(g)(1atm,400K)`A. `qgt0,deltaG=0,Wgt0`B. `deltaH=0M,deltaG=0,deltaS=0`C. `deltaG=0M,deltaHgt0,deltaSgt0`D. `deltaGlt0,deltaHgt0,deltaSgt0` |
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Answer» Correct Answer - C |
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