InterviewSolution
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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.
| 2051. |
Heat of hydronization of ethene is `x_(1)` and that of benzene is `x_(2)`. Hence, resonance energy is :A. `x_(1)-x_(2)`B. `x_(1)+x_(2)`C. `3x_(1)-x_(2)`D. `x_(1)-3x_(2)` |
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Answer» Correct Answer - C |
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| 2052. |
Heat of hydrogenation of ethene is `x_(1)` and that of benzene is `x_(2)`. Hence, resonance energy is:A. `x_(1)-x_(2)`B. `x_(1)+x_(2)`C. `3x_(1)-x_(2)`D. `x_(1)-3x_(2)` |
| Answer» Correct Answer - C | |
| 2053. |
The free energy change due to a reaction is zero whenA. The reactants are initially mixedB. A catalyst is addedC. The system is at equilibriumD. The reactants are completely consumed |
| Answer» Correct Answer - C | |
| 2054. |
Calculate the increase in internal energy of `1 kg` of water at `100^(@)C` when it is converted into steam at the same temperature and at `1atm` (`100 kPa`). The density of water and steam are `1000kg m^(-3)` and `0.6 kg m^(-3)` respectively. The latent heat of vaporization of water `=2.25xx10^(6) J kg^(1)`. |
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Answer» Correct Answer - `2.08xx10^(6)J` Volume of `1 kg` of water `=1/(1000)m^(3)= 0.001m^(3)` Volume of `1 kg` of steam `=1/(0.6)m^(3)= 1.7m^(3)` `:.` Increase in volume, `DeltaV= 1.7-0.001= 1.7,^(3)` Work done by the system `=pDeltaV` `DeltaW=(100xx10^(3))xx1.7= 1.7xx10^(5)J` `:. DeltaU= DeltaQ-DeltaW= mL- DeltaW` `= 1xx2.25xx10^(6)-1.7xx10^(5)` `=2.08xx10^(6)J` |
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| 2055. |
In which of the following cases entropy increases?A. Solid changing to liquidB. Expansion of gasC. Crystal dissolvesD. Boiling of an egg |
| Answer» Correct Answer - A::B::C::D | |
| 2056. |
In which of the following change entropy decreases?A. Crystallization of sucrose from solutionB. Dissolving sucrose in waterC. Melting of iceD. Vaporization of camphor |
| Answer» Correct Answer - A | |
| 2057. |
Which of the following are endothermic processes?A. Combustion of glucoseB. Decomposition of waterC. Dehydrogenation of ethane to etheneD. Conversion of graphite to diamond |
| Answer» In endithermic reaction, heat is absorbed by the system form the surroundings. | |
| 2058. |
The value of entropy of the universe is alwaysA. ConstantB. DecreasingC. IncreasingD. Zero |
| Answer» Correct Answer - C | |
| 2059. |
When a solid melts, there is/are:A. an increase in entropyB. an increase in enthalpyC. a decrease in internal energyD. a decrease in enthalpy |
| Answer» Correct Answer - A::B | |
| 2060. |
The Entropy of the universeA. Tends towards a maximumB. Tends towards a minimumC. Tends to zeroD. Temains constant |
| Answer» Correct Answer - A | |
| 2061. |
Which are the intensive properties?A. TemperatureB. Refractive indexC. VolumeD. Enthalpy reduces to zero |
| Answer» Correct Answer - A::B | |
| 2062. |
In which of the following cases, the reaction is spontaneous at all temperaturesA. `DeltaH gt 0, Delta S gt 0`B. `DeltaH lt 0, Delta S gt 0`C. `DeltaH lt 0, Delta S lt 0`D. `DeltaH gt 0, Delta S lt 0` |
| Answer» Correct Answer - B | |
| 2063. |
Which of the following statements is not correct ?A. For a spontaneous process, `DeltaG` must be negative.B. Enthalpy, entropy, free energy etc. are state variables.C. A spontaneous process is reversible in nature.D. Total of all possible kinds of energy of a syste is called its internal energy. |
| Answer» Correct Answer - C | |
| 2064. |
One word answer is given for the following definitions, Mark the one which is incorrect.A. The process in which temperature remains constant : IsobaricB. The process in which volume remains constant : IsochoricC. The relation between `DeltaH and DeltaH` when all the reactants and products are solid : `DeltaH=DeltaU`D. The relation between `DeltaG, DeltaH and DeltaS:` `DeltaG=DeltaH-TDeltaS` |
| Answer» Correct Answer - A | |
| 2065. |
Mathc the following columns and mark the appropriate choice. A. (A) `rarr` (ii), (B) `rarr` (iii), (C) `rarr` (i), (D) `rarr` (iv)B. (A) `rarr` (iv), (B) `rarr` (i), (C) `rarr` (iii), (D) `rarr` (ii)C. (A) `rarr` (i), (B) `rarr` (ii), (C) `rarr` (iv), (D) `rarr` (iii)D. (A) `rarr` (iii), (B) `rarr` (iv), (C) `rarr` (i), (D) `rarr` (ii) |
| Answer» Correct Answer - D | |
| 2066. |
The thermal power of density `omega` is generated uniformly inside a uniform sphere of radius `R` and heat conductivity coefficient `x`. Find the temperature distribution in the sphere provided the steady-state temperature at its surface is eqal to `T_0`. |
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Answer» Here again `nabla^2 T = -(w)/(k)` So in spherical polar coordinates, `(1)/(r^2) (del)/(del r) (r^2 (del T)/(del r)) = -(w)/(k)` or `r^2 (del T)/(del r) = -(w)/(3 k) r^3 + A` or `T = B - (A)/( r) - (w)/(6 k) r^2` Again `A = 0` and `B = T_0 + (w)/(6k) R^2` so finally `T = T_0 + (w)/(6k)(R^2 - r^2)`. |
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| 2067. |
Let `alpha dt` be the probability of a sgas molecule experiencing a collision during the time interval `dt , alpha` is a constant. Find : (a) the probability of a molecule experiencing no collisions during the time interval `t` , (b) the mean time interval between successive collisions. |
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Answer» (a) Let `P(t)` = probability of no collision in the interval `(0, t)`, Then `P(t + dt) = P (t) (1 - alpha dt)` or `(dP)/(dT) = - alpha (t)` or `P (t) = e^(- alpha t)` where we have used `P(0) = 1` (b) The mean interval between collision is also the mean interval of no collision. Then `lt t gt = (int_0^oo te^(- alpha t) dt)/(int_0^oo e^(-alpha t)) = (1)/(alpha) (Gamma (2))/(Gamma(1)) = (1)/(alpha)`. |
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| 2068. |
As a result of some process the pressure of an ideal gas increases `n-fold`. How many times have the mean free path `lamda` and the number of collisions of each molecule per unit time `v` changed and how, if the process is (a) isochoric , (b) isothermal ? |
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Answer» (a) In an isochoric process `lamda` is constant and `v alpha sqrt(T) alpha sqrt(p V) alpha sqrt(p) alpha sqrt(n)` (b) `lamda = (kT)/(sqrt(2) pi d^2)` must decrease `n` times in an isothermal process and `v` must increase `n` times because `ltvgt` is constant in an isothermal process. |
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| 2069. |
An ideal gas consisting of rigid diatimic molecules goes through an adiabatic process process. How do the mean free path `lamda` and the number of collisions of each molecule per second `v` depend in this process on (a) the volume`V` , (b) the pressure `p` , ( c) the temperature `T` ? |
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Answer» (a) `lamda alpha(1)/(n) = gt (1)/(N//V) = (V)/(N)` Thus `lamda alpha V` and `v alpha (T^(1//2))/(V)` But in an adiabatic process `(gamma = (7)/(5) "here")` `TV^(gamma - 1) = "constant so" TV^(2//5) =` constant or `T^(1//2) alpha V^(-1//5)` Thus `v alpha V^(-6//5)` (b) `lamda alpha (T)/(p)` But `p((T)/(p))^gamma =` constant or `(T)/(p) alpha p^(-1//gamma)` or `T alpha p^(1-1//gamma)` Thus `lamda alpha p^(-1//gamma) = p^(-5//7)` `v = (lt v gt )/(lamda) alpha(p)/(sqrt(T)) alpha p^(1//2 + (1)/(2 gamma)) =p^((gamma + 1)/(2 gamma) = p^(6//7))` ( c) `lamda alpha V` But `TV^(2//5) =` constant or `V alpha T^(-5//2)` Thus `lamda alpha T^(-5//2)` `v alpha(T^(1//2))/(V) alpha T^3`. |
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| 2070. |
One mole of water being in equilibrium with a negligible amount of its saturated vapour at a temperature `T_1` was completely converted into saturated vapour at a temperature `T_2`. Find the entropy increment of the system. The vapour is assumed to be an ideal gas, the specific volume of the liquid is negligible on comparsion with that of the vapour. |
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Answer» The required entropy change can be calculated along a process in which the water is heated from `T_1` to `T_2` and then allowed to evaporate. The entropy change for this is `Delta S = C_p 1n (T_2)/(T_1) + (q M)/(T_2)` where `q` = specific latent heat of vaporization. |
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| 2071. |
A narrow molecular beam makes its way into a vessel filled with gas under low pressure. Find the mean free path of molecules if the beam intensity decreases `eta-fold` over the distance `Delta l`. |
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Answer» From the formula `(1)/(eta) = e^(- Delta l//lamda)` or `lamda = (Delta l)/(1n eta)`. |
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| 2072. |
An ideal gas goes through a polytropic process with exponent `n`. Find the mean free path `lamda` and the number of collisions of each molecule per second `v` as a function of (a) the volume `V` , (b) the pressure `p` , ( c) the temperature `T`. |
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Answer» In the polytropic process of index `n` `pV^n =` constant, `TV^(n -1) =` constant and `p^(1 - n) T^n =` constant (a) `lamda alpha V` `v alpha (T^(1//2))/(V) = V^((1 - n)/(2)) V^-1 = V^((-n + 1)/(2))` (b) `lamda alpha (T)/(p), T^n alpha p^(n -1)` or `T alpha p^(1 - (1)/(n)` so `lamda alpha p^(-1//n)` `v = (lt v gt)/(lamda) alpha (p)/(sqrt(T)) alpha p^(1 -(1)/(2) + (1)/(2n)) = p^((n + 1)/( 2n))` ( c) `lamda alpha (T)/(p), p alpha T^((n)/(n - 1))` `lamda alpha T^(1 -(n)/(n 1)) = T^(-(1)/(n -1)) = T^((1)/(1 - n))` `v alpha (p)/(sqrt(T)) alpha T^((n)/(n -1) -(1)/(2)) =T^((n + 1)/(2(n -1)))`. |
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| 2073. |
How does the mean free path `lamda` and the number of collisions of each molecule per unit time `v` depend on the absolute temperature of an ideal gas undergoing (a) an isochoric process , (b) an isobaric process ? |
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Answer» (a) `lamda = (1)/(sqrt(2) pi d^2 n)` `d` is a constant and `n` is a constant for an isochoric process so `lamda` is constant for an isochoric process. `v = (lt v gt)/(lamda) = (sqrt(8 RT)/(M pi))/(lamda) alpha sqrt(T)` (b) `lamda = (1)/(sqrt(2) pi d^2)(kT)/(p) alpha T` for an isobaric process. `v = (lt v gt)/(lamda) alpha (sqrt(T))/(T) = (1)/(sqrt(T))` for an isobaric process. |
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| 2074. |
Heat of `20` K cal is supplied to the system and `8400 J` of external work is done on the system so that its volume decreases at constant pressure. The change in internal enregy is `(J=4200 J//kcal)`A. `9.24xx10^(4) J`B. `7.56xx10^(4) J`C. `8.4xx10^(4) J`D. `10.5xx10^(4) J` |
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Answer» Correct Answer - A `dU=dQ+dW` |
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| 2075. |
Assertion: Air quickly leaking out of a balloon becomes coolers. Reason: The leaking air undergoes adiabatic expansion.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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| 2076. |
For a spontaneous process the correct statement is -A. Entropy of the system always increaseB. Free energy of the system always increasesC. Total entropy change is always negativeD. Total entropy change is always positive |
| Answer» Correct Answer - D | |
| 2077. |
For a phase change: `H_(2)O(l)hArrH_(2)O(s)` `0^(@)C`, 1 barA. `Delta G = 0`B. `Delta S = 0`C. `Delta H = 0`D. `Delta U = 0` |
| Answer» Correct Answer - A | |
| 2078. |
When you make ice cubes, the entropy of waterA. Does not changeB. IncreasesC. DecreasesD. Many either increase or decrease depending on the process used |
| Answer» Correct Answer - C | |
| 2079. |
Assertion: The increase in internal energy `(DeltaE)` for the vaporisation of 1 mole of water at 1 atm and `373K` is zero. Reason: For all isothermal processes `DeltaE=0` .A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false. |
| Answer» Correct Answer - D | |
| 2080. |
Assertion :- For an isolated system q is zero. Reason :- In an isolated system change in U and V is zero.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false. |
| Answer» Correct Answer - C | |
| 2081. |
Assertion. Chlorine whensolidifies doe not have zero entropy even at absolute zero. Reason. Chlorine is a pungent smelling gas and it is difficult to solidify it.A. If both A and R are true,andR is the true explanation of A.B. If both A and R are true,but R is not the true explanation of A.C. If A is true, but R is falseD. If both A and R are false. |
| Answer» Correct Answer - b | |
| 2082. |
Assertion . Enthalpy of graphite is lowr than that of diamond. Reason. Entropy of graphite is greater than that of diamond.A. If both A and R are true,andR is the true explanation of A.B. If both A and R are true,but R is not the true explanation of A.C. If A is true, but R is falseD. If both A and R are false. |
| Answer» Correct Answer - b | |
| 2083. |
Define molar specific heat at constant pressure, Cp and molar specific heat at constant volume, Cv. |
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Answer» Molar specific heat at constant pressure, Cp, is the amount of heat required to raise the temperature of 1 mol of a gas through 1K at constant pressure. Molar specific heat at constant volume, Cv, is the amount of heat required to raise the temperature of 1 mol of a gas through 1K at constant volume. |
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| 2084. |
`|Delta_(f)H|" of "(H_(2)O,l)gt|Delta_(f)H|" of "(H_(2)O,g)` `DeltaH_("condensation")` is negative.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-13B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-13C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
| Answer» Correct Answer - A | |
| 2085. |
Statement-1: Decrease of free energy during the process under constant temperature and pressure provides a measure of its spontaneity. Statement -2: A spontaneous change must have +ve sign of `DeltaS_("system").`A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-12B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-12C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
| Answer» Correct Answer - C | |
| 2086. |
Which of the following Process is non-spontaneousA. Heat flow from hot end to cold endB. Water flow from higher level to lower levelC. Gas flow from lower pressure region to higher pressure regionD. Gas flow from higher pressure region to lower pressure region |
| Answer» Correct Answer - C | |
| 2087. |
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant `K_(c)` is:A. `DeltaG^(@)=RT ln K_(c)`B. `-DeltaG^(@)=RT lnK_(c)`C. `DeltaG=RT ln K_(c)`D. `-DeltaG=RT lnK_(c)` |
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Answer» Correct Answer - b `DeltaG=DeltaG^(@)+RTln Q`, Where `Q` is reaction quotient , At equibrium `DeltaG=0` and `Q=K_(c) :. -DeltaG^(@)=RT ln K_(c)` |
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| 2088. |
The most random state of `H_(2)O` system isA. IceB. `H_(2)O (l)` at `80^(@)C`, 1 atmC. SteamD. `H_(2)O (l)` at `25^(@)C`, 1 atm |
| Answer» Correct Answer - C | |
| 2089. |
Assertion (A): May endothermic reactions that are not spontaneous at room temperature become spontaneous at high temperature. Reason (R ) : Entropy of the system increases with increase in temperature.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-11B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-11C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
| Answer» Correct Answer - B | |
| 2090. |
Assertion (A): May endothermic reactions that are not spontaneous at room temperature become spontaneous at high temperature. Reason (R ) : Entropy of the system increases with increase in temperature.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false. |
| Answer» Correct Answer - B | |
| 2091. |
Assertion (A): May endothermic reactions that are not spontaneous at room temperature become spontaneous at high temperature. Reason (R ) : Entropy of the system increases with increase in temperature.A. If both (A) and (R ) are correct, and (R ) is the correct explanation for (A).B. If the both (A) and (R ) are correct, but(R ) is not a correct explanation for (A).C. If (A) is correct, but (R ) is incorrect.D. If (A) is incorrect, but (R ) is correct. |
| Answer» On increasing temperature, the endothermic reaction becomes spontaneous and entropy also increases. | |
| 2092. |
Thermodynamic parameter which is a state function and also used to measure disorder of the system isA. EntropyB. FugacityC. ViscosityD. Periodicity |
| Answer» Correct Answer - A | |
| 2093. |
Which thermodynamic parameter is not a state function :-A. q at constant pressureB. q at constant volumeC. w at adiabaticD. w at isothermal |
| Answer» Correct Answer - D | |
| 2094. |
Statement: In the case of an ideal gas the changes in Gibbs and Helmholtz free energies are equal to each other (`DeltaG=DeltaA`) for isothermal reversible process. Explanation: There is no change in internal energies and enthalpies for ideal gas at constant temperature.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 explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 2095. |
Assertion:Phase transition involves change in internal energy only. Reason:Phase transition occurs at constant pressure.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 explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - D | |
| 2096. |
Enthalpy change for the reaction , `4H(g) rarr 2H_(2)(g)`is`-869.6 kJ`. The dissociation energy of `H-H` bondisA. `+217 .4kJ`B. `-434.8 kJ`C. `-869.6kJ`D. `+ 434.8kJ` |
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Answer» Correct Answer - D `2H_(2)(g) rarr 4H(g), DeltaH = +869.6kJ` or `H_(2)(g) rarr 2H(g), DeltaH= + ( 869.6)/( 2) Kj` `=434.8 kJ` |
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| 2097. |
The work done by `1` mole of ideal gas during an adiabatic process is (are ) given by :A. `(P_(2)V_(2) - P_(1)V_(1))/(gamma-1)`B. `(nR(T_(1)-T_(2)))/(gamma-1)`C. `(P_(2)V_(2)-P_(1)V_(1))/(gamma)`D. None of these |
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Answer» Correct Answer - A `DeltaU=w = (P_(2)V_(1)-P_(1)V_(1))/(gamma-1)` |
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| 2098. |
Gibbs Helmholtz equation relates the enthalpy, entropy and free energy change of the process at constant pressure and temperature as `DeltaG=DeltaH-TDeltaS " (at constant P, T)"` In General the magnitude of `DeltaH` does not change much with the change in temperature but the terms `TDeltaS` changes appreciably. Hence in some process spontaneity is very much dependent on temperature and such processes are generally known as entropy driven process. When `CaCO_(3)` is heated to a high temperature it decomposes into CaO and `CO_(2)`, however it is quite stable at room temperature. It can be explained by the fact thatA. `Delta_(r )H` dominates the term `TDeltaS` at high temperatureB. the term `TDeltaS` dominates the `Delta_(r )H` at high temperatureC. at high temperature both `Delta_(r )S` and `Delta_(r )H` becomes negativeD. thermodynamics can not say anything about spontaneity |
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Answer» Correct Answer - B `CaCO_(3)rarrCaO+CO_(2)" "DeltaH+ve` Reaction becomes spontaneous at high temperature because `TDeltaS` dominates over `DeltaH_(rxn)`. |
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| 2099. |
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio `C_P//C_V` for the gas isA. `(3)/(2)`B. `(5)/(3)`C. `(7)/(2)`D. `(4)/(3)` |
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Answer» Correct Answer - A `T^((gamma)/(1-gamma))."P = constant or "PpropT^((gamma)/(gamma-1))` `because " "PpropT^(3)` `therefore " "(gamma)/(gamma-1)=3, gamma=(3)/(2)` |
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| 2100. |
For an exothermic reaction to be spontaneous (`Delta`S = negative)A. Temperature must be highB. Temperature must be zeroC. Temperature may have any magnitudeD. Temperature must be low |
| Answer» Correct Answer - D | |