4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 21 InterviewSolution

1001.	Which of the following is intensive property -A. EnthalpyB. EntropyC. specific volumeD. volume
Answer» Correct Answer - C

Discussion

1002.	At equilibrium which is correct :-A. `Delta G = 0`B. `Delta S = 0`C. `Delta H = 0`D. `Delta G^(@) = 0`
Answer» Correct Answer - A

Discussion

1003.	In which of the following reactions,standard reaction entropy change`(DeltaS^(@))`is positive and standard Gibb,s energy change`(DeltaG^(@))`decreases sharply with increasing temperature?A. `Mg(s)+(1)/(2)O_(2)(g)rarr MgO(s)`B. `(1)/(2)` C graphite `+(1)/(2)O_(2)(g)rarr (1)/(2)CO_(2)(g)`C. C graphite `+(1)/(2)O_(2)(g)rarr CO (g)`D. `CO(g)+(1)/(2)O_(2)(g)rarr CO_(2)(g)`
Answer» Correct Answer - C

Discussion

1004.	What are Reversible Process? State the conditions for Reversible Process.
Answer» Reversible process : A reversible process is one which can be reversed in such a way that all changes occuring in the direct process are exactly repeated in the opposite order and inverse sense. The conditions for reversibility are : There must be complete absence of dissipative forces. (friction, viscosity etc.) The speed of process should be infinitely slowly. The temperature of the system must not differ appreciably form its surroundings.

1004.

What are Reversible Process? State the conditions for Reversible Process.

Answer»

Reversible process : A reversible process is one which can be reversed in such a way that all changes occuring in the direct process are exactly repeated in the opposite order and inverse sense.

The conditions for reversibility are :

There must be complete absence of dissipative forces. (friction, viscosity etc.)
The speed of process should be infinitely slowly.
The temperature of the system must not differ appreciably form its surroundings.

Discussion

1005.	Define Free expansion.
Answer» Free expansion is adiabatic process in which no work is performed on or by the system. The final and initial energies are equal in free expansion.

Discussion

1006.	Mention the Essential conditions for adiabatic process.
Answer» Essential conditions for adiabatic process : (i) All walls of the container and the piston must be perfectly insulating. (ii) The speed of process should be fast.

1006.

Mention the Essential conditions for adiabatic process.

Answer»

Essential conditions for adiabatic process :

(i) All walls of the container and the piston must be perfectly insulating.

(ii) The speed of process should be fast.

Discussion

1007.	State the Second Law of Thermodynamics.
Answer» (1) Clausius statement : It is impossible for a self acting machine to transfer heat from a colder body to a hotter one without the aid of an external agency. (2) Kelvin’s statement : It is impossible for a body or system to perform continuous work by cooling it to a temperature lower than the temperature of the coldest one of its surroundings. (3) Kelvin-Planck’s statement : It is impossible to design an engine that extracts heat and fully utilises into work without producing any other effect.

1007.

State the Second Law of Thermodynamics.

Answer»

(1) Clausius statement : It is impossible for a self acting machine to transfer heat from a colder body to a hotter one without the aid of an external agency.

(2) Kelvin’s statement : It is impossible for a body or system to perform continuous work by cooling it to a temperature lower than the temperature of the coldest one of its surroundings.

(3) Kelvin-Planck’s statement : It is impossible to design an engine that extracts heat and fully utilises into work without producing any other effect.

Discussion

1008.	Define (a) Surroundings(b) Thermodynamic variables
Answer» (a) Surroundings: Everything outside the system which can have a direct effect on the system is called its surroundings. (b) Thermodynamic variables: The quantities like pressure (P) , volume (V) and temperature (T) which help us to study the behaviour of a thermodynamic system are called thermodynamic variables.

1008.

Define (a) Surroundings(b) Thermodynamic variables

Answer»

(a) Surroundings: Everything outside the system which can have a direct effect on the system is called its surroundings.

(b) Thermodynamic variables: The quantities like pressure (P) , volume (V) and temperature (T) which help us to study the behaviour of a thermodynamic system are called thermodynamic variables.

Discussion

1009.	What is Thermodynamic system ?
Answer» An assembly of a very large number of particles having a certain value of pressure, volume and temperature is called a thermodynamic system.

Discussion

1010.	State and explain the first law of thermochemistry.
Answer» If enthalpy of formation of a compound is -x kJ, enthalpy of its decomposition intoits elements is + xkJ. Example: C(s) + O₂(g) → CO₂; ΔH = -xkJ According to the law (Lavoisier & Laplae law) CO₂(g) → C(s) + O₂ (g); ΔH = +x kJ

1010.

State and explain the first law of thermochemistry.

Answer»

If enthalpy of formation of a compound is -x kJ, enthalpy of its decomposition intoits elements is + xkJ.

Example: C(s) + O₂(g) → CO₂; ΔH = -xkJ According to the law (Lavoisier & Laplae law) CO₂(g) → C(s) + O₂ (g); ΔH = +x kJ

Discussion

1011.	Calculate ΔH for the process at 25°C of dissolving 1.00 mol of KCl in an excess of water. Does this process represent an ionization reaction? Explain.ΔH°f[K+(aq)] = -251.2 kJ mol-1ΔH°f[Cl-(aq)] = -167.08 kJ mol-1ΔH°f[KCl] = -437.6 kJ mol-1
Answer» ΔH = ΔH°_fK⁺(aq) + ΔH°_fCl^-(aq) - ΔH°_fKCl(s) = -251.2 - 167.08 - (-437.6) = -418.28 + 437.6 = 19.32 kJ The reaction represents a solution process and not an ionisation. The KCl solid is ionic before being dissolved as well as after.

1011.

Calculate ΔH for the process at 25°C of dissolving 1.00 mol of KCl in an excess of water. Does this process represent an ionization reaction? Explain.ΔH°f[K+(aq)] = -251.2 kJ mol-1ΔH°f[Cl-(aq)] = -167.08 kJ mol-1ΔH°f[KCl] = -437.6 kJ mol-1

Answer»

ΔH = ΔH°_fK⁺(aq) + ΔH°_fCl^-(aq) - ΔH°_fKCl(s)

= -251.2 - 167.08 - (-437.6)

= -418.28 + 437.6

= 19.32 kJ

The reaction represents a solution process and not an ionisation. The KCl solid is ionic before being dissolved as well as after.

Discussion

1012.	Which one of the following is a state property/function?A. `U +PV`B. `q +w`C. `(q_(rev))/(T)`D. `q`
Answer» `q` is not state function.

Discussion

1013.	How much work can be done by `100` calories of heat?
Answer» `1cal = 4.184 J`. `:. 100 cal = 418.4 J` `:.` Work done `= 418.4J`

Discussion

1014.	Which of the following is an extensive property?A. VolumeB. Surface tensionC. ViscosityD. Density
Answer» Volume is an extensive property.

Discussion

1015.	The heat produced from the complete combustion of 1 g of fuel or food is called its` "…............."`
Answer» Correct Answer - calorific value

Discussion

1016.	A strip of magnesium of mass `15 g` is droped into an opean beaker of dilute hydrochloric acid. Calculate the work done by the system as a result of reaction. The atmospheric pressure is `1.0 atm` and temperature is `25^(@)C`. Also calculate the work done if the reaction is carried out in closed beaker.
Answer» Correct Answer - `-1.548 kJ,zero`

Discussion

1017.	For the reaction `K(g)+F(g) +K^(o+)+F^(Theta)` (separated ions `DeltaH = 19 kcal mol^(-1)`), if the ionisation potential of `K` and the electron affinity of `F^(Theta)` have a geometric means of `3.88eV` and `IP gt EA`, calculate the values fo ionisation potential and electron affinity.
Answer» `K(g) rarr K^(o+) (g) +e^(-) …(i)IE =x` `F(g) +e^(-) rarr F^(Theta)(g) …(ii) EA =- y` Add equations (i) and (ii), `K(g) +F(g) rarr F^(o+)(g) +F^(Theta)(g)` `DeltaH = 19 kcal mol^(-1) =0.82 eV atom^(-1)` `[{:(19kcal =19xx10^(3)cal =19 xx10^(3)xx4.18J),(=(19xx10^(3)xx4.18)/(1.6xx10^(-19))eVmol^(-1)),(=(19xx10xx4.18)/(1.6xx10^(-19))xx(1)/(6.23xx10^(23)atm)eVat om^(-1)),(=0.82eV):}]` `:. x-y = 19 kcal mol^(-1) = 0.82 eV` ...(ii) `sqrt(xy) = (3.88) eV` (given) `xy = (3.88)^(2)` `xy = 15.0544` `x = (15.0544)/(y)` Substituting the value of `x` in equation (iii), we get `(15.0544)/(y) -y = 19` Solving, we get `y = 3.48 eV` `x+y = 7.48 eV` `x - y = 0.82 eV` `IE = x = 4.313` `EA = y = 3.48 eV`

Discussion

1018.	Change in internal energy, when 4 kJ of work is done on the system and 1 kJ of heat is given out by the system is ……(a) +1 kJ(b) -5 kJ (c) +3 kJ(d) -3 kJ
Answer» (c) +3 kJ ∆U = q + w ∆U = – 1 kJ + 4 kJ ∆U = + 3 kJ

1018.

Change in internal energy, when 4 kJ of work is done on the system and 1 kJ of heat is given out by the system is ……(a) +1 kJ(b) -5 kJ (c) +3 kJ(d) -3 kJ

Answer»

(c) +3 kJ

∆U = q + w

∆U = – 1 kJ + 4 kJ

∆U = + 3 kJ

Discussion

1019.	Calcualte the work down when 11.2 g of iron dissolves in hydrochloric acid in (i) a closed vessel (ii) an open beaker at `25^(@)C` ( Atomic meass of `Fe = 56u)`
Answer» Iron reacts with HCl acid to produce `H_(2)` gas as `Fe(s) + 2HCl(aq) rarr FeCl_(2) (aq) + H_(2)(g)` Thus, 1 mole of Fe, i.e., 56 g Fe produce `H_(2)` gas `= 1 `mole `:. 11.2 g` Fe will produce `H_(2)` ga `= (1)/( 56) xx 11.2 = 0.2 `mole. (i) If the reaction is carried out in a closed vessel , `Delta V = 0` `:. w = - P_(ext)DeltaV = 0` (ii) If the reaction is carried out in open beaker ( external pressure being 1 atm) Final volume occupied by 0.2 moles of `H_(2) ` at `25^(@)C` and 1 atm pressure can be calculated as follows `:` `PV = nRT` `:. V = ( nRT)/( P) = ( 0.2 mol xx 0.0 821L atm K^(-1) xx 298 K)/( 1 atm) = 4.89 L` `:. DeltaV = V _("final") - V_("initial") = 4.89 L` `w= - P_(ext)DeltaV = -1 atm xx 4.89 L atm` `= - 4.89 L atm` `= - 4.89 xx 101.3 J = - 495.4 J`

Discussion

1020.	A system is provided `50J` of heat and work done on the system is `20J`. What is the change in the internal enegry?
Answer» `q = 50J` `w = +20J` (work done on the system) `:. DeltaU = q +w = 50 +20 = 70J`

Discussion

1021.	`A` gas expands from `40` liters to `90` liters at a constant pressure of `8` atmospheres. Work done by the gas during the expansion isA. `4xx10^(-4)J`B. `4xx10^(4)J`C. `4xx10^(3)J`D. `4xx10^(2)J`
Answer» Correct Answer - B `W=P(V_(2)-V_(1))`

Discussion

1022.	Match the column I with column II and mark the appropriate choice. A. (A) `rarr` (iv), (B) `rarr` (iii), (C) `rarr` (i), (D) `rarr` (ii)B. (A) `rarr` (ii), (B) `rarr` (i), (C) `rarr` (iv), (D) `rarr` (iii)C. (A) `rarr` (i), (B) `rarr` (ii), (C) `rarr` (iii), (D) `rarr` (iv)D. (A) `rarr` (iii), (B) `rarr` (iv), (C) `rarr` (ii), (D) `rarr` (i)
Answer» Correct Answer - D (A) : `CH_(4(g))+2O_(2(g))rarrCO_(2(g))+2H_(2)O` `" shows combustion reaction."` (B) : `H_(2(g)) rarr 2H_((g))` show bond dissociation. (C) : `NaCl_((s)) rarr Na_((g))^(+) +Cl_((g))^(0)` `" shows combustion reaction."` (D) : `NaCl_((s)) rarr Na_((aq))^(+)+Cl_((aq))^(-)` `" shows combustion reaction."`

Discussion

1023.	Which of the following is not a correct statement about enthalpy of solution?A. For most ionic compounds, `DeltaH_("soln.")^(@)` is positive and the dissociation process is endothermic.B. Solubility of most salts increases with increase in temperature.C. If the lattice enthalpy is very high, the dissolution of compound becomes very easy.D. Enthalpy of solution is determined by the selective values of the lattice enthalpy and hydration enthalpy.
Answer» Correct Answer - C Dissolution on compound may not take place if lattice enthalpy is very high.

Discussion

1024.	A sample of ideal gas `(gamma = 1.4)` is heated at constant pressure. If an amount of 85 J of heat is supplied to gas, find `Delta U`.
Answer» Correct Answer - 9 Gas is diatomic as `gamma = 1.4`, thus, `C_(V) = (5)/(2)R` and `C_(P) = (7)/(2)R`, Given, `Delta H = 85J` at constant pressure Also, `Delta T (85)/(nC_(P)) = (140 xx 2)/(7 n xx R) = (40)/(n)` Now, `W = - nRT Delta T = - n xx 2 xx (40)/(n) = - 80 J` Also, `q_(p) = DeltaH = Delta U + (-W)` `Delta U = Delta H + W = 85 - 80` `Delta U = 5J`

Discussion

1025.	To a system `300` joules od heat is given and it does `60` joules of work. How much does the internal energy of the system change in this process? (in joule)A. `240`B. `156.5`C. `-300`D. `-528.2`
Answer» Correct Answer - A `dU=dQ-dW`

Discussion

1026.	Identify the state functions and path functions out of the following `:` enthalpy , entropy , heat, temperature, work , free energy.
Answer» State function `:` Enthalpy, Entropy , Temperature, Free energy, Path functions`:` Heat, work.

Discussion

1027.	Calculate `DeltaU` for a gas, if enthalpy change is 40 atm-L for the state change (5 atm, 10 L) to (3 atm,15 L) :A. 45 atm-LB. 35 atm-LC. 30 atm-LD. 40 atm-L
Answer» Correct Answer - A

Discussion

1028.	Maximum heat absorbed during isothermal expansion of an ideal gas from (10 atm, 1 L) to (1 atm, 10 L) is :A. 90 atm-LB. 10 atm-LC. 9 atm-LD. 23.03 atm-L
Answer» Correct Answer - D

Discussion

1029.	The change in entropy, `Delta S` is positive for an endothermic reaction, if enthalpy change `Delta H` occurs at the same temperature T, then the reaction is feasibleA. At all temperaturesB. When `Delta H gt T Delta S`C. When `Delta H lt T Delta S`D. Not feasible at all
Answer» Correct Answer - C `Delta G = -ve` if `DeltaH lt T Delta S`

Discussion

1030.	The allotropic form of sulphur for which the standard enthalpy of formation is taken as zero is `"…..................."`
Answer» Correct Answer - rhombic sulphur

Discussion

1031.	The enthalpy of solution of sodium chloride is `4 kJ mol^(-1)` and its enthalpy of hydration of ions is ` -784 kJ mol^(-1)` . Then the lattice enthalpy of NaCl ( in `kJ mol^(-1))` isA. `+ 788`B. `+4`C. `+398`D. `+780`
Answer» Correct Answer - A `DeltaH_("solution")=DeltaH _("lattice")+DeltaH_("hydration")+4= DeltaH _("lattice") +( -784)` or `DeltaH _("lattice") =+ 788 kJmol^(-1)`

Discussion

1032.	The enthalpy of solution of sodium chloride is `4 kJ mol^(-1)` and its enthalpy of hydration of ion is `-784 kJ mol^(-1)`. Then the lattice enthalpy of `NaCl` (in `kJ mol^(-1)`) isA. `+"780 kJ mol"^(-1)`B. `+"394 kJ mol"^(-1)`C. `+"788 kJ mol"^(-1)`D. `+"398 kJ mol"^(-1)`
Answer» Correct Answer - C `DeltaH_("sol")=DeltaH_("lattice")+DeltaH_("hyd")` `4=DeltaH_("lattice")+(-784)` `DeltaH_("lattice")=4-(-784)=+"788 kJ mol"^(-1)`

Discussion

1033.	One kg of graphite is burnt in a closed vessel. The same amount of the same sample is burnt in an open vessel. Will the heat evolved in the two cases be same ? If not, in which case it would be greater?
Answer» Same in both cases because `Deltan_(g) = 0`.

Discussion

1034.	A gas under constant pressure of `4.5 xx 10^(5) Pa` when subjected to `800 kJ` of heat, changes the volume from `0.5 m^(3) to 2.0 m^(3)`. The change in internal energy of the gas isA. `6.75xx10^(5) J`B. `2.25xx10^(5) J`C. `3.25xx10^(5) J`D. `1.25xx10^(5) J`
Answer» Correct Answer - D `dU=dQ-dW`

Discussion

1035.	For a spontaneous process, the change in Gibbs function is equal to theA. Heat content of the systemB. Entropy change of the systemC. Work of expansionD. Useful work
Answer» Correct Answer - D

Discussion

1036.	For an endothermic reaction, `Delta S` is positive. Then the reaction isA. Feasible when `T Delta S gt Delta H`B. Feasible when `Delta H gt T Delta S`C. Feasible at all temperaturesD. Not feasible at all
Answer» Correct Answer - A

Discussion

1037.	A sample of an ideal gas in a cyclinder is compressed adiabatically to `1/3rd` of its volume. Will final pressure be more or less than `3 times` the initial pressure?
Answer» Change is pressure will be more than 3 times the initial pressure.

Discussion

1038.	Three moles of an ideal gas kept at a constant temperature of `300K` are compressed from a volume of 4 litre to 1 litre. Calculate work done in the process. `R= 8.31 J mol e^(-1)K^(-1)`.
Answer» Here, `n=3, T= 300K, V_(1)= 4` litre, `V_(2)= 1 litre, W=?` In an isothermal process, work done is `W= 2.3026 nRT"log"_(10)(V_(2))/(V_(1))` `=2.3026xx3xx8.31xx300 log_(10)(1//4)` `W= -1.037xx10^(4)J`

Discussion

1039.	A gas is suddenly compressed to `1/4th` of its original volume. Caculate the rise in temperature when original temperature is `27^(@)C. gamma= 1.5`.
Answer» Here, `V_(2)=1/4V_(1)`, `T_(1)= 27^(@)C= 27+273=300K`, `gamma=1.5` `(T_(2)-T_(1))=?` As `T_(2)V_(2)^(gamma-1)= T_(1)V_(1)^(gamma-1)` `:. T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1)= 300(4)^(1.5-1)= 600K` `T_(2)-T_(1)= 600-300= 300k= 300^(@)C`

Discussion

1040.	Air in the cyclinder of a diesel engine is compressed to `1/15` of its initial volume. If initial temperature is `300K` and initial pressure is `10^(5)Pa`, find the final temperature and final pressure. Take `gamma= 1.4`.
Answer» Here, `(V_(2))/(V_(1))= 1/15, P_(1)= 10^(5)Pa`, `T_(1)=300K, T_(2)=?, P_(2)=?` In an adiabatic process, `P_(2)V_(2)^(gamma)= P_(1)V_(1)gamma` `:. P_(2)=P_(1)((V_(1))/(V_(2)))^(gamma)` `P_(2)=10^(5)(15)^(1.4)= 44.3xx10^(5)Pa` Also, `T_(2)V_(2)^(gamma-1)= T_(1)V_(1)^(gamma-1)` `T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1) = 300(15)^(1.4-1)= 886K`

Discussion

1041.	`C_(P) -C_(V)` for an ideal gas is………….. .
Answer» `C_(P) - C_(V)` for na ideal gas is equal to `R`.

Discussion

1042.	The total energy of `1 mol` of an ideal monatomic gas at `27^(@)C` is……. .
Answer» The total energy of `1mol` of an ideal monatomic gas at `27^(@)C` is `900`. ltbRgt `E = (3)/(2)RT = (3)/(2) xx2xx 300 = 900 cal`

Discussion

1043.	A system is said to be………..if it can neither exchange matter nor energy with the surroundigs.
Answer» A system is said to be isoalated if it can neither exchange matter nor energy with the surroundigs.

Discussion

1044.	Consider a heat engine as shown in figure. `Q_(1)and Q_(2)` are heat added both to `T_(1)` and heat taken from `T_(2)` in one cycle of engine. W is the mechanical work done on the engine. If `Wgt0`, then possibilities areA. `Q_(1)gtQ_(2)gt0`B. `Q_(2)gtQ_(1)gt0`C. `Q_(2)ltQ_(1)lt0`D. `Q_(1)lt0, Q_(2)gt0`
Answer» Correct Answer - A::C Consider the figure we can write `Q_(1)=W+Q_(2)` `rArr" "W=Q_(1)-Q_(2)gt0 " "("By question")` `rArr" "Q_(1)gtQ_(2)gt0" "("If both "Q_(1)andQ_(2)" are positive")` ltBrgt We can also, write `Q_(2)ltQ_(1)lt0 ("If both "Q_(1)andQ_(2)" are negative")`.

Discussion

1045.	A container that suits the occurrence of an isothermal process should be made ofA. CopperB. GlassC. WoodD. cloth
Answer» Correct Answer - A

Discussion

1046.	Write two limitation of the first law of thermodynamics.
Answer» Limitation of the first law of thermodynamics : (i) It does not give the direction of flow of heat. (ii) It does not explain why heat cannot be spontaneously converted into work.

1046.

Write two limitation of the first law of thermodynamics.

Answer»

Limitation of the first law of thermodynamics :

(i) It does not give the direction of flow of heat.

(ii) It does not explain why heat cannot be spontaneously converted into work.

Discussion

1047.	Heat cannot flow itself from a body at lower temperature to a body at higher temperature is a statement or consequence of which law of thermodynamics ?
Answer» Second law of thermodynamics.

Discussion

1048.	Can we increase the coefficient of performance of a refrigerator by increasing the amount of working substance?
Answer» Correct Answer - No, we cannnot.

Discussion

1049.	Assertion: In adiabatic compression, the internal energy and temperature of the system get decreased. Reason: The adiabatic compression is a slow process.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
Answer» Correct Answer - D

Discussion

1050.	The initial pressure and volume of a gas enclosed in a cylinder are respectively 1 × 105 N/m2 and 5 × 10-3 m3 . If the work done in compressing the gas at constant pressure is 100 J. Find the final volume of the gas.
Answer» Data : P = 1 × 10⁵ N/m² , V = 5 × 10^-3 m³ , W= -100 J W = P(Vf – V_i) ∴ V_f – V_i = \(\frac{W}{P}\) ∴ V_f= V_i + \(\frac{W}{P}\) = 5 x 10^-3 + \(\frac{(-100)}{(1\times10^5)}\) = 5 × 10^-3 -1 × 10^-3 = 4 × 10^-3 m³ This is the final volume of the gas.

1050.

The initial pressure and volume of a gas enclosed in a cylinder are respectively 1 × 105 N/m2 and 5 × 10-3 m3 . If the work done in compressing the gas at constant pressure is 100 J. Find the final volume of the gas.

Answer»

Data : P = 1 × 10⁵ N/m² , V = 5 × 10^-3 m³ ,

W= -100 J

W = P(Vf – V_i) ∴ V_f – V_i = \(\frac{W}{P}\)

∴ V_f= V_i + \(\frac{W}{P}\) = 5 x 10^-3 + \(\frac{(-100)}{(1\times10^5)}\)

= 5 × 10^-3 -1 × 10^-3 = 4 × 10^-3 m³

This is the final volume of the gas.

Discussion

Explore topic-wise InterviewSolutions in .

Which of the following is intensive property -A. EnthalpyB. EntropyC. specific volumeD. volume

At equilibrium which is correct :-A. `Delta G = 0`B. `Delta S = 0`C. `Delta H = 0`D. `Delta G^(@) = 0`

What are Reversible Process? State the conditions for Reversible Process.

Define Free expansion.

Mention the Essential conditions for adiabatic process.

State the Second Law of Thermodynamics.

Define (a) Surroundings(b) Thermodynamic variables

What is Thermodynamic system ?

State and explain the first law of thermochemistry.

Calculate ΔH for the process at 25°C of dissolving 1.00 mol of KCl in an excess of water. Does this process represent an ionization reaction? Explain.ΔH°f[K+(aq)] = -251.2 kJ mol-1ΔH°f[Cl-(aq)] = -167.08 kJ mol-1ΔH°f[KCl] = -437.6 kJ mol-1

Which one of the following is a state property/function?A. `U +PV`B. `q +w`C. `(q_(rev))/(T)`D. `q`

How much work can be done by `100` calories of heat?

Which of the following is an extensive property?A. VolumeB. Surface tensionC. ViscosityD. Density

The heat produced from the complete combustion of 1 g of fuel or food is called its` "…............."`

For the reaction `K(g)+F(g) +K^(o+)+F^(Theta)` (separated ions `DeltaH = 19 kcal mol^(-1)`), if the ionisation potential of `K` and the electron affinity of `F^(Theta)` have a geometric means of `3.88eV` and `IP gt EA`, calculate the values fo ionisation potential and electron affinity.

Change in internal energy, when 4 kJ of work is done on the system and 1 kJ of heat is given out by the system is ……(a) +1 kJ(b) -5 kJ (c) +3 kJ(d) -3 kJ

Calcualte the work down when 11.2 g of iron dissolves in hydrochloric acid in (i) a closed vessel (ii) an open beaker at `25^(@)C` ( Atomic meass of `Fe = 56u)`

A system is provided `50J` of heat and work done on the system is `20J`. What is the change in the internal enegry?

`A` gas expands from `40` liters to `90` liters at a constant pressure of `8` atmospheres. Work done by the gas during the expansion isA. `4xx10^(-4)J`B. `4xx10^(4)J`C. `4xx10^(3)J`D. `4xx10^(2)J`

A sample of ideal gas `(gamma = 1.4)` is heated at constant pressure. If an amount of 85 J of heat is supplied to gas, find `Delta U`.

To a system `300` joules od heat is given and it does `60` joules of work. How much does the internal energy of the system change in this process? (in joule)A. `240`B. `156.5`C. `-300`D. `-528.2`

Identify the state functions and path functions out of the following `:` enthalpy , entropy , heat, temperature, work , free energy.

Calculate `DeltaU` for a gas, if enthalpy change is 40 atm-L for the state change (5 atm, 10 L) to (3 atm,15 L) :A. 45 atm-LB. 35 atm-LC. 30 atm-LD. 40 atm-L

Maximum heat absorbed during isothermal expansion of an ideal gas from (10 atm, 1 L) to (1 atm, 10 L) is :A. 90 atm-LB. 10 atm-LC. 9 atm-LD. 23.03 atm-L

The change in entropy, `Delta S` is positive for an endothermic reaction, if enthalpy change `Delta H` occurs at the same temperature T, then the reaction is feasibleA. At all temperaturesB. When `Delta H gt T Delta S`C. When `Delta H lt T Delta S`D. Not feasible at all

The allotropic form of sulphur for which the standard enthalpy of formation is taken as zero is `"…..................."`

The enthalpy of solution of sodium chloride is `4 kJ mol^(-1)` and its enthalpy of hydration of ions is ` -784 kJ mol^(-1)` . Then the lattice enthalpy of NaCl ( in `kJ mol^(-1))` isA. `+ 788`B. `+4`C. `+398`D. `+780`

The enthalpy of solution of sodium chloride is `4 kJ mol^(-1)` and its enthalpy of hydration of ion is `-784 kJ mol^(-1)`. Then the lattice enthalpy of `NaCl` (in `kJ mol^(-1)`) isA. `+"780 kJ mol"^(-1)`B. `+"394 kJ mol"^(-1)`C. `+"788 kJ mol"^(-1)`D. `+"398 kJ mol"^(-1)`

One kg of graphite is burnt in a closed vessel. The same amount of the same sample is burnt in an open vessel. Will the heat evolved in the two cases be same ? If not, in which case it would be greater?

A gas under constant pressure of `4.5 xx 10^(5) Pa` when subjected to `800 kJ` of heat, changes the volume from `0.5 m^(3) to 2.0 m^(3)`. The change in internal energy of the gas isA. `6.75xx10^(5) J`B. `2.25xx10^(5) J`C. `3.25xx10^(5) J`D. `1.25xx10^(5) J`

For a spontaneous process, the change in Gibbs function is equal to theA. Heat content of the systemB. Entropy change of the systemC. Work of expansionD. Useful work

For an endothermic reaction, `Delta S` is positive. Then the reaction isA. Feasible when `T Delta S gt Delta H`B. Feasible when `Delta H gt T Delta S`C. Feasible at all temperaturesD. Not feasible at all

A sample of an ideal gas in a cyclinder is compressed adiabatically to `1/3rd` of its volume. Will final pressure be more or less than `3 times` the initial pressure?

Three moles of an ideal gas kept at a constant temperature of `300K` are compressed from a volume of 4 litre to 1 litre. Calculate work done in the process. `R= 8.31 J mol e^(-1)K^(-1)`.

A gas is suddenly compressed to `1/4th` of its original volume. Caculate the rise in temperature when original temperature is `27^(@)C. gamma= 1.5`.

Air in the cyclinder of a diesel engine is compressed to `1/15` of its initial volume. If initial temperature is `300K` and initial pressure is `10^(5)Pa`, find the final temperature and final pressure. Take `gamma= 1.4`.

`C_(P) -C_(V)` for an ideal gas is………….. .

The total energy of `1 mol` of an ideal monatomic gas at `27^(@)C` is……. .

A system is said to be………..if it can neither exchange matter nor energy with the surroundigs.

A container that suits the occurrence of an isothermal process should be made ofA. CopperB. GlassC. WoodD. cloth

Write two limitation of the first law of thermodynamics.

Heat cannot flow itself from a body at lower temperature to a body at higher temperature is a statement or consequence of which law of thermodynamics ?

Can we increase the coefficient of performance of a refrigerator by increasing the amount of working substance?

The initial pressure and volume of a gas enclosed in a cylinder are respectively 1 × 105 N/m2 and 5 × 10-3 m3 . If the work done in compressing the gas at constant pressure is 100 J. Find the final volume of the gas.