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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.
| 1901. |
For a substance A, vapour pressure of liquid and solid state at temperature 400K is `0.1 "bar"` and `0.16 "bar"` respectively. What will be Gibbs free energy change for the following process. `A(s)[0.1"bar",400K,n=1]toA(l)` `[0.16"bar",400K,n=1]` |
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Answer» Correct Answer - C |
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| 1902. |
What will be change in molar Gibbs free energy of `H_(2)O(l)` at 300 K constant temperature if it is compressed from 10 bar to 20 bar : [1 bar L=100 J]A. `36 J//mol-K`B. `28 J//mol-K`C. `18 J//mol-K`D. `10 J//mol-K` |
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Answer» Correct Answer - C |
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| 1903. |
Which of the following options correctly reagarding the following statements? Statements-1: `DeltaS_(f)^(@)NH_(3)(g)and DeltaS_(f)^(@) PCl_(5)(g)ltO` Statement-2: On heating a metal , entorpy of metal increases. Statement-3: `DeltaS_(f)^(@)NH_(2)(l)=0`A. All the statement are incorrectB. Only statement-3 are incorrectC. only statement-1 is correctD. All the statement are incorrect |
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Answer» Correct Answer - B |
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| 1904. |
In which of the following cases, entropy of the system is increasing.A. `DeltaS_(f)^(@)NH_(2)(l)=0`B. `C_("graphite")to C("diamond")`C. Ideal monotomic gas subjected to change on state from `1"atm" 300K to 2"atm" 600K`D. Dimerisation of acetic acid |
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Answer» Correct Answer - C |
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| 1905. |
From a source of infinite heat capacity 300 kcal heat is extracted at `727^(@)` C temperature then change in its entropy (`"cal"//"kelvin"`) will be ?A. `-300 ln 2`B. `-300`C. `-500`D. `-500 ln 2` |
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Answer» Correct Answer - B |
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| 1906. |
Statement-1 : The thrid law of thermodynamics implies that absolte zero cannot be reached. Statement-2 : `Delta G^(@)` for an ideal gas reaction is a function of temperature. Statement-3 : The adiabatic expansion of a gas into a vacuum is spontaneous.A. F F FB. T T TC. T F FD. F T F |
| Answer» Correct Answer - B | |
| 1907. |
Statement-1 : Entropy of all elements is zero at zero Kelvin. Statement-2 : Standard entropy of all elements is greater than zero. Statement-3 : Entropy of all elements and compounds decreases with decrease of temperature and becomes zero at absolute zero temperature.A. F F FB. T T TC. T F FD. F T F |
| Answer» Correct Answer - B | |
| 1908. |
The free energy change for a reversible reaction at equilibrium isA. ZeroB. PositiveC. NegativeD. Cannot say |
| Answer» Correct Answer - A | |
| 1909. |
Statement-1 : Neutralisation reaction is an Endothermic Process. Statement-2 : Standard Enthalpy of Neutralisation for different pairs of strong acid and strong base are different. Statement-3 : Standard Enthalpy of Neutralisation for a pair of strong acid and strong base is higher than that of weak acid and weak base.A. F F TB. F T TC. T F TD. F T F |
| Answer» Correct Answer - A | |
| 1910. |
The chemical reaction : A `rarr` P, `Delta H^(@)` = 2.8 kJ is spontaneous only above 400 K. Therefore `Delta S` of reaction must be at leat `(JK^(-1))`. |
| Answer» Correct Answer - 7 | |
| 1911. |
Assertion: Mass and volume are extensive properties. Reason: Mass/volume is also an extensive parameter.A. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) area true and (R) is not the correct explanation of (A)C. (A) is true but (R) is falseD. (A) is false but (R) is true |
| Answer» Correct Answer - C | |
| 1912. |
`1 L N_(2)` (g) and `3 L H_(2)` (g) at 1 bar pressurre is allowed to react at constant pressure. The temperature of 100 gm water surrounding reaction vessel rose by `((1)/(14))"^(@)C` find the magnitude of change in internal energy in joules for the process in reaction vessel. [Specific heat of `H_(2)O(l) =4.2J//"^(@)C//g,` 1litre atm=100 J] |
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Answer» Correct Answer - 100 |
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| 1913. |
For an adiabatic process, which of the following relations is correct?A. `Delta U = 0`B. `P DeltaV = 0`C. `q =0`D. `q = +w` |
| Answer» `q = 0` (Adiabatic process) | |
| 1914. |
For an adiabatic process, which of the following relations is correct?A. `Delta E = 0`B. `P Delta V = 0`C. q = 0D. `q = + W` |
| Answer» Correct Answer - C | |
| 1915. |
Which of the following is an extensive property?A. MassB. EnthalpyC. EnergyD. All of these |
| Answer» Correct Answer - D | |
| 1916. |
Identify the intensive quantity from the followingA. Enthalpy and temperatureB. Volume and temperatureC. Enthalpy and volumeD. Temperature and refractive index |
| Answer» Correct Answer - D | |
| 1917. |
A well stoppered thermo flask containing some ice cubes is an example ofA. Closed systemB. Open systemC. Isolated systemD. Non-thermodynamic system |
| Answer» Correct Answer - C | |
| 1918. |
The first law of thermodynamics is concerned with the conservation ofA. Total energy of a systemB. Energy changes in a systemC. Rate of a chemical changeD. Mass changes in nuclear reactions |
| Answer» Correct Answer - B | |
| 1919. |
A reaction from unknown reactants (R) : `R rarr P` is molar entropy of the products (P) is measured to be 460 J/mol K, and 71.3 kJ of heat was given off in the process to the surroundings when 1 mol reacted, then the molar entropy of the reactants has to beA. at least -230 J/mol KB. at the most 690 J/mol KC. at the most 230 J/mol KD. at the least 690 J/mol K |
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Answer» Correct Answer - B Let us solve this at boundary condition of `Delta G^(@) = 0` `Delta S^(@) = (Delta H^(0))/(T) = (-71.3 xx 1000)/(310) = -230 JK^(-1)` `rArr -230 JK^(-1) = S_(m)^(0) (R) = 460 JK^(-1) - S_(m)^(0) (R)` `rArr S_(m)^(0) (R) = 690 JK^(-1)`. It is the maximum value of entropy which the reactant can have. If `S_(m)^(0) gt 690 JK^(_1) mol^(-1), Delta G^(0) gt 0` |
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| 1920. |
One mole of a monatomic ideal gas initially at a pressure of 2.00 bar and a temperature of 273 K is taken to a final pressure of 4.00 bar by a reversible path defined by `p//V`= constant. Taking `C_(V)` to be equal to 12.5 K `mol^(-1) K^(-1)`, the value of `(Delta U)/(w)` for this process is calculated to beA. `-3.0`B. `-1.5`C. `+1.5`D. `+3.0` |
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Answer» Correct Answer - A Given, `(P)/(V) = K` (constant) also, `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` `rArr T_(2) = T_(1) ((p_(2)V_(2))/(p_(1)V_(1))) = T_(1) ((p_(2))/(p_(1)))`......(i) `rArr Delta U = C_(V) Delta t = 3C_(V) T_(1) rArr -dw = pdV = kVdV` `rArr w = - (K)/(2) xx (3T_(1)R)/(K) = (3RT_(1))/(2)` `rArr (DeltaU)/(w) = - (2C_(V))/(R) = -3` |
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| 1921. |
Thermodynamics is not concerned about….A. energy changes involved in a chemical reactionB. the extent to which a chemical reaction proceedsC. the rate at which a reaction proceedsD. the feasibility of a chemical reaction |
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Answer» Correct Answer - C Thermodynamics is not concened with the rate at which a reaction proceeds. Thermodynamics deals with the energy change, feasibility and extent of a reaction, but not with the rate and mechanism of a process. |
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| 1922. |
Which of the following is correct for the change shown below ? `underset(298K)(aA+bB) overset(Delta_(r)S_(1)^(@))rarrcunderset(298)(C+dD)` `Delta_(r)S_(1)^(@)darr " " uarr Delta_(r)S_(4)^(@)` `underset(0K)(aA+bB) overset(Delta S_(3)^(@))rarr cunderset(0K)(C+dD)`A. `Delta_(r)S_(1)^(@) = cS_(C)^(@) + dS_(D)^(@) - (aS_(A)^(@)+bS_(B)^(@))`B. `Delta_(r)S_(2)^(@) = -[aS_(A)^(@)+bS_(B)^(@)]`C. `Delta_(r)S_(3)^(@) = 0`D. `Delta_(r)S_(4)^(@) = -[cS_(C)^(@)+dS_(D)^(@)]` |
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Answer» Correct Answer - A::B::C `Delta_(r)S_(4)^(@) = sumS_(P) - sum S-(R) = CS_(C)^(@)+dS_(D)^(@) - 0 - 0` |
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| 1923. |
Molar heat capacity of water in equilibrium with ice at canstant pressure isA. zeroB. `oo`C. `40. 45 kJ K^(-1) "mol"^(-1)`D. `75.48 JK^(-1) "mol"^(-1)` |
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Answer» Correct Answer - B `C_(P) = (q)/(nDeltaT) rArr DeltaT = 0 rArr C_(P) = oo` |
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| 1924. |
When 20 ml of `0.1 M HCl` is mixed with 20 ml of `0.1 M NaOH`, the rise in temperature is `T_(1)`. When the experiment is repeated using 20 ml of `0.2 M HCl` and `0.2 M NaOH` solutions, the rise in temperature is `T_(2)`. ThenA. All are correctB. i and ii are correctC. ii and iii are correctD. i and iii are correct |
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Answer» Correct Answer - B 0.05 mol `H^(+) + 0.05` mol `OH^(-)` |
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| 1925. |
Given, `2Fe_2O_(3(s)) to 4Fe_((s)) + 3 O_(2(g)), Delta_r G_1^@ = +1487kJmol^(-1)` `6CO_((g)) +3O_(2(g)) to 6CO_(2(g)), Delta_r G_2^@ = -1543.2kJmol^(-1)` Select the correct statements:A. `Delta_(r)G^(@)` for reduction of iron oxide by CO is `+56.2 kJ mol^(-1)`B. `Fe_(2)O_(3)` can be reduced by CO spontaneouslyC. `Fe_(2)O_(3)` cannot be reduced by CO spontaneouslyD. The reduction of `Fe_(2)O_(3)` takes part in higher part of blast furnace |
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Answer» Correct Answer - B `Delta G^(@)` for `2Fe_(2)O_(3)+600 rarr 4Fe+6CO_(2) + Delta_(r)G_(1)^(@) + Delta_(r)G_(2)^(@) = +1487 - 1543.2 = -56.2 kJ mol^(-1)`. The reduction occurs spontaneously in lower part of blast furnace. |
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| 1926. |
Molar heat capacity of water in equilibrium with the ice at constant pressure is `:`A. ZeroB. Infinity `(oo)`C. `40.45 kJ K^(-1)mol^(-1)`D. `75.48 J K^(-1)mol^(-1)` |
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Answer» By definition `C_(P,m) = (dq_(p))/(dT)` For `H_(2)O(l) hArr H_(2)O(s)`, temperature does not change if some heat is given to the system. Hence `C_(P,m) =(+ve)/(Zero) =oo` |
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| 1927. |
Standard molar enthalpy of formation of `CO_(2)` is equal to `:`A. ZeroB. The standard molar enthalpy of combustion of gaseous carbon.C. The sun of standard molar enthalpies of formation of `CO` and `O_(2)`.D. The standard molar enthalpy of combustion of carbon (graphite) |
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Answer» Standard molare enthalpy of formation `CO_(2)` and the standard molar enthalpy of combustion of carbon (graphite) refer to the same chemical equation: `C("graphite") +O_(2)(g) rarr CO_(2)(g)` |
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| 1928. |
Thermodynics is no concerned about `"…............"`.A. energy changes involved in a chemical reaction.B. the extent to which a chemical reaction proceeds.C. the rate at which a reaction proceeds.D. the feasibility of a chemical reaction. |
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Answer» Correct Answer - C Thermodynamics is not concerned with the rate at which a reaction proceeds. |
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| 1929. |
How many calories are required to heat 40 gram of argon from 40 to`100^(@)C` at constant volume ?( `R= 2 cal // molK)`A. 120B. 2400C. 1200D. 180 |
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Answer» Correct Answer - D For a monoatomic gas like argon, `C_(v) = 3cal mol^(-1), C_(p) = 5 cal mol^(-1)`. Molar heat capacity is the heat required to raise the temperature of 1 mole of the gas by `1^(@)C `. 40 g of argon `=` 1 mole. Hence,heat required to raiseto raise the temperature from `40^(@)` to `100^(@)C` at constant volume `= 3 xx ( 100 - 40) = 180 cal` |
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| 1930. |
Comment on the thermodynamic stability of `NO(g)`, given `1/2N_(2)(g)+1/2O_(2)(g)rarr NO(g), Delta_(r)H^(Θ)=90 kJ mol^(-1)` `NO(g)+1/2O_(2)(g) rarr NO_(2)(g), Delta_(r)H^(Θ)=-74 kJ mol^(-1)` |
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Answer» The positive value of `Delta_(r)H` indicates that heat is absorbed during the formation of `NO_((g))`. This means that `NO_((g))` has higher energy than the reactants `(N_(2) and O_(2))` Hence, `NO_((g))` is unstable. The negative value of `Delta_(r)H` indicates that heat is evolved during the formation of `NO_(2(g)` from `NO_((g)) and O_(2(g))`. The product, `NO_(2(g))` is stabilized wiht minimum energy. Hence, unstable `NO_((g))` changes to unstable `NO_(2(g)).` |
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| 1931. |
For an ideal gas expanding adiabatically in vacuum,A. `Delta H lt 0`B. `Delta H = 0`C. `Delta H gt 0`D. none of these |
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Answer» Correct Answer - B For an ideal gas expanding adiabatically, `Delta U = W` because `q = 0. W = - P Delta V = 0` because for expansion against vacuum, `P_("ext") = 0`. According to thermodynamics, `Delta H = Delta U + P Delta V` Since both `Delta U` and `P Delta V` are zero, we have `Delta H = 0`. |
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| 1932. |
The stadard heats of formation of `NO_(2)(g)` and `N_(2)O_(4) (g)` are `33.5` and `8.4 kJ mol^(-1)` respectively. The heat of dimerisation of `NO_(2)` in kJ isA. 41.9B. `-25.1`C. `-52.2`D. `-58.6` |
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Answer» Correct Answer - D `2NO_(2) rarr N_(2)O_(4)` |
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| 1933. |
For the reaction `CO (g) + (1)/(2) O_(2) (g) rarr CO_(2) (g)` `Delta H` and `Delta S` are `283 kJ` and `-87 J K^(-1)`, respectively. It was intended to carry out this reaction at 1000,1500,3000, and 3500 K. At which of these temperatures would this reaction be thermodynamically spontaneous?A. 3000 and 3500 KB. 1500 and 3000 KC. 1500 , 3000, and 3500 KD. 1000, 1500, and 3000 K |
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Answer» Correct Answer - D According to thermodynamics, `Delta G = Delta H - T Delta S` Since both `Delta H` and `Delta S` are negative, the reaction will be spontaneous at a temperature below equilibrium temperature so that the `Delta H` term prdominates. At equilibrium, `Delta G = 0`. Thus, `T = (Delta H)/(Delta S) = (- 283 xx 10^(3) J)/(- 87 xx 10 J K^(-1)) = 3252 K` This implies that the reaction will be spontaneous at any temperature below `3252 K`. |
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| 1934. |
The heats of combustion for `C, H_(2)` and `CH_(4)` are `-349, -241.8` and `-906.7 kJ` respectively. The heat of formation of `CH_(4)` isA. 174.1 kJB. 274.1 kJC. 374.1 kJD. 74.1 kJ |
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Answer» Correct Answer - D `DeltaH=H_(R)-H_(P)" "C+2H_(2) rarr CH_(4)` |
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| 1935. |
`Delta G^(@)` for a reaction is `46.06 kcal mol^(-)`. `K_(P)` for the reaction at `300 K` isA. `10^(-22.22)`B. `10^(-8)`C. `10^(-44.55)`D. `10^(-35.54)` |
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Answer» Correct Answer - D According to thermodynamics, `Delta G^(@) = - 2.303 RT log k_(P)` We have `Delta G^(@) = 46.06 kcal mol^(-1)` `= (46.06) (1000) (4.184) J mol^(-1)` `R = 8.314 J K^(-1) mol^(-1)` `T = 300 K` `:. Log K_(p) = (Delta G^(@))/(-2.303 RT)` `= ((46.06)(1000)(4.184))/(-(2.303)(8.314)(300))` `= (192715.04)/(5744.14)` `= 33.54` or `K_(P) = 10^(-33.54)` |
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| 1936. |
Given standard enthalpy of formation of `CO(-110 "KJ mol"^(-1))` and `CO_(2)(-394 "KJ mol"^(-1))`. The heat of combustion when one mole of graphite burns isA. `-110` KJB. `-284` KJC. `-394` KJD. `-504` KJ |
| Answer» Correct Answer - C | |
| 1937. |
The heat of combustion of carbon and monoxide are -394 and -285 KJ `mol^(-1)` respectively. The heat of formation of CO in KJ `mol^(-1)` is :-A. `+109`B. `-109`C. `+218`D. `-218` |
| Answer» Correct Answer - B | |
| 1938. |
The value of `Delta H` and `Delta S` for five different reaction are given below. `|{:("Reaction",Delta H(kJ mol^(-)),Delta S(JK^(-)mol^(-))),(I,+98.0,+14.8),(II,+55.5,+14.8),(III,+28.3,-84.8),(IV,-40.5,+24.6),(V,+34.7," 0.0"):}|` On the basis of these values,k predict whihc one of these will be spontaneous at all temperature?A. Reaction IB. Reaction IIC. Reaction IIID. Reaction IV |
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Answer» Correct Answer - D According to thermodynamics, `Delta G = Delta H = T Delta S` If `Delta H` is negative and `Delta S` is positive, then `Delta G` will always be negative regardless of the temperature. |
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| 1939. |
The enthalpy combustion of a substanceA. Always positiveB. Always negativeC. Numerically equal to the heat of formationD. 1 and 3 both |
| Answer» Correct Answer - B | |
| 1940. |
Calculate the maximum work down when pressure on `10g` of hydrogen is reduced from `20` to `1atm` at a constant temperature of `273K`. The gas behaves ideally. Will there be any change in internal energy. Also calculate `q. (R = 2 cal K^(-1) mol^(-1))` |
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Answer» `w =- 2.303 nRT log.(P_(1))/(P_(2))`, mole of `H_(2) = (10)/(2)` `= 2.303 xx (10)/(2) xx 2xx 273 log.(20)/(1) = 8180 cal` Since the change is taking place at constant temperature, internla enegry will not change, i.e., `DeltaU = 0` `q = DeltaU - w = 0 - 8180 =- 8180cal` |
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| 1941. |
Calculate the lattice energy of the reaction `Li^(o+)(g) +CI^(Theta) (g) rarr LiCI(s)` from the following data: `Delta_("sub")H^(Theta)(Li) = 160.67 kJ mol^(-1), (1)/(2)D(CI_(2)) = 122.17 kJ mol^(-1) IP(Li) = 520.07 kJ mol^(-1),E_(A)(CI) = - 365.26 kJ mol^(-1)` and `Delta_(f)H^(Theta)(LiCI) =- 401.66 kJ mol^(-1)` |
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Answer» Applying the reaction `-Q = DeltaH^(Theta) +(1)/(2)D +IP -EA +U` and substituting the respective values. `-401.66 = 160.67 +122.17 +520.07 - 365.25 +U` `U =- 839.31 kJ mol^(-1)` |
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| 1942. |
Find the electron affinity of chlorine from the following data. Enthalpy of formation of `LiCI` is `-97.5 kcal mol^(-1)`, lattice energy of `LiCI =- 197.7 kcal mol^(-1)`. Dissociation energy of chlirine is `57.6 kcal mol^(-1)`, sublimation enthalpy of lithium `=+ 38.3 kcal mol^(-1)`, ionisation energy of lithium `=123.8 kcal mol^(-1)`. |
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Answer» `CI +e^(-) rarr CI^(Theta) DeltaH =?` Given: `Li(s) +(1)/(2)CI_(2)(g) rarr LiCI(s), DeltaH_(1) =- 97.5 kcal mol^(-1)` `Li^(o+) +CI^(Theta) (g)rarr LiCI(s), DeltaH_(2) =- 197.7 kcal mol^(-1)` `CI(g) rarr 2CI(g), DeltaH_(3) = 57.6 kcal mol^(-1)` `Li(s) rarr Li(g), DeltaH_(4) = 38.3 kcal mol^(-1)` `Li(g) rarr Li^(o+)(g) +e^(-), DeltaH_(5) = 123.8 kcal mol^(-1)` Rewritting equations: `LI(s) +(1)/(2)CI(2)(g) rarr LiCI(s), DeltaH_(1) = - 97.5 kcal mol^(-1)` `LICI(s) rarr Li^(o+)(g) +CI^(Theta)(g), DeltaH_(2) = 197.7 kcal mol^(-1)` `(1)/(2)[2CI_(2)(g) rarr CI_(2)(g)], DeltaH_(3) =(1)/(2)xx-57.6 kcal mol^(-1)` `Li(g) rarr Li(s), DeltaH_(4) =- 38.3 kcal mol^(-1)` `Li^(o+)(g) +e^(-) rarr Li(g), DeltaH_(5) =- 123.8 kcal mol^(-1)` `{:ulbar(CI(g)+e^(-)CI^(Theta)(g)):}` `DeltaH = - 97.5 +197.7 -(1)/(5) xx 57.6 - 38.3 - 123.8` `=- 90.7 kcal mol^(-1)` or Operating: `DeltaH = DeltaH_(1) - DeltaH_(2) - (1)/(2)DeltaH_(3)-DeltaH_(4)-DeltaH_(5)` `=97.5 -(-197.7) -(1)/(2)(57.6)-38.3 -123.8` `=- 90.7 kcal mol^(-1)` |
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| 1943. |
A sample of ideal gas undergoes isothermal expansion in a reversible manner from volume `V_(1)` to volume `V_(2)`. The initial pressure is `P_(1)` and the final pressure is `P_(2)`. The same sample is then allowed to undergoes reversible expansion under adiabatic conditions from volume `V_(1) to V_(2)`. The initial pressure being same but final pressure is `P_(2)`. The work of expansion in adiabatic process `(w_(adi))` is related to work of expansion in isothermal process `(w_(iso))` isA. `w_(adi) = w_(iso)`B. `w_(adi) lt w_(iso)`C. `w_(adi) = 2w_(iso)`D. `w_(adi) gt w_(iso)` |
| Answer» Work in reversible isothermal expansion is greater than work done in adiabatic expansion. | |
| 1944. |
`1mol` of an ideal gas undergoes reversible isothermal expansion form an initial volume `V_(1)` to a final volume `10V_(1)` and does `10kJ` of work. The initial pressure was `1xx 10^(7) Pa`. c. Calculate `V_(2)`. b. If there were `2mol` of gas, what must its temperature have been? |
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Answer» `w =- 2.303 nRt "log" (V_(2))/(V_(1))` where, `w` is work done by the system under isothermal reversible conditions, not that work done by the system is negative `-10 xx 10^(3) =- 2.303 xx 1xx 8.314 xx T "log"(P_(1))/(P_(2)) ……(i)` Also, `P_(1)V_(1) = P_(2)V_(2)` at constant temperature `1 xx 10^(7) xx V_(1) = P_(2) xx 10 V_(1)` `,. P_(2) = (1xx10^(7))/(10) = 10^(6) Pa` `:.` By equation (i), `- 10 xx 10^(3) =- 2.303 xx1xx 8.314 xxT "log" (10^(7))/(10^(6))` `:. T = 522.27K` Now, using `PV = nRT` for `1`mole of gas, `P = 1xx 10^(7) Pa = 10^(7) Nm^(-2)` `1 xx 10^(7) xx V_(1) = 1xx 8.314 xx 522.27` `V = 4.34 xx 10^(-4)m^(3)` b. If `2` mole of gas have been used, the temperature would have been `= (522.27)/(2) = 261.13K` |
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| 1945. |
`1mol` of an ideal gas undergoes reversible isothermal expansion form an initial volume `V_(1)` to a final volume `10V_(1)` and does `10kJ` of work. The initial pressure was `1xx 10^(7) Pa`. a. Calculate `V_(2)`. b. If there were `2mol` of gas, what must its temperature have been? |
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Answer» `W=-2.303 nRT"log"(V_(2))/(V_(1))` (i) Where `W` is work done by the system under isothermal reversible condition, note that work done by the system is negative `=-10xx10^(3)=-2.303xx1xx8.314xxT"log"(P_(1))/(P_(2))`....(1) Also `P_(1)V_(1)=P_(2)V_(2)` at constant temperature `1xx10^(7)xxV_(1)=P_(2)xx10V_(1)` `:. P_(2)=(1xx10^(7))/10=10^(6) Pa` `:.` By eq. (i) `-10xx10^(3)=-2.303xx1xx8.314xxT"log"(10^(7))/(10^(6))` `T=522.27 K` Now using, `PV=nRT` for `1` mole of gas, `P=1xx10^(7) Pa=10^(7) Nm^(-2)` `1xx10^(7)xxV_(1)=1xx8.314xx522.27` `V_(1)=4.34xx10^(-4) m^(3)` (ii) If `2` mole of gas have been used, the temperature would have been `522.27/2=261.13 K` |
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| 1946. |
1 mol of `H_(2) SO_(4)` in mixed with 2 mol of `NaIH`. The heat evolved will beA. `57.3 kJ`B. `2 xx 57.3 KJ`C. `57.3//2 kJ`D. cannot be predicted |
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Answer» Correct Answer - B During neutralization process, `H^(+) (aq.) + OH^(-) (aq.) rarr H_(2) O(1)` `57.3 kJ` energy (heat) is released for energy `1 mol H_(2) O` formation. When `1 mol` of `H_(2) SO_(4)` is mixed with 2 mol of `NaOH`, we get 2 mol of `H_(2) O`: `H_(2) SO_(4) + 2 NaOH rarr Na_(2) SO_(4) + 2 H_(2) O` Thus, the amount of energy released is `2 xx 57.3 kJ`. |
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| 1947. |
In an isothermal expansion of a gaseous sample the correct relation is (consider `w`(work) with sign according to new `IUPAC` convention) [The reversible and irreversible processes are carried out between same initial and final states]A. `W_(rev) gt W_(irrev)`B. `W_(irrev) gt W_(rev)`C. `q_(rev) lt q_(irrev)`D. can not be predicted |
| Answer» Correct Answer - B | |
| 1948. |
Heat of neutralisation of a strong dibasic acid in dilute solution by NaOH is nearly :A. `-27.4` Kcal/eq.B. `-13.7` Kcal/eqC. `13.7` Kcal/eq.D. `-13.7 Kcal/mol |
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Answer» Correct Answer - B Heat of neutralisation of strong acid and strong base is constant, i.e., `-13.7 KCal eq^(-1)` |
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| 1949. |
The energy required to break 76 gm gaseous fluorine into free gaseous atom is 180 kcal at `25^(@)`C. The bond energy of F - F bond will beA. 180 kcalB. 90 kcalC. 45 kcalD. 104 kcal |
| Answer» Correct Answer - B | |
| 1950. |
In the reaction , `CO_(2)(g)+H_(2)(g)rarrCO(g)+H_(2)O(g), " "DeltaH= 2.8kJ, DeltaH` representsA. heat of reactionB. heat of combustionC. heat of formationD. heat of solution |
| Answer» Correct Answer - A | |