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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.
| 3701. |
Enthalpy of combustion of carbon to CO2 is -393.5 kJ mol-1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas. |
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Answer» Formation of CO2 from carbon and dioxygen gas is represented as: |
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| 3702. |
For an isolated process, U = 0, what will be the value of S for this? |
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Answer» Since ∆U = 0, there is no exchange of energy with the surroundings. In the process, As will be positive and the reaction will be spontaneous. That means As is greater than zero. |
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| 3703. |
A mixture of hydrocarbons containing acetylene and ethane, when burned under controlled consitions produced `16.20g` of water and `139.7 kcal` of heat. Given the molar composition of the mixture. Standard heats of combustion for acetylene and ethane are, respectively, `-310.6` and `-373.8 kcal`. |
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Answer» `CHunderset(x)-=CH+(5)/(2)O_(2)rarrunderset(2x)2CO_(2)+H_(2)ODeltaH_(1)=-310.6kcal` `C_(2)underset(y)(H_(6))+(7)/(2)O_(2)rarrunderset(2y)2CO_(2)+underset(3y)3H_(2)O DeltaH_(2)=-372.kcal` Total mole of water `=(x +3y) = (16.20)/(18)` `x = (0.9 -3y)` `x (-310.6) +y (-373.8) =- 139.7` `-310.6x - 372.8y =- 139.7` `-310.6 (0.9-3y) - 372.8y =- 139.7` `-310.6 xx 0.9 +3 xx310.6y - 373.8y =- 139.7` `-279.54 +559y =- 139.7` `559y = 139.84` `y = 0.25 mol` `x +3y = 0.9` `x +3 xx 0.25 = 0.9` `x = 0.15 mol` Acetylene `= 0.15 mol` `C_(2)H_(6) = 0.25 mol` |
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| 3704. |
In a process, `701 J` of heat is absorbed by a system and `394 J` of work is done by the system. What is the change in internal energy for the process? |
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Answer» Correct Answer - q=+701 J w=-394 J, since work is done by the ststem `DeltaU=307 J` According to the first law of thermodynamics, `DeltaU=q+W(i)` Where, `DeltaU`= change in internal energy for a process q=heat W=work Given, q=+701 (Since heat is absorbed) W=-394 J (Since work is done by the system) Substituting the values in expression, (i), we get `DeltaU=701 J +(-394 J)` `DeltaU=307 J` Hence, the change in internal energy for the given process is 307 J. |
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| 3705. |
In thermodynamics, a process is called reversible whenA. surroundings and sysytem change of the each otherB. there is no boundary between system and surroundingsC. the surroundings are always in equilibrium with the systemD. the System changes into the boundary surroundings spontaneously |
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Answer» Correct Answer - c |
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| 3706. |
The enthalpy change for chemical reaction is denoted aas `DeltaH^(Theta)` and `DeltaH^(Theta) = H_(P)^(Theta) - H_(R)^(Theta)`. The relation between enthalpy and internal enegry is expressed by equation: `DeltaH = DeltaU +DeltanRT` where `DeltaU =` change in internal enegry `Deltan =` change in number of moles, `R =` gas constant. Which of the following equations corresponds to the definition of enthalpy of formation at `298K`?A. `C("graphite") +2H_(2)(g) +(1)/(2)O_(2)(l) rarr CH_(3) OH(g)`B. `C("diamond") +2H_(2)(g)+(1)/(2)O_(2)(g) rarr CH_(3)OH(l)`C. `2C("graphite") +4H_(2)(g) +O_(2)(g) rarr 2CH_(3)OH(l)`D. `C("graphite") +2H_(2)(g) +(1)/(2)O_(2)(g) rarr CH_(3)OH(l)` |
| Answer» Graphite is a standard state of carbon and `CH_(3)OH` in liquid state is also standard state. | |
| 3707. |
A sample of organ gas at `1atm` pressure and `27^(@)C` expands reversibly and adiabatically from `1.25 dm^(3)` to `2.50 dm^(3)`. Calculate the enthalpy change in this process. `C_(vm)` for orgon is `12.48J K^(-1) mol^(-1)`. |
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Answer» `DeltaH = n xx C_(P) xx DeltaT` and `C_(P) = C_(V) +R = 12.48 + 8.134 = 20.794 J K^(-1)` `n = (PV)/(RT) = (1xx1.25)/(0.0821xx300) = 0.05` Also for reversible adiabatic change `(TV)^(gamma -1) =` constant or `T_(2)V_(2)^(gamma -1) = T_(1)V_(1)^(gamma-1)` or `T_(2) = T_(1) xx ((V_(1))/(V_(2)))^(gamma -1) (gamma = 1.66` for argon) `= 300 xx ((1.25)/(2.50))^(1.66-1) = 300 xx ((1)/(2))^(0.66)` `T_(2) = 189.85K` `DeltaT = T_(2) - T_(1) = 189.85 - 300 =- 110.15` Thus, `DeltaH = 0.05 xx 20.794 xx (-110.15) =- 114.52J` |
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| 3708. |
For an ideal gas, an illustratio of three different paths `A(B+C)` and `(D+E)` from an initial state `P_(1), V_(1), T_(1)` to a final state `P_(2), V_(2),T_(1)` is shown in the given figure. Path `A`represents a reversible isothermal expansion form `P_(1),V_(1)` to `P_(2),V_(2)`, Path `(B+C)` represents a reversible adiabatic expansion `(B)` from `P_(1),V_(1),T_(1)to P_(3),V_(2),T_(2)` followed by reversible heating the gas at constant volume `(C)`from `P_(3),V_(2),T_(2)` to `P_(2),V_(2),T_(1)`. Path `(D+E)` represents a reversible expansion at constant pressure `P_(1)(D)` from `P_(1),V_(1),T_(1)` to `P_(1),V_(2),T_(3)` followed by a reversible cooling at constant volume `V_(2)(E)` from `P_(1),V_(2),T_(3) to P_(2),V_(2),T_(1)`. What is `DeltaS` for path `A`?A. `nR In (V_(2))/(V_(1))`B. `P(V_(2)-V_(1))`C. `-P(V_(2)-V_(1))`D. `nR(V_(2)-V_(1))` |
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Answer» For path `A` `DeltaS = (q_(rev))/(T) = (nRT)/(T_(1)) In (V_(2))/(V_(1)) = nR In (V_(2))/(V_(1))` |
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| 3709. |
A carnot engine absorbs `1000J` of heat energy from a reservoir at `127^(@)C` and rejecs `600J` of heat energy during each cycle. Calculate (i) efficiency of the engine, (ii) temperature of sink, (iii) amount of useful work done per cycle. |
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Answer» Here, `Q_(1)= 1000J, Q_(2)=600J` `T_(1)= 127^(@)C= 127+273= 400K`, `eta=? T_(2)=? W=?` From `(Q_(2))/(Q_(1))= (T_(2))/(T_(1)), T_(2)= (Q_(2))/(Q_(1))xxT_(1)` `T_(2)= (600)/(1000)xx400= 240K= 240-273= -33^(@)C` `eta=1 -(T_(2))/(T_(1))=1 -(240)/(400)= 0.4= 40%` Also, `W=Q_(1)-Q_(2)= 1000-600= 400J` |
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| 3710. |
One mole of a non-ideal gas undergoes a change of state `(2.0atm,3.0L,95K)rarr(4.0atm,5.0L,245K)` With a change in internal energy `DeltaE=30L` atm. The change in enthalpy `(DeltaH)` in the process in `L`-atm isA. `40.0`B. `42.3`C. `44.0`D. not defined because pressure is not constant |
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Answer» Correct Answer - c `H=U+PV` or `H_(2)-H_(1)=DeltaU+(P_(2)V_(2)-P_(1)V_(1))` `=30+(4xx5-2xx3)` `=44 L atm` |
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| 3711. |
Which of the following statements is flase?A. Work is a state functionB. Temperature is a state functionC. Change in the state is completely defined when the initial and final states are specifiedD. Work appears at the boundary of the system |
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Answer» Correct Answer - a |
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| 3712. |
One mole of a non-ideal gas undergoes a change of state `(2.0atm,3.0L,95K)rarr(4.0atm,5.0L,245K)` With a change in internal energy `DeltaE=30L` atm. The change in enthalpy `(DeltaH)` in the process in `L`-atm isA. 42.3B. 40C. 44D. not defined, because |
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Answer» Correct Answer - C In this process, all the three variables `p,V` and `T` are changing. According to thermodynamics, `H = U + PV` `:. H_(1) = U_(1) + P_(1) V_(1)` and `H_(2) = U_(2) + P_(2) V_(2)` where the subscripts describe the initial and final states. `Delta H = H_(2) - H_(1)` `= (U_(2) + P_(2) V_(2)) - (U_(1) + P_(1) V_(2))` `(U_(2) + U_(1)) + (P_(2) V_(2) - P_(1) V_(1))` `30 + (4 xx 5 - 2 xx 3)` `30 + 14` `44 L atm` |
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| 3713. |
One moles of an ideal gas which `C_(V) = 3//2 R` is heated at a constant pressure of `1 atm` from `25^(@)C` to `100^(@)C`. Calculate `DeltaU, DeltaH` and the entropy change during the process. |
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Answer» As `(C_(P) - C_(V)) = R rArr C_(P) = C_(V) +R` `rArr C_(P) = (3)/(2)R+R =(5)/(2)R` Heat given at constnat pressure `(DeltaH) = nC_(P) DeltaT =1 xx (5)/(2)R xx (373-298)` `rArr (DeltaH) = 1xx (5)/(2) xx 1.987 xx 75 = 372.56 cal` Work done in the process `=- P DeltaV = - P (V_(2)-V_(1))` `=- P ((nRT_(2))/(P)-(nRT_(1))/(P))` as `(PV = nRT)` `=nRT (T_(2)-T_(1)) =- 1 xx 1.987 xx (373-298) =- 149.05 cal` From the first law of thermodynamics, `DeltaU = q +w = 372.56 - 149.05 = 223.51 cal` |
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| 3714. |
A heated irron block at `127^(@)C` loses `300J` of heat to the surroundings which are at a temperature of `27^(@)C`. This process is `0.05 J K^(-1)`. Find the value of `x`. |
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Answer» `Delta_(sys)S = (q_(sys))/(T_(sys)) =- (300)/(273+127)` `= (-300)/(400) (-3)/(4) J K^(-1)` `Delta_(surr)S = (-q_(sys))/(T_(surr)) =- (300)/(273+27)` `= (300)/(300) = + 1JK^(-1)` `Delta_(total)S or Delta_("universe")S = Delta_(sys)S + Delta_(surr)S` `= (-3)/(4) +1 = (1)/(4) = 0.25 J K^(-1)` `:. 0.05x = 0.25` `x = 5` |
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| 3715. |
One mole of a non-ideal gas undergoes a change of state `(2.0atm,3.0L,95K)rarr(4.0atm,5.0L,245K)` With a change in internal energy `DeltaE=30L` atm. The change in enthalpy `(DeltaH)` in the process in `L`-atm isA. 40B. 42C. 44D. non defined , becouse pressure is not system |
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Answer» Correct Answer - c |
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| 3716. |
A sample of argon gas at 1 atm pressure and `27^(@)C` expands reversibly and adiabatically from `1.25dm^(3)` to `2.50 dm^(3)`. Calculatethe enthalpy changein this process. `C_(v,m)` for argonis `12.48 JK^(-1) mol^(-1)` |
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Answer» For adiabatic process , `T_(1)V_(1)^(gamma-1)= T_(2)V_(2)^(gamma-1) ` or `((V_(1))/(V_(2)))^(gamma-1) = (T_(2))/(T_(1))` But `gamma=( C_(p))/(C_(v))` `:gt gamma -1 = (C_(p))/(C_(v))-1= ( C_(p)-C_(v))/(C_(v))= (R)/(C_(v))` ,brgt Hence `((V_(1))/(V_(2)))^(R//C_(v))= (T_(2))/(T_(1))` or `ln.(T_(2))/(T_(1))=(R)/(C_(v))ln.(V_(1))/(V_(2)) ` or `log.(T_(2))/(T_(1))= (R)/(C_(v)) log. (V_(1))/(V_(2))` i.e., `log. (T_(2))/(300) = ( 8.314)/( 12.48) log. ( 12.5)/(25)` or ` logT_(2) = log 300 + 0.666 log . ( 1)/(2)` `= 2.4771 + 0.666 ( - 0.3010) = 2.2766` or `T_(2) =` Antilog 2.2766 `=189 K` `C_(p)=(DeltaH)/( DeltaT)` or `DeltaH = C_(p) Delta T ` or for n moles `Delta H = n C_(p) Delta T`. Taking the gas as ideal. `PV = nRT ` or ` 1 xx 1.25 = n xx 0.0821 xx 300` or `n = 0.05` mole Further, `C_(p) = C_(v) + R = 12.48 + 8.314 = 20.794J` `:. Delta H = 0.05 xx 20.794 xx ( 189 - 300) = - 115 . 4 J` |
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| 3717. |
When a gas is compressed adiabatically and reversibly, the final temperature is -A. Higher than the initial temperature is -B. Lower than the initial temperatureC. The same as initial temperatureD. Dependent upon the rate of compression |
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Answer» Correct Answer - A For adiabatic process `-Delta E = W` `nC_(V).Delta T = -ve` `rArr Delta T = -ve` `rArr T_(2)lt T_(1)` |
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| 3718. |
One mole of a monoatomic ideal gas is heated at constant pressure from `25^(@)C` to `300^(@)C`. Calculate the `DeltaH,DeltaU`, work done and entropy change during the process. Given `C_(v)=3/2R`. |
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Answer» Correct Answer - `DeltaH=1375 cal,DeltaU=825 cal,` Work done `=-550 cal`, `DeltaS=3.269 calK^(-1) mol^(-1)` |
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| 3719. |
One mole of non`-` ideal gas undergoes a change of state `(1.0 atm, 3.0L, 200 K )` to `(4.0 atm, 5.0L,250 K)` with a change in internal energy `(DeltaU)=40 L-atm`. The change in enthalpy of the process in `L-atm`,A. 43B. 57C. 42D. None of these |
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Answer» Correct Answer - B When both P and V are changing `" "DeltaH=DeltaU +Delta(PV)` `" "DeltaU+(P_(2)V_(2)-P_(1)V_(1))` `" "DeltaH=40+(20-3)` = 57 L-atm |
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| 3720. |
A sample of `3.0` mole of perfect gas at `200 K` and `2.0 atm` is compressed reversibly and adiabatically until the temperature reaches `250 K`. Given that molar heat capacity at `27.5 J K^(-1) mol^(-1)` at constant volume calculate `q,W,DeltaU,DeltaH` and the final pressure and volume. |
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Answer» Correct Answer - `q=zero, W=+4.157 kJ,DeltaU=+4.157 kJ`, `DeltaH=5.372 kJ,5.2 atm,11.8 litre` |
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| 3721. |
One mole of a non-ideal gas undergoes a change of state (1.0 atm,3.0 L,200 K) to (4.0 atm,5.0 L,250 K) with a change in internal energy (`DeltaU`)=40 L-atm. The change in enthalpy of the process in L-atm :A. 43B. 57C. 42D. None of these |
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Answer» Correct Answer - B |
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| 3722. |
`DeltaH=30kJmol^(-1), DeltaS=75J//K//mol. "find boiling temperature at" 1atm: `A. `400K`B. `300K`C. `150K`D. `425K` |
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Answer» Correct Answer - A |
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| 3723. |
Find `DeltaG^(@)andDeltaH^(@)` for the reaction `CO(g)+(1)/(2)O_(2)(g) toCO_(2)(g)at 300K` when the standard entropy is `-0.094kJ mol^(_1)K^(-1).` The standard Gibbs free energiers fo formation for `CO_(2)andCOare-394.4 and-137.2kJ mol^(-1) respectively.`A. `DeltaG^(@)=257.2kJ//mol, DeltaH^(@)=285.4kJ//mol`B. `DeltaG^(@)=514.4kJ//mol, DeltaH^(@)=-570.8kJ//mol`C. `DeltaG^(@)=514.4kJ//mol, DeltaH^(@)=-570.8kJ//mol`D. `DeltaG^(@)=-257.2kJ//mol, DeltaH^(@)=-285.4kJ//mol` |
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Answer» Correct Answer - D |
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| 3724. |
The bond dissociation energy of gaseous `H_(2), Cl_(2)` and `HCl` are `104, 58` and `103 kcal mol^(-1)` respectively. The enthalpy of formation for `HCl` gas will beA. `-44.0 kcal`B. `-22.0 kcal`C. `22.0 kcal`D. `44.0 kcal` |
| Answer» Correct Answer - B | |
| 3725. |
The bond dissociation energy of gaseous `H_(2),Cl_(2)` and HCl are 104,58,103 kcal `mol^(-1)` respectively. The enthalpy of formation for HCl gas will be :A. `-44.0 kcal`B. `-22.0 kcal`C. 22.0 kcalD. 44.0 kcal |
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Answer» Correct Answer - B |
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| 3726. |
The heat of formation of CO and CO2 are -26.4 kcal and -94 kcal, respectively. Heat of combustion of carbon monoxide will be (a) +26.4 kcal (b) -67.6 kcal (c) -120.6 kcal (d) +52.8 kcal |
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Answer» (b) -67.6 kcal CO(g) + O2(g) → CO2(g) ∆HC0 (CO) = [∆Hf (CO2) – ∆Hf(CO) + ∆Hf(O2)] ∆HC0 (CO) = -94 KCal – [- 26.4 KCal + 0] ∆HC0 (CO) = -94 KCal + 26.4 KCal ∆HC0 (CO) = -67.4 KCal |
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| 3727. |
When enthalpy and entropy change for a chemical reaction are `-2.5 xx10^(3)` cals and `7.4` cals `deg^(-1)` respectively. Predict that reaction at 298 K isA. SpontaneousB. ReversibleC. IrrevesibleD. Non- spontaneous |
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Answer» Correct Answer - A `DeltaG=DeltaH-T Delta S` |
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| 3728. |
`CO_((g))+1/2 O_(2) (g) rarr CO_(2(g))` `DeltaH=-67.37` K.cal at `25^(@)C` the change in entropy accompanying process is `-20.7` cal.`deg^(-1) mol^(-1)` then at `25^(@)C` the change in free energy isA. `-10` K. cal `mol^(-1)`B. `-15` K. cal. `"mole"^(-1)`C. `-32` K. cal `mol^(-1)`D. `-61.2` K.cal `mol^(-1)` |
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Answer» Correct Answer - D `DeltaG=DeltaH-T Delta S` |
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| 3729. |
1 g of graphiteis burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation C ( graphite ) `+ O_(2)(g) rarr CO_(2)(g)` During the reaction, temperature rises from 298K to 299 K. If the heat capacity ofthe bomb calorimeter is `20.7 kJ // K `, what is the enthalpy change for the above reaction at 298K and 1 atm ? |
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Answer» Rise in temperature of the calorimeter `= 299- 298K = 1K` Heat capacity of the calorimeter `= 20.7 k J K^(-1)` `:. `Heat absorbed by the calorimeter `= C_(v) xxDelta T = (20.7 k J K^(-1)) (1K) = 20.7 kJ` This s the heat evolved in the combustionof1 g of graphite. `:. ` Heat evolved inthe combustionof 1 mole of graphite , `i.e., 12g` of graphite `= 20.7 xx 12 kJ = 248.4 kJ` As this is the heat evolved and the vessel is closed, therefore, enthalpy change of the reaction `( Delta U )` `= - 248.4 kJ mol^(-1)` |
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| 3730. |
A 1.250 g sample of octane (C18H18) is burnt in excess of oxygen in a bomb calorimeter. The temperature of calorimeter rises from 294.05 to 300.78 K. If heat capacity of the calorimeter is 8.93 kJ/K, find the heat transferred to calorimeter. |
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Answer» Mass of octane, m = 1.250 g = 0.00125 Heat capacity, c = 8.93 kJ/K Rise in temp, ΔT = 300.78 - 294.05 = 6.73 K Heat transferred to calorimeter = m x c x ΔT = 0.00125 x 8.93 x 6.73 = 0.075 kJ |
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| 3731. |
A 1.25g sample of octane (C18 H18) is burnt in excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rises from 294.05 to 300.78K. If heat capacity of the calorimeter is 8.93 KJ/K. find the heat transferred to calorimeter. |
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Answer» Mass of octane, M = 1.250g. = 0.00125. Heat capacity, c = 8.93 kJ/k Rise in temp, ΔT = 300.78 - 294.05 = 6.73K Heat transferred to calorimeter = m x c x ΔT = 0.00125 x 8.93 x 6.73 = 0.075 kJ. |
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| 3732. |
Concerete is produced from a mixture of cement , water, sand and small stones. It consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In later steps of cement production a small amount of gypsym , `CaSO_(4).2H_(2)O` is added to improve subsequent hardening of concrete . The use of elevated temperature during the final production may lead to formation of unwanted hemihydrate, `CaSO_(4).1//2H_(2)O.` Consider the following reaction: `CaSO_(4).2H_(2)O(s) rarr CaSO_(4).(1)/(2) H_(2)O(s)+(3)/(2)H_(2)O(g)` The following themodynamic data apply at `25^(@)C` standard pressure : 1 bar `{:("Compound","DeltaH_(f(KJ//mol))^(@),S^(@)(JK^(-1)mol^(-1))),(CaSO_(4).2H_(2)O(s),-2021.0,194.0),(CaSO_(4).(1)/(2)H_(2)O(s),-1575.0,130.5),(H_(2)O(g),-242.8,188.6):}` R=8.314 `JK^(-1)mol^(-1)` Equillibrium pressure (in bar) of water vapour in closed vessel containing `CaSO_(4).2H_(2)O(s),CaSO_(4).(1)/(2) H_(2)O(s),H_(2)O(g)` at `25^(@)`CA. `17.35 xx 10^(-4)` barB. `2.15 xx 10^(-4)` barC. `8.10 xx 10^(-3)` barD. `7.00xx10^(-4)` bar |
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Answer» Correct Answer - C `DeltaH^(@)=DeltaH^(@) -DeltaS^(@)=17920 J "mol"^(-1) " "implies " "DeltaG^(@) =-2.303 RT " log" K_(p)` log `K_(p)=(DeltaG^(@))/(2.303 RT) = 7.22 xx 10 ^(-4) (P_(H_(2)O))^(3//2) " "implies " " P_(H_(2)O) = 8.1 xx 10^(-3)` atm |
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| 3733. |
Concerete is produced from a mixture of cement , water, sand and small stones. It consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In later steps of cement production a small amount of gypsym , `CaSO_(4).2H_(2)O` is added to improve subsequent hardening of concrete . The use of elevated temperature during the final production may lead to formation of unwanted hemihydrate, `CaSO_(4).1//2H_(2)O.` Consider the following reaction: `CaSO_(4).2H_(2)O(s) rarr CaSO_(4).(1)/(2) H_(2)O(s)+(3)/(2)H_(2)O(g)` The following themodynamic data apply at `25^(@)C` standard pressure : 1 bar `{:("Compound","DeltaH_(f(KJ//mol))^(@),S^(@)(JK^(-1)mol^(-1))),(CaSO_(4).2H_(2)O(s),-2021.0,194.0),(CaSO_(4).(1)/(2)H_(2)O(s),-1575.0,130.5),(H_(2)O(g),-242.8,188.6):}` R=8.314 `JK^(-1)mol^(-1)` `DeltaH^(@)` for the formation of 1.00 kg of `CaSO_(4).(1)/(2) H_(2)O(s)` from `CaSO_(4).2H_(2)O(s)` isA. `+446 KJ`B. `+484 KJ`C. `-446 KJ`D. `-484 KJ` |
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Answer» Correct Answer - B `DeltaH^(@) = H[CaSo_(4).(1)/(2)H_(2)O(s)] + H[(3)/(2)H_(2)O(g)] implies -H[CaSO_(4).2H_(2)O(s)] = 833 KJ "mol"^(-1) =+ 484 KJ " for " 1 Kg` |
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| 3734. |
For the reaction `C_(2)H_(4)(g)+3O_(2)(g) rarr 2CO_(2) (g) +2H_(2)O(l)`, Delta E=-1415 kJ`. The `DeltaH` at `27^(@)C` isA. `-1410 kJ`B. `-1420 kJ`C. `+1420 kJ`D. `+1410 kJ` |
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Answer» Correct Answer - B `DeltaH=DeltaU+Delta nRT` |
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| 3735. |
A weightless piston divides a thermally insulated cylinder into two equal parts. One part contains one mole of an ideal gas with adiabatic exponent `gamma`, the other is evacuated. The initial gas temperature is `T_0`. The piston is released and the gas fills the whole volume of the cylinder. Then the piston is slowly displaced back to the initial position. Find the increment of the internal energy and entropy of the gas was resulting from these two processes. |
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Answer» The process consists of two parts. The first part is free expansion in which `U_f = U_i`. The second part is adiabatic compression in which work done results in change of internal energy. Obviously, `0 = U_F - U_f + int_(V_f)^(V_0) pdV, V_f = 2 V_0` Now in the first part `p_f = (1)/(2) p_0, V_f = 2 V_0`, because there is no change of temperature. In the second part, `p V^gamma = (1)/(2) p_0 (2 V_0)^gamma = 2^(gamma -1) p_0 V_0^gamma` `int_(2 V_0)^(V_0) pdV = int_(2 V_0)^(V_0) (2^(gamma - 1)p_0 V_0^gamma)/(V^gamma) dV = [(2^(gamma -1) p_0 V_0^gamma)/(- gamma + 1) V^(1 - gamma)]_(2 V_0)^(V_0)` =`2^(gamma -1) p_0 V_0^gamma V_0^(-gamma + 1) (2^(- gamma +1) -1)/(gamma -1) = -((2^(gamma -1) -1))/(gamma -1) RT` Thus `Delta U = U_F - U_i = (R T_0)/(gamma - 1) (2^(gamma -1) - 1)` The entropy change `Delta S = Delta S_1 + Delta S_(II)` `Delta S_I = R 1n 2` and `Delta S_(II) = 0` as the process is reversible adiabatic. Thus `Delta S = R 1n 2`. |
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| 3736. |
The heat of neutralisation of oxalic acid is `-25.4 kcal mol^(-1)` using strong base, `NaOH`. Hence, the enthaly change of the process is `H_(2)C_(2)O_(4) hArr 2H^(o+) +C_(2)O_(4)^(2-)` isA. `2.0 kcal`B. `-11.8 kcal`C. `1.0 kcal`D. `-1.0 kcal` |
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Answer» Oxalic acid has two ionisable `H^(o+)`. Hence, expected heat of neutralisation, if it behaves as a strong acid would have been `=- 13.7 xx2=- 27.4 kcal mol^(-1)` But experimetal value `=- 25.4 kcal mol^(-1)` `:.` Heat of ionisation `= 2.0 kcal mol^(-1)` |
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| 3737. |
In a given process on an ideal gas, `dW=0` and `dQ` is negative, then for the gas:A. The temperature will decreaseB. The volume will increaseC. The pressure will remain constantD. The temperature will increase |
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Answer» Correct Answer - A `q=dU+(-dW)` if `dW=0, q=-ve , dU` will be `-ve` or `T` will decrease. |
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| 3738. |
`C_(P) - C_(V) = R`. This `R` isA. Change in `KE`.B. Change in rotation energyC. Work done which system can do on expanding the gas per mol per degree increases in temperature.D. All correct |
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Answer» `PV =RT` at temperature `T` for one mol `P(V+DeltaT) = R (T +1)` at temperature `(T +1)` for one mole `:. P DeltaV = R` |
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| 3739. |
An ideal mono-atomic gas follows the path `ABDC`. The work done during the complete cycle is A. `-PV`B. `-2PV`C. `-1//2 PV`D. zero |
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Answer» Correct Answer - A Net work done during a complete cycle is equal to area under the cycle. `(-ve` if cycle is clockwise, i.e. work done by the gas and `+ve` when cycle is anticlockwise, i.e., work done on the gas) |
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| 3740. |
A metal block of density `5000kg//m^(3)` and mass `2kg` in suspended by a spring of force constant `200 N//m`. The spring block system is submerged in water vessel. Total mass of water in vessel is `300 gm` and in equilibrium the block is at a height `40` cm above the bottom of vessel. The specific heat of material of block is `250J//kg//k` and that of water is `4200J//kg//k`. Neglect the heat capacities of vessel and the spring. If the support is broken the rise in temperature of water, when block reaches botton of vessel isA. `0.0012^(@)C`B. `0.0049^(@)C`C. `0.0028^(@)C`D. `0.0^(@)C` |
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Answer» Correct Answer - B When the block is in equilibrium un the water the spring stretches by x `Kx+` weight of liquid displaced `=` weight of block `200x+(0)/(5000)xx1000xx10=2xx10 :. x=0.08m` In equilibrium energy stored in spring `U=(1)/(2)kx^(2)` `U=(1)/(2)xx200xx(0.08)^(2)= 0.64J` when support in booken & block reaches bottom of the vessel `U+Mgh=M_(W)S_(W) DeltaT+MxxS_(B) DeltaT` Where `M` is mass of block `:. 0.64+2xx10xx0.4=[0.3xx4200+2xx250] DeltaT` `:. DeltaT=0.0049^(@)C` |
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| 3741. |
One mole of Argon undergoes a process given by `PV^(3//2)`= const. If heat obtained by gas is `Q` and molar specific of the following in the process is `C` then which of the following is correct if temperature of gas changes by `-26 K` (assusme Argon as an ideal gas)A. `C=0.5 R, Q=13 R`B. `C= -0.5 R, Q=1.3 R`C. `C= -0.5 R, Q=13 R`D. `C= 0, Q=13 R` |
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Answer» Correct Answer - C If a gas under goes a themodynamic process `PV^(x)` =const, then molar specific heat `C=(R)/(gamma-1)-(R)/(x-1)` for Arogn `r=5//3` `:. C=(R)/(5/3-1)-(R)/(3/2-1)=-0.5R` `Q=nCDeltaT=1xx-0.5Rxx-26=13R` |
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| 3742. |
Which expansion will produce more change in entropy during reversible and isothermal process?A. `1 mol H_(2)` at `300K` from `2 L` to `20L`B. `1 mol N_(2)` at `400 K` from `1L` to `10L`C. `1mol O_(3)` at `500K` form `3L` to `30L`D. All have same `DeltaS` |
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Answer» `DeltaS = 2.303 nR "log" (V_(2))/(V_(1))` `= 2.303 xx 1xx 8.314 log(10)` `= 19.15 J K^(-1) mol^(-1) ((V_(2))/(V_(1))=10 "for all")` |
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| 3743. |
Heat is supplied to a certain homogeneous sample of amtter, at a uniform rate. Its temperature is plotted against time, as shown in the figure below. Which of the following conclusions can be drawn? A. Its specific heat capacity is greater in the solid state than in the liquid state.B. Its specific heat capacity is smaller in the solid state than in the liquid state.C. Its latent heat of vaporisation is greater than its latent heat of fusion.D. Its latent heat of vaporisation is smaller than its latent heat of fusion. |
| Answer» The latent heat of vaporisation is greater them its latent heat of fusion. | |
| 3744. |
Statement -1. If free energy change of the system is negative, the process is spontaneous even if the total free energy change of the system and the surroundings is positive. Statement -2. The spontaneity of a process depends only on the free enegy change of the system.A. Statement -1 is True, Statement-2 is True, Statement-2 is a correct explanation of Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement -1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - a | |
| 3745. |
Statement -1. A compound is formed from its elements only if free energy of formation of the compound is positive . Statement -2. Many photochemical reactions have positive valeu for the free energy change.A. Statement -1 is True, Statement-2 is True, Statement-2 is a correct explanation of Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement -1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - d | |
| 3746. |
Assertion. At 298 K and 1 atm pressure, solids do not have any entropy but liquids and gases have definite values of entropy. Reason . At 298 K and 1 atm pressure, particle in a solid are fixed and do not have translatory, rotatory or vibraory motion whereas in liquids and gases, molecules have all the three types of motion.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 - d | |
| 3747. |
50mL of 0.10 B `Ba(OH)_(2)` is added to 50 mLof `0.10 M H_(2)SO_(4)`. The rise in temperature is`DeltaT_(1)` . If experiment is replaced by taking 100 mL of each solution, the rise in temperature is`DeltaT_92)`. ThenA. `DeltaT_(1)= 2DeltaT_(1)`B. `DeltaT_(2)= 2DeltaT_(1)`C. `DeltaT_(2)= DeltaT_(1)`D. `DeltaT_(2)= 4DeltaT_(1)` |
| Answer» Correct Answer - C | |
| 3748. |
The efficiency of a carnot cycle is `1//6` . On decreasing the tempertaure of the sink by `65^(@)C`, the efficiency increases to `1//3` .Calculate the tempretaure of source and sink . |
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Answer» Correct Answer - `117^(@)C, 52^(@)` `(1)/(6)=1 - (T_(L))/(T_(H))` `T_(H)=(6)/(5)T_(L)` `(1)/(3)= 1-((T_(L) - 65))/(T_(H))` `T_(L) = 325 K = 52 ^(@)C` |
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| 3749. |
A Carnot cycle has an efficiency of `40%` . Its low temperature resrvoir is at `7^(@)C` . What is the temperature of source ? |
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Answer» Correct Answer - `193.66^(@)C` `eta= (T_(2) - T_(1))/(T_(2))` `0.4 = (T_(2)- 280)/(T_(2))` `T_(2)= 466.67 K = 193.66 ^(@)` |
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| 3750. |
One mole of NaCl(s) on melting observed 30.5 kJ of heat and its entropy is increased by `28.8 JK^(-1)`. What is the melting point of sodium chloride ? |
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Answer» Correct Answer - T= 1059 K `Deltas= (DeltaH_(vap))/(T_(b))= (30.5)/(T_(b))` `T_(b)= 1059 K` |
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