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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.
| 801. |
A carnot engine is working in such a temperature of sink that its efficiency is maximum and never changes with any non-zero temperature of source. The temperature of sink will most likely to beA. 0 KB. `0^(@)C`C. `0^(@)F`D. Data insufficient |
| Answer» Correct Answer - A | |
| 802. |
Efficiency of carnot engine working between ice point and steam point isA. 0.249B. 0.257C. 0.268D. 0.288 |
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Answer» Correct Answer - C Here ice point ,`T_(2)=0^(@)C = 0 +273 =273 K` and steam point, `T_(1)=100^(@)C=100+273=373 K` Efficiency of the cannote engine `eta=1-(T_(2))/(t_(1))=1-(273)/(373)=(100)/(373)=(100)/(373)xx100%=26.8%` |
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| 803. |
How does second law of thermodynamics explain expansion of gas? |
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Answer» Since from second law,. d S ≥ O d S = change in entropy During the expansion of gas, the thermodynamic probability of gas is larger and hence its entropy is also very large. Since form second law, entropy cannot decrease ∴ following the second law, gas molecules move from one partition to another. |
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| 804. |
Define extensive properties. |
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Answer» Properties which depends upon amount of substance called extensive properties. Volume, enthalpy, entropy. |
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| 805. |
If 150 kJ of energy is needed for muscular work to walk a distance of one km , then how much of glucose one has to consumeto walk a distanceof five km, provided only `30%` energy is available for muscular work. The enthalpy of combustion of glucose is `3000kJ mol^(-1)`A. 75gB. 30gC. 180gD. 150g |
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Answer» Correct Answer - d Energy consumed for walking a distance of5 km `= 150 xx5kJ =750 kJ` 1 mole of glucose , `C_(6)H_(12)O_(6) ( 180g)` gives theoretically energy `= 3000kJ` Actual energy available`= ( 30)/( 100) xx 3000 = 900 kJ` Thus, for 900kJ of energy , glucose required `= 180 g` `:. `For 750 kJ of energy , glucose required `= ( 180)/( 900) xx750 g = 150 g` |
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| 806. |
Acetic acid and hydrochloric acid react With KOH solution. The enthalpy of neutralization of acetic acid is -55.8 kJ mol-1 while that of hydrochloric acid is -57.3 kJ mol-1. Why? |
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Answer» It is because HCl is strong acid, it ionises completely in aqueous solution whereas acetic acid is weak acid which does not ionise completely and some energy is used in its complete ionisation. |
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| 807. |
The temperature of the surface of sun is about 6000 K.Can we produce a temperature is 7000 K by converging sun's rays using a large convex lens? |
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Answer» No, as according to the second law of thermodynamics, heat cannot be transferred on its own, from a body at lower temperature to another at higher temperature. The maximum temperature we can produce by this method is 6000 K. |
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| 808. |
Enthalpies of solution of `BaCl_(2)(s)` and `BaCl_(2).2H_(2)O(s)` are `-20 kJ//"mole"` and `8.0kJ//"mole"` respectively. Calculate heat of hydration of `BaCl_(2)(s)`. |
| Answer» Correct Answer - `-28kJ//"mole"` | |
| 809. |
Standard enthalpy of formation of `C_(3)H_(7)NO_(2)(s),CO_(2)(g)` and `H_(2)O(l)` are `133.57,-94.05` and `-68.32kcal mo1^(-1)` respectively Standard enthalpy combustion of `CH_(4)` at `25^(@)C` is `-212.8 kcal mo1^(-1)` Calculate `DeltaH^(Theta)` for the reaction: `2CH_(4)+CO_(2)+1//2N_(2)rarrC_(3)H_(7)NO_(2)(s)+1//2H_(2)` Calculate `DeltaU` for combustion of `C_(3)H_(7)NO_(2)(s)`. |
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Answer» `2CH_(4) +CO_(2) +1//2N_(2) rarr C_(3)H_(7)NO_(2), DeltaH^(Theta) = ?` First find `Delta_(f)H^(Theta) of CH_(4)` Given `CH_(4) +2O_(2) rarr CO_(2) +2H_(2)O, DeltaH^(Theta) =- 212.8` Using the definition of `DeltaH`, `DeltaH^(Theta) = [Delta_(f)H^(Theta)underset((CO_(2))).+2Delta_(f)H^(Theta)2(H_(2)O)]-Delta_(f)H^(Theta)(CH_(4))` (Note that `Delta_(f)H^(Theta)O_(2)=0)` `rArr -212.8 =[-94.05 +2)-68.32) -Delta_(f)H^(Theta)(CH_(4))]` ` rArr Delta_(f)H^(Theta)(CH_(4)) =- 17.89 kcal//mol` Now find the `Deltah` of the required equation using `Delta_(f)H^(Theta) (CH_(4))`. `[Delta_(f)H^(Theta)(C_(3)H_(7)NO_(2))-0] -[2xxDelta_(f)H^(Theta)(CH_(4))+Delta_(f)H^(Theta)(CO_(2)) +0]` `rArr DeltaH =(-133.57) -2(-17.89) -(-68.32)` `=- 374 kcal mol^(-1)` Now calculate `DeltaH` (combustion) of `C_(3)H_(7)NO_(2)`. `C_(3)H_(7)NO_(2)(s) +15//4O_(2) rarr 3CO_(2)(g) +1//2N_(2)(g) +7//2H_(2)O(g)` `Delta_("comb")H^(Theta) =3Delta_(f)H^(Theta) (CO_(2)) +0 +7//2Delta_(f)H^(Theta) (H_(2)O) -Delta_(f)H^(Theta) (C_(3)H_(7)NO_(2)) -0` `=3(-96.05) +7//2 (-68.32) -(-133.57)` `=- 387.70 kcal mol^(-1)` Find `DeltaU^(Theta)` using `DeltaU = DeltaH^(Theta) -DeltanRT` `DeltaU =- 387.70 -(1//4) xx2xx 10^(-3) (298)` `=- 387.72 kcal mol^(-1)` |
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| 810. |
Which has larger absolute entropy per mole? a. `H_(2)O(l) at 298 K or H_(2)O(l) at 350K`. b. `N_(2)`or No both at` 298K` |
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Answer» a. `H_(2)O(l)` at `350K` has larger absolute entropy per mole. b. `NO` at `298K` has greater entropy. |
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| 811. |
In order to get maximum calorific output, a burner should have an optimum fuel value to oxygen ratio which corresponds to 3 times as much oxygen as it required theoretically for complete combustion of the fuel. A burner which has been adjusted for methane as fuel ( with x litre `//` hour of `CH_(4)` and 6x litre `//` hour of `O_(2))` is to be readjusted for butane, `C_(4)H_(10)`. In order to get the same calorific output, what should be the rate of supply of butane and oxygen?Assume that the losses due to incomplete combustion etc. are the same for both the fuels and that the gases behave ideally. Heats of combustion `:` `CH_(4) = 809 kJ // mol,C_(4) H_(10) =2878 kJ // mol` |
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Answer» Calculation of `C_(4)H_(10)` `CH_(4)+ 2O_(2) rarr CO_(2) + 2H_(2)O, DeltaH = - 809 kJ mol^(-1)` Initial volume `//` hr (L) x 6x Suppose the volume of 1 mole of the gas under the given conditions is V litre. Thus, V litre (1 mole)of `CH_(4)` give energy on combustion `= 809 kJ` `:. ` x litre of `CH_(4)` will give energy on combustion `= (809 x)/( V ) kJ` Now, 2878 kJ of energy is obtained from 1 moleor V litre of `C_(4) H_(10)` `:. ( 809x)/( V) kJ` of energy will be obtained from `C_(4)H_(10) = ( 809 x xx V )/( V xx 2878) = 0.281 x ` litre Thus, butane supplied for same calorific output `= 0.281 x` litre `//` hr. Calculation of`O_(2)` `C_(4)H_(10) + ( 13)/( 2) O_(2) rarr 4CO_(2) + 5H_(2)O , DeltaH = - 2878 kJ mol^(-1)` Volume of `O_(2)` required `= 2 xx` volume of `O_(2)` for combustion of `C_(4)H_(10)` (Given ) `= 3 xx ( 13)/(2) xx ` Vol. of `C_(4) H_(10)= 3 xx ( 13)/(2) xx 0.28 x = 5.48 x `litre `//` hr |
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| 812. |
In order to get maximum calorific output a burner should have an optimum fuel to oxygen ratio which corresponds to three times as much oxygen as is required theorectically for complete combusion of the fuel A burner which has been adjused for methane as fuel (with `x L h^(-1)` of `CH_(4)` and `6x Lh^(-1) of CO_(2))` is to be readjusted for butane `C_(4)H_(10)` in order to get the same calorific output what should be the rate of supply to butane and oxygen? Assume that losses due to incomplete combustion etc are the same for both fuels and that the gases behave ideally Heats of combusion `CH_(4) =809 kJ mol^(-1),C_(4)H_(10) =2878 kJmol^(-1)` . |
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Answer» In `1h, xL` of `CH_(4)` required `6x L` of `O_(2)` `DeltaH` (combustion) of `CH_(4) = 809 kJ mol^(-1)` `rArr (809)/(24.48) kJ mol^(-1) at 1.0 atm` and `25^(@)C` `DeltaH` (combustion) of `C_(4)H_(10) = 2878 kJ mol^(-1)` `rArr (2878)/(24.48) kJ L^(-1) at 1.0 atm` and `25^(@)C` `xL of CH_(4)` produces `809. x//24.48 kJ` Now this much enegry has to be provided by burning of `C_(4)H_(10)`. `rArr (809)/(24.48) (x) kJ` will be provided by `((809)/(24.48)(x)) xx ((24.48)/(2878)) - (0.28x)L of C_(4)H_(10)` `C_(4)H_(10) +13//2O_(2) rarr 4CO_(2) +5H_(2)O` `rArr 1mol C_(4)H_(10) = 13//2 mol O_(2)` `rArr (3xx13//2)` times of `O_(2)` is required per mol. `rArr` Rate of `O_(2)` per hour `=(0.28x) xx (39//2)` `=(5.48x) L O_(2)` |
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| 813. |
For the reactions, `N_(2)(g) +3H_(2)(g) rarr 2NH_(3)(g)` predict, whether the work is done on the system or by the system. |
| Answer» Volume is decreasing, therefore, work is done by the system. | |
| 814. |
Using the sata (all values are in kilocalorie per mole at `25^(@)C`) given below, calculate the bond enegry of `C-C` and `C-H` bonds. `DeltaH^(Theta)` combustion of ethane `=- 372.0` `DeltaH^(Theta)` combustion of propane `=- 530.0` `DeltaH^(Theta)` for `C`( garphite) `rarr C(g) =+ 172.0` Bond enegry of `H-H` bond `=+ 104.0` `Delta_(f)H^(Theta) of H_(2)O(l) =- 68.0` `Delta_(f)H^(Theta) of CO_(2)(g) =- 94.0` |
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Answer» `C_(2)H_(6)(g) +(7)/(2)O_(2) rarr 2CO_(2)(g) +3H_(2)O(l), DeltaH^(Theta) =- 372.0` `Delta_(f)H_((C_(2)H_(6)))^(Theta) = 2 xx (-94.0) +3 xx (-68.0) + 372.0` `=- 20 kcal` `C_(3)H_(8)(g) +5O_(2) rarr 3CO_(2)(g) +4H_(2)O (l), DeltaH^(Theta) =- 530.0` `Delta_(f)H_((C_(3)H_(8)))^(Theta)=2xx(-94.0) +4xx (-68.0) +530.0` `=- 24 kcal` `2C(s) +3H_(2)(g) rarr C_(2)H_(6)(g), DeltaH^(Theta) =- 20.0` `2C(g)rarr 2C(s),DeltaH^(Theta) =- 344.0` `4H(g) rarr 3H_(2)(g), DeltaH^(Theta) =- 312.0` `"Adding"ulbar(2C(g)+6H(g)rarrC_(2)H_(6)(g),DeltaH=-676kcal)` So, the enthalpy of formation of six `(C-H)` and one `(C-C)` bonds is `-676.0 kcal`. `3C(s) +4H_(2)(g) rarr C_(3)H_(8)(g), DeltaH^(Theta)=- 24.0` `3C(g) rarr 3C(s),DeltaH^(Theta) =- 516.0` `8H (g)rarr 4H_(2)(g),DeltaH^(Theta) =- 416.0` `"Adding"ulbar(3C(g)+8H(g)rarrC_(3)H_(8)(g),DeltaH=-956.0kcal)` So, the enthalpy of formation of eight `(C-H)` and two `(C-C)` bond is `-965kcal`. Let the bond enegry of `C-C` be `x` and of `C-H` be `y` kcal. In ethane, `x+6y = 676` In propane, `2 x +8y = 956` On solving, `x = 82` and`y =99` Thus, bond energy of `C-C =82` kcal and bond enegry of `C-H = 99 kcal`. |
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| 815. |
Using the data (all values are in kilocalorie per mole at `25^(@)C`) given below, calculate the bond enegry of `C-C` and `C-H` bonds. `DeltaH^(Theta)` combustion of ethane `=- 372.0` `DeltaH^(Theta)` combustion of propane `=- 530.0` `DeltaH^(Theta)` for `C`( garphite) `rarr C(g) =+ 172.0` Bond enegry of `H-H` bond `=+ 104.0` `Delta_(f)H^(Theta) of H_(2)O(l) =- 68.0` `Delta_(f)H^(Theta) of CO_(2)(g) =- 94.0` |
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Answer» Correct Answer - `C - C = 82 k cal// C - H = 99 k cal` For formation of `C_(3)H_(8) 3C + 4H_(2) rarr C_(3)H_(8) Delta H_(1) = ?` For formation of `C_(2)H_(6) 2C + 3H_(2) rarr C_(2) H_(6), Delta H_(2) = ?` Thus (i) `Delta H_(1) = - [2 (C - C) + 8 (C - H)] + [3C_(s - g) + 4 (H - H)]` (ii) `Delta H_(2) = [1 (C - C) + 6 (C - H) ] + [2 C_(s - g) + 3 (H - H) ]` Let the bond energy of `C - C` and `C - H` bonds be x kcal and y kcal respectively. Then, we have (iii) `Delta H_(1) = - (2 x + 8y) + [3 xx 1724 xx 104]` and (iv) `Delta H_(2) = - (x + 6y) + [2 xx 172 + 3 xx 104]` Given (v) `C + O_(2) rarr CO_(2) , Delta H = - 94.0 kcal` (vi) `H_(2) + (1)/(2) O_(2) rarr CO , Detlta H = -68.0 kcal` (vii) `C_(2)H_(6) + (7)/(2)O_(2) rarr 2CO_(2) + 3H_(2) O , Delta H = - 372.0 kcal` (viii) `C_(2)H_(8) + 5O_(2) rarr 3 CO_(2) + 4H_(2) O, Delta H = - 530 kcal` (vii) we get (ix) (ix) `2 C + 3H_(2) rarr C_(2)H_(6), Delta H_(2) = - 20.0 kcal` Again `3-(v) + 4 xx (vi) (viii)` gives, (x) `3C + 4 H_(2) rarr C_(3)H_(8), Delta H_(1) = - 20.0 kcal` solving equation (iii), (iv), (ix) and (x) we get `x + 6y = 676, 2 x + 8y = 956` or `x = 82` kcal and `y = 99 kcal` hence bond energy of `C - C` bond = 82 kcal and bond energy of `C - H` bond = 99 kcal |
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| 816. |
Determine the enthalpy change of the reaction. `C_(3)H_(5) (g) + H_(2) (g) rarr C_(2)H_(6) (g) + CH_(4) (g)` at `25^(@)C`, using the given heat of combustion value under standard conditions: Compound `H_(2) (g) CH_(4) (g) C_(2) H_(6) (g) C` (graphite) `Delta H^(@) (kJ//mol) -285.8 =890.0 - 1560.0 -393.5` The standard heat of formation of `C_(3)H_(8) (g)` is `-103.8 kJ//mol` |
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Answer» Correct Answer - `-55.7 kJ//mol` From the given data, we can get following equations. (i) `H_(2) + (1)/(2) O_(2) rarr H_(2) O, Delta H_(1) = - 285.8 kJ` (ii) `CH_(4) + 2O_(2) rarr CO_(2) + 2H_(2) O, Delta H_(2) = -890 kJ` (iii) `C_(2)H_(6) + (7)/(2)O_(2) rarr 2CO_(2) + 3H_(2) O, Delta H_(3) = - 1560 kJ` (iv) `C (s) + O_(2) rarr CO_(2), Delta H_(4) = - 393.5 kJ` (v) `3C (s) + 4H_(2) rarr C_(3) H_(8) , Delta H_(4) = - 103.8 kJ` The required equation is `C_(3)H_(8) (g) + H_(2) (g) rarr C_(2) H_(6) (g) + CH_(4) (g) Delta H = ?` We can get the desired equation using the manipulations given below `[3 xx (iv) + 5 xx (i) ] - [(v) + (iii) + (ii)]` `Delta H = (3 Delta H_(4) + 5 Delta H_(1)) - (Delta H_(5) + Delta H_(3) + Delta H_(2))` `= [3 xx (-393.5) + 5 xx (-285.8)] - [-103.8 - 1560 - 590]` `= -55.7 kJ` mole |
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| 817. |
The table given below lists the bond dissociation energy `(E_("diss"))` for single covalent bonds formed between C and atoms A, B, D, E. `{:("Bond",,,E_("diss")("kcal mol"^(-1))),(C-A,,," "240),(C-B,,," "382),(C-D,,," "276),(C-E,,," "486):}` Which of the atoms has smallest size ?A. DB. EC. AD. D |
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Answer» Correct Answer - B `B.E.uarr"size"darr` |
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| 818. |
Calculate `P-Cl` bond enthalpy`P(s)+(3)/(2)Cl_(2)(g)rarr PCl_(3)(g)`Given `: Delta_(f)H(PCl_(3),g)=306kJ //mol` `DeltaH_("atomization")(P,s)=314kJ//mol`, `Delta_(r)H(Cl,g)=1231kJ//mol`A. 123.66 kJ/molB. 371 kJ/molC. 19 kJ/molD. None of these |
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Answer» Correct Answer - A `P(s)+(3)/(2)Cl_(2)(g)rarrPCl_(3)(g)` `306 = (314 + 3 xx 121) - [B.E.(P-Cl)xx3],` `B.E.(P-Cl)=123.66 " kJ"//"mol"` |
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| 819. |
Carbon monoxide is carried around a closed cyclic processes `abc`, in which `bc` is an isothermal process, as shown in Fig. The gas absorbs `7000 J` of heat as its temperature is increased from `300 K` to `1000 K` in going from `a` to `b`. The quantity of heat ejected by the gas during the process `ca` is A. `4200j`B. `5000j`C. `9000j`D. `9800j` |
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Answer» Correct Answer - D |
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| 820. |
One mole of an ideal monoatomic gas is mixed with 1 mole of an ideal diatomic gas. The molar specific heat of the mixture at constant volume isA. 4 calB. 6 calC. 8 calD. 3 cal |
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Answer» Correct Answer - A `C_(v)= (3)/(2) R` formonoatomic and `C_(v)= ( 5)/(2) R` for diatomic gases. Hence, for the mixture, `C_(v) = ((3)/(2)R+(5)/(2)R)/(2)= 2R= 4 cal` |
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| 821. |
An amount `Q` of heat is added to a mono atomic ideal gas in a process in which the gas performs a work `Q//2` on its surrounding. Find equation of the process.A. `TV^(1//3)=` constantB. `TV^(-1//4)=` constantC. `TV^(1//4)=` constantD. `TV^(-1//3)=` constant |
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Answer» Correct Answer - A `DeltaU=DeltaW` (polytropic process) `(nRDeltaT)/(gamma-1)=(nRDeltaT)/(r-1)impliesr=(1)/(3)` |
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| 822. |
System in which there is no exchange of matter, work or energy from surrounding is (a) closed (b) isolated (c) adiabatic(d) isothermal |
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Answer» Answer: (b) isolated |
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| 823. |
For reaction `2A(g)+3B(g)to4C(g)+D(s)` Calulate work involved during system reaction if reaction occurs at constant pressure and 300K:A. `600kccal`B. `300kcal`C. `150cal`D. `1200kacl` |
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Answer» Correct Answer - A |
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| 824. |
A Carnot engine, whose efficiency is `40%`, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency `60%`. Then, the intake temperature for the same exhaust (sink) temperature must be:A. 1200KB. 750 KC. 600 KD. 800 K |
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Answer» Correct Answer - B Efficiency of carnot engien `eta=1-(T_(2))/(T_(1))` where `T_(1)` is the temperature of the source and `T_(2)` is the temperature of the sink For the `1^(st)` case `eta=40%,T_(1)=500K` `therefore(40)/(100)=1-(T_(2))/(500)` `(T_(2))/(500)=1-(40)/(100)=(3)/(5)` `T_(2)=(3)/(5)xx500=300 K` For the `2^(nd)` case `eta=60%,T_(2)=300 K` `therefore (60)/(100)=1-(300)/(T_(1))` `(300)/(T_(1))=1-(60)/(100)=(2)/(5)` `t_(1)=(5)/(2)xx300=750 K` |
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| 825. |
An ideal gas is made go to through a cyclic thermodynamics process in four steps. The `Q_(2)= 300J, Q_(3)= - 400J, Q_(4)= -100J` respectively. The corresponding works involved are `W_(1)= 300J, W_(2)= -100J, W_(3)= 100J and W_(4)`. What is the value of `W_(4)`?A. `100 J`B. `500 J`C. `-700 J`D. `-400 J` |
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Answer» Correct Answer - D According to first law of thermodyanmics, `SigmadQ=SigmadU+SigmadW` As the process is cycle, `SigmadU=0` `:. SigmadW= SigmadQ` `W_(1)+W_(2)+W_(3)+W_(4)= Q_(1)+Q_(2)+Q_(3)+Q_(4)` `W_(4)= Q_(1)+Q_(2)Q_(3)+Q_(4)-(W_(1)+W_(2)+W_(3))` `=100+300-400-100-(300-100+100)` `= -100-300= -400J` |
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| 826. |
The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle isA. `P_(0)V_(0)`B. `(13/2)P_(0)V_(0)`C. `(11/2)P_(0)V_(0)`D. `4P_(0)V_(0)` |
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Answer» Correct Answer - B In the (figure)shown, suppose temp. of monoatomic gas at A is `T_(0)` In going A to B , volume is constant. As pressure increases from `P_(0)` to `2P_(0)` temp. increases from `T_(0)` to `2T_(0)` `:.` Heat extracted from source `= 1xx C_(v).dT` `Q_(1)=3/2R(2T_(0)-T_(0))= 3/2RT_(0)= 3/2P_(0)V_(0)` In going from B to C, pressure is constant. As volume increases from `V_(0) to 2 V_(0)`, temp. increases from `2T_(0) to 4T_(0) ( :. T prop V)` `:.` Heat extracted from source `=1xxc_(p).dT` `Q_(2)=1xx5/2R(4T_(0)-2T_(0))= 5RT_(0)= 5P_(0)V_(0)` In going from C to D, volume is constant. Pressure decreases from `2p_(0) to P_(0)`. Temp. decreases from `4T_(0) to 2T_(0)`. No heat is extracted from the source. In going to D to A, pressure is constant Volume decreases from `2v_(0) to V_(0)`. No heat is extracted from the source. Hence in single cycle, heat extracted from the source `=Q_(1)+Q_(2)= 3/2P_(0)V_(0)+5P_(0)V_(0)= (13)/2p_(0)v_(0)` |
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| 827. |
A Carnot engine, whose efficiency is `40%`, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency `60%`. Then, the intake temperature for the same exhaust (sink) temperature must be:A. efficiency of carnot engine cannot be made larger than `50%`B. `1200K`C. `750K`D. `600 K` |
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Answer» Correct Answer - C Here, `eta_(1)= 40%, T_(1)= 500K, T_(2)= T` `eta_(2)= 60%, T_(1)=?, T_(2)=T` From `eta= 1-(T_(2))/(T_(1))` `(40)/(100)=1 -T/(500), T= 300K` `(60)/(100)= 1 -(300)/(T_(1)), T_(1)= 750 K` |
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| 828. |
The two ends of a metal rod are maintained at temperature `100^(@)C` and `110^(@)C`. The rate of heat flow in the rod is found to be `4.0 J//s`. If the ends are maintained at temperature s `200^(@)C` and `210^(@)C`. The rate of heat flow will beA. `16.8 J//s`B. `8.0 J//s`C. `4.0 J//s`D. `44.0 J//s` |
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Answer» Correct Answer - C We know that the rate of flow of heat is proportional to temperature different between two ends. So `(dQ)/(dt) prop (T_(2)-T_(1)) or (dQ)/(dt)= K (T_(2)-T_(1))` Case (i), `(dQ_(1))/(dt)=k[110-100]= kxx10`….(i) Case (ii), `(dQ_(2))/(dt)=k[210-200]= kxx10`....(ii) From (i) and (ii), `(dQ_(1))/(dt)=(dQ_(2))/(dt)= 4.0 J//s` |
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| 829. |
Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure Efficiency of this cycle is nearly: (Assume the gas to be close to ideal gas) A. `15.4%`B. `9.1%`C. `10.5%`D. `12.5%` |
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Answer» Correct Answer - A Helium is monoatomic gas, for which `C_(v)3/2R` and `C_(p)= 5/2R` Work done by the gas in one complete cycle `W= area ABCDA=P_(0)V_(0)` (figure) From A to B, heat given to gas `= nC_(v)DeltaT` `=1xx(5/2R) DeltaT= 5/2 (2P_(0)) Delta V= 5P_(0)V_(0)` From C to D and from D to A, heat is rejected by the gas, `Efficiency= ("work done by the gas"//"cycle")/("total heat given to gas"//"cycle")` `=(P_(0)V_(0))/(3/2P_(0)V_(0)+5P_(0)V_(0))=2/(13)xx100= 15.4%` |
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| 830. |
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. `2/3`C. `4/3`D. `5/3` |
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Answer» Correct Answer - A In an adiabatic process, `P propT^(gamma/(gamma-1)) propT^(3)` (given) `:. (gamma)/(gamma-1)=3` `3gamma-3=gamma` `2gamma=3, gamma=3/2` |
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| 831. |
An ideal monoatomic gas is taken the cycle `ABCDA` as shown in following `P-V` diagram. The work done during the cycle is A. `PV`B. `2PV`C. `4PV`D. `3PV` |
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Answer» Correct Answer - C In cycle process `W`=area of cycle in `P-V` diagram |
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| 832. |
An ideal monoatomic gas is taken the cycle `ABCDA` as shown in following `P-V` diagram. The work done during the cycle is A. PVB. 2 PVC. 4 PVD. Zero |
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Answer» Correct Answer - C |
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| 833. |
An ideal gas is taken through the cycle `AtoBtoCtoA,` as shown in the figure, If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process CtoA is A. `-5J`B. `-10J`C. `-15J`D. `20J` |
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Answer» Correct Answer - A |
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| 834. |
Two moles of an ideal gas is expanded irreversibly and isothermally at `37^(@)C` until its volume is doubled and `3.41 KJ` heat is absorbed from surrounding. `DeltaS_("total")("system +surrounding")` is:A. `-0.52J//K`B. `0.52J//K`C. `22.52J//K`D. 0 |
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Answer» Correct Answer - B |
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| 835. |
When `10g` of `AI` is used for reduction in each of the following alumino thermic reactions. Which reaction would generate more heat and by hwo much? a. `2AI + Cr_(2)O_(3) rarr AI_(2)O_(3) +2Cr` b. `2AI +Fe_(2)O_(3) rarr AI_(2)O_(3) +2Fe` Standard heat of formation of `AI_(2)O_(3), Cr_(2)O_(3)`, and `Fe_(2)O_(3)` are `-1676 kJ, 1141 kJ`, and `-822.2 kJ`, respectively. |
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Answer» Given, `{:(2AI+(3)/(2)O_(2)rarrAI_(2)O_(3),,,DeltaH_(1)=-1676kJ..(i)),(2Cr+(3)/(2)O_(2)rarrCr_(2)O_(3),,,DeltaH_(2)=-1141kJ..(ii)),(2Fe+(3)/(2)O_(2)rarrFe_(2)O_(3),,,DeltaH_(3)=-822.2 kJ..(iii)):}` To calculate ltbr. `DeltaH_(a) = ?` For `2AI+ Cr_(2)O_(3)rarr AI_(2)O_(3) +2Cr ...(iv)` Operate: `DeltaH_(a) = DeltaH_(1)-DeltaH_(2)` `=- 1676 -(-1141) =- 535 kJ` `DeltaH_(b) = ?` For `2AI +Fe_(2)O_(3) rarr AI_(2)O_(3) +2Fe ......(v)` Operate: `DeltaH_(b) = DeltaH_(1) -DeltaH_(3)` `=- 1676 -(-822.2) =- 853.8 kJ` For reaction (iv) `2 xx 27 g` of AI produces` = 535 kJ` of heat `10g` of AI produces `=(535xx10)/(2xx27) = 99.07 kJ` For reaction (v) `2 xx 27g` of AI produces `= 853.8 kJ` of heat `10g` of AI produces `= (853.8xx10)/(54) = 158.11kJ` Reaction (ii) produces more heat `= 158.11 -99.07` `= 59.04 kJ` |
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| 836. |
Determine the heat of reaction for this process `Fe(s)+O_(2)(g)to2FeO(s) DeltaH^(@)=-544.0KJ` `4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s) DeltaH=-1648.4Kj` `Fe_(3)O_(4)(s)to 3Fe(s)+2O_(2)(g) DeltaH^(@)=-+1118.4Kj`A. `-1074.1KJ`B. `-22.2KJ`C. `+249.8KJ`D. `+2214.6KJ` |
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Answer» Correct Answer - b |
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| 837. |
Two moles of an ideal gas is expanded isothermally and irreversibly at `27^(@)C` from volume `V_(1)" to" 2.5 V_(1)"and" 4.17KJ` heat is absorbed from surroundings. Determine `DeltaS_("sys")`? |
| Answer» Correct Answer - `DeltaS_("system")=8 Cal` | |
| 838. |
Which of the following statement are always correct ?A. The entropy changes of a system particulaing in adiabatic process is always positiveB. The entropy changes of a system particulaing in adiabatic irrversible process is always positive .C. The entroy change of surrounding is always zero in abiabatic process.D. The entrophy change of a system participating in adiabatic process is always zero . |
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Answer» Correct Answer - b,c |
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| 839. |
Metal ions are activators and increase catlytic activity of enzyme molecular . If water vapours is assumed to be a perfect gas, molar enthalpy change for vaporisation of 1 mole of water 1 bar `100^(@)`C is `41 kJ//mol`. Choose the correct statement(s). `(Taken R= 8.3 J//"mole"//K)`A. `DeltaU_("vaporisation")` of 1 mole of water at 1 bar and `100^(@)C= 37.904 kJ//"mol"`B. `(DeltaU=DeltaH)` for conversion of the water into ice at `0^(@)C`C. In the isothermal process of (b), `DeltaH=0`D. `(DeltaH=DeltaU)` for conversion of 1 mole of water into steam `100^(@)`C |
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Answer» Correct Answer - a,b |
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| 840. |
A heat engine carries one mole of an ideal mono-atomic gas around the cycle as shown in the figure below. Process `1 rarr 2` takes place at constant volume, process `2 rarr 3` is adiabatic and process `3 rarr 1` takes place at constant pressure. Then the amount of heat added in the process `1 rarr2` is A. `3740J`B. `-3740J`C. `2810J`D. `3228J` |
| Answer» Correct Answer - `3740 J` | |
| 841. |
A gold ring that weighting 3.81g is heated to `84^(@)C`and placed in 50.0g of `H_(2)O` at `22.1^(@)C`. What is the final temperature? A. `22.2^(@)C`B. `24.0^(@)C`C. `26.5^(@)C`D. `53.1^(@)C` |
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Answer» Correct Answer - A |
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| 842. |
Calculate work done by `1` mole of ideal gas expand isothermally and irreversibly from pressure of `5` atm to `2` atm against a constant external pressure of `1` atm at `300 K` temperature. |
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Answer» `int dW = -int P_("ext")dv` `W_("irr") = -P_("ext")[V_(2)-V_(1)] = -P_("ext")((nRT)/(P_(2))-(nRT)/(P_(1))) = -P_("ext") xx nRT((1)/(P_(2))-(1)/(1P_(1)))` `= -1 xx 1 xx .082 xx 300 ((1)/(2)-(1)/(5)) = -1 xx .082 xx 300 xx (3)/(10) = -7.38 L. atm = -747.8 J` `W_("irr") = -0.7478 KJ` |
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| 843. |
Consider a classroom that is roughly `5mxx10mxx3m`. Initially `T=27^(@)`C and p=1 atm. There are 50 people in a insulated class losing energy to the room at the average rate of 150 Watt per person. How long can they remain in class if the body temperature is `42^(@)` C and person feels uncomfortable above this temperature. Heat capacity of air=`(7//2)R`A. `4.34` minutesB. `5.91` minutesC. `6.86` minutesD. `7.79` minutes |
| Answer» Correct Answer - B | |
| 844. |
`Fe_(2)O_(2)(s)+(3)/(2)C(s)to(3)/(2)CO_(2)(g)+2Fe(s)` `DeltaH^(@)=+234.12KJ ` `C(s)+O_(2)(g)toCO_(2)(g) DeltaH^(@)=-393.5KJ` Use these equations and `DeltaH^(@)` value to calculate `DeltaH^(@)` for this reaction : `4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s)`A. `-1228.7Kj `B. `-1255.3KJ`C. `-1021.2KJ`D. `-129.4KJ` |
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Answer» Correct Answer - a |
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| 845. |
A heat engine carries one mole of an ideal monoatomic gas around the cycle as shown in the figure, the amount of heat added in the process AB and heat removed in the process CA are : A. `q_(AB)=450R` and `q_(CA)= -450R`B. `q_(AB)=450R` and `q_(CA)= -225R`C. `q_(AB)=450R` and `q_(CA)= -375R`D. `q_(AB)=375R` and `q_(CA)= -450R` |
| Answer» Correct Answer - C | |
| 846. |
For the following reaction, `C_("diamond")+O_(2)toCO_(2)(g), DeltaH=-94.3" "kcal` `C_("graphite")+O_(2)toCO_(2)(g), DeltaH=-97.6" "kcal` the heat require to change 1 g of `C_("diamond")toC_("graphite")` is:A. 1.59 kcalB. 0.1375 kcalC. 0.55 kcalD. 0.275 kcal |
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Answer» Correct Answer - D |
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| 847. |
The rectangular box shown in Fig has partition which can slide without friction along the length of the box. Initially each of the two chambers of the box has one mole of a mono-atomic ideal gas `(lambda=5//3)` at a pressure `P_0`, volume `V_0` and temperature `T_0`. The chamber on the left is slowly heated by an electric heater. The walls of the box and the lead wires of the heater is negligible. The gas in the left chamber expands pushing the partition until the final pressure in both chambers becomes `243P_0//32`. Determine (i) the final temperature of the gas in each chamber and (ii) the work done by the gas in the right chamber. A. `2.25 T_(0)`B. `4.5 T_(0)`C. `8.75 T_(0)`D. `12.93 T_(0)` |
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Answer» Correct Answer - A For the left chamber, the process of heating is slow initially parameters are `P_(0),V_(0),T_(0)` Finally prarameters are `(243)/(32)P_(0),V_(1),T_(1)` `(P_(0)V_(0))/(T_(0))=((243P_(0))/(32))(V_(1))/(T_(1))` `:. T_(1)=(243)/(32)(V_(1))/(V_(0))T_(0)............(i)` Adiabatic compression ocurs in right chamber initial parameters are `P_(0),V_(0),T_(0)` Final parameters are `(243)/(32)P_(0),V_(2),T_(2)` `P_(0)V_(0)^(gamma)=(P_(0)243)/(32)V_(2)^(gamma) (V_(2)/V_(0))^(5//3)=(32)/(243)` `(V_(2))/(V_(0))=(8)/(27)V_(0) , V_(1)+V_(2)=2V_(0)` or ` V_(1)=2V_(0)-2V_(0)-(8)/(27)V_(0)=(46)/(27)V_(0)....(ii)` (i)and (ii)`impliesT_(1)=12.93T_(0)` For right chamber `(T_(0)/(T_(2)))^(gamma) ((P_(0)xx32)/(243 P_(0)))^(gamma-1)(T_(0))/(T_(2))=((2)/(3))^(2)=(4)/(9)T_(2)=2.25T_(0)` Work done by gas in right chamber `W=(P_(2)V_(2)-P_(0)V_(0))/(1-gamma)=-(3)/(2)[(9)/(4)-1]P_(0)V_(0)` `=-(3)/(2)xx(5)/(4)RT_(0)=-(15)/(8)xx8.3 T_(0)J=-15.58T_(0)J` |
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| 848. |
A flask of `1L` having `NH_(3)(g)` at `2.0atm` and `200K` is connected with the another flask of volume `800 mL` having `HCI(g)` at `8atm` and `200K` through a narrow tube of negligible volume. The two gases react to form `NH_(4)(CI(s)` with evolution of `43kJ mol^(-1)` heat. if heat capacity of `HCI(g)` at constant volume is `20 JK^(-1) mol^(-1)` and neglecting heat capacity of flask, and volume of solid `NH_(4)CI` formed, calculate the final temperature, and final pressure in the flasks. (Assume `R = 0.08 L atm K^(-1) mol^(-1))` |
| Answer» Correct Answer - `14.39 atm`; | |
| 849. |
The heat of combustion of ethane and benzene is `-1560` and `-3268" kJ mol"^(-1)` respectively. Which two has higher efficiency as fuel per gram and the amount of heat produced per gram?A. Benzene, `41.9" kJ g"^(-1)`B. Ethane, `"52 kJ g"^(-1)`C. Benzene, `"78 kJ g"^(-1)`D. Ethane, `"30 kJ g"^(-1)` |
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Answer» Correct Answer - B Heat produced by ethane `=(1560)/(30)="52 kJ g"^(-1)` Heat produced by benzene `=(3268)/(78)="41.9 kJ g"^(-1)` |
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| 850. |
Using the data given below, calculate the value of equilibrium constant for the reaction. `underset("Acetylene")(3HC-=CH(g)) hArr underset("Benzene")(C_(6)H_(6)(g))` at 298 K , assuming ideal gas behaviour. `Delta_(f)G^(@) ` for `HC-= CH(g) = 2.09 xx 10^(5) J mol^(-1), Delta_(f)G^(@) ` for`C_(6)H_(6)(g) =1.24xx 10^(5) J mol^(-1)` `R = 8.314 J K^(-1) mol^(-1)` Based on your calculated value, comment whether this process can be recommended as a practical method for making benzene. |
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Answer» Correct Answer - `K = 1.43 xx 10^(88) ,` Yes `Delta_(r) G^(@) =Delta_(f)G^(@) (C_(6)H_(6)) - 3 Delta_(f)G^(@) (HC -= CH) = 1.24 xx 10^(5) - 3 ( 2.9 xx 10^(5)) J = - 5.03 xx 10^(5)J` `Delta_(r)G^(@) = - 2.303 RT log K `. Calculate K `Delta _(r)G^(@) ` is negative and K is very high . Hence, the process can occur easily. |
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