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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.
| 701. |
Calculate the standard enthalpy of formation of acetylena from the following data: `C(s)+O_(2)(g)to CO_(2)(g), DeltaH^(@)=-393.5KJmol^(-1)` `H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(l),DeltaH^(@)=-285.8KJmol^(-1)` `2C_(2)H_(2)(g)+50_(2)(g)to 4Co_(2)(g)+2H_(2)O(l0,``DeltaH^(@)=-2598.8KJmol^(-1)`A. `226.6KJmol^(-1)`B. `230.5KJmol^(-1)`C. `233.8KJmol^(-1)`D. none of these |
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Answer» Correct Answer - a |
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| 702. |
The enthalpy of neutralisation of HCl and NaOH is `-57KJmol^(-1)` .The heat evolved at constant pressure (in KJ when 0.5 mole of `H_(2)SO_(4)` reacts with 0.75 mole of NaOH is equal to :A. `57xx(3)/(4)`B. `57xx0.5`C. 57D. `57xx0.25` |
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Answer» Correct Answer - a |
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| 703. |
A reaction has `DeltaH^(@) gt0 and DeltaG^(@) gt0 ` `25^(@)C`. This reaction:A. is at equillibruim at `25^(@)C`.B. could not be spontaneous under standard conditions at any temperature.C. could be spontaneous under standard conditions at temperatures above `25^(@)C`D. could be spontaneous under standard conditions at temperature below `25^(@)C`. |
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Answer» Correct Answer - C |
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| 704. |
One mole of an ideal gas is supplied 2KJ of heat. If the temperature of the gas rises from 0°C to 200°C, calculate work done by the gas and change in its internal energy if 1. The process Is isobaric 2. The process is isochoric. |
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Answer» number of moles, n = 1 Initial temperature, Ti = (0 + 273) = 273 K Final temperature, Tf = (200 + 273) = 473 K Net heat supplied, H = 2000 J 1. Isobaric process : work done, W = P - ∆ V or W = n R∆T ∴ W= 1 × 8.314 × (473-273) ∴ W= 1662.8J = 1.66KJ ∴ Change in internal energy, ∆U = H - W= 2000 - 1662.8 = 337.2J 2. isochoric Process: Work done in an isochoric process , W =0. Therefore change in internal energy ∆U = H = 2 kJ. |
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| 705. |
The equilibrium state is attained when the reversible reaction is carried out in ………… space.A. `q_(p) lt q_(v)`B. `q_(p) gt q_(v)`C. `q_(p) = q_(v)`D. `q_(v) = 0` |
| Answer» Correct Answer - `q_(P) = q_(V)` | |
| 706. |
An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 joules per second, at what rate is the internal energy increasing? |
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Answer» Rate of heat supplied to the system is 100 W = 100 Js-1 Rate of work done by the system is 75Js-1 . From the first law of thermodynamics, we have ∆ Q = ∆ U + ∆ W ……..(1) If the above parameters are observed for a time‘∆ t’, …….(2) where \(\frac{ΔQ}{Δ t}\) is the rate of heat supplied to the system for the given time,\(\)\(\frac{ΔU}{Δ t}\) is the rate of increase in internal energy of the system for the given time and \(\frac{ΔW}{Δ t}\) is the rate of work done by the system in the given time. ∴ From equation (2), we have 100 = \(\frac{ΔU}{Δ t}\) + 75 \(\frac{ΔU}{Δ t}\) = 25 Js-1 Therefore the internal energy increases at rate of 25 Joules per second. |
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| 707. |
One mole of an ideal gas is carried through the reversible cyclic process as shown in figure. The max. temperature attained by the gas duing the cycle A. 7/6RB. 12/49 RC. 49/12 RD. 12/7 R |
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Answer» Correct Answer - C Max temp attained by gas in between B to C. According to equation of straight line `(P-4)/(1-4) = (V-1)/(2-1) rArr P_4 = -3V + 3` ` rArr P = 7-3V` For 1 mole gas `(RT)/(V) = 7-3V, RT = 7V - 3V^(2)`, `R(dT)/(dV) = 7-6V = 0, V = (7)/(6)` substituting in eq (1) `RT = (7-3 xx (7)/(6)) xx (7)/(6) rarr T = (49)/(12R)` |
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| 708. |
An electric heater supplies heat to a system at a rate of `100W`. If sustem performs work at a rate `74 Joul es` per second, at what rate is the internal energy increasing? |
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Answer» Heat is supplied to the system at a rate of 100 W. `therefore"Heat supplied", Q = 100 J//s` The system performs at a rate of 75 `J//s.` `therefore"Work done", W = 75 J//s` From the first law of thermodynamics, we have: `Q=U+W` Where, U=Internal energy `thereforeU=Q-W` `= 100 – 75` `= 25 J//s` `= 25 W` Therefore, the internal energy of the given electric heater increases at a rate of 25 W. |
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| 709. |
One mole of ideal monoatmic gas is carried throught the reversible cyclic process as shown in figure. Calculate net heat absorbed by the gas in the path BC A. `(1)/(2)P^(@)V^(@)`B. `(7)/(2)P^(@)V^(@)`C. `2P^(@)V^(@)`D. `(5)/(2)P^(@)V^(@)` |
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Answer» Correct Answer - A `DeltaE=q+w` `W_(BC) =(1)/(2) (2V^(@) -V^(@))(P^(@)-3P^(@))(0-P^(@))=-2P^(@)V^(@)` `DeltaE=nC_(v)DeltaT=1xx(3)/(2)R((P^(@)2V^(@))/(R)-(3P^(@)V^(@))/(R)) =-(3)/(2) P^(@)V^(@)` `q_("BC")=DeltaE-W=-(3)/(2) P^(@)V^(@)+2P^(@)V^(@)=(1)/(2)P^(@)V^(@)` |
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| 710. |
Assertion: The isothermal curves intersect each other at a certain point. Reason: The isothermal changes takes place rapidly, so the isothermal curves have very little slope.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 assertion is false but reason is true. |
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Answer» Correct Answer - D |
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| 711. |
The relation between the slope of isothemal curve and slope of adiabatic curveA. slope of adiabatic curve =`gamma` times slope of isotharmal curveB. slope of isothermal curve =`gamma` times slope of adiabatic curveC. slope of adiabatic curve =`gamma^(2)` times slope of isothermal curveD. slope of isothermal curve =`gamma^(2)` times slope of adiabatic curve |
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Answer» Correct Answer - A For isothermal process , PV= constant Differentiating both side `PdV+VdP=0 or (dP)/(dV)=(-P)/(V)` Again for adiabatic process, `PV^(gamma)`=constant Again differentiating both side `dPV^(gamma)+gammaV^(gamma-1)dVP=0 or (dP)/(dV)=-(P)/(V)xxgamma` slope of diabatic curve = `gammaxxslope` of isothermal curve |
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| 712. |
Which thermodynamic variable is defined by Zeroth law of thermodynamics? |
| Answer» Temperature. | |
| 713. |
In the diagrams (i) to (iv) of variation of volume with changing pressure is shown. A gas is taken along the path ABCD . The change in internal energy of the gas will be A. Positive in all cases (i) to (iv)B. Positive in cases (i), (ii) and (iii) but zero in (iv) caseC. Negative in cases (i), (ii) and (iii) but zero in (iv) caseD. Zero in all four cases |
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Answer» Correct Answer - D |
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| 714. |
A system is taken through a cyclic process represented by a circle as shown in the figure. The heat absorbed by the system isA. `pixx10^(3)j`B. `(pi)/(2)j`C. `4pixx10^(2)j`D. `pij` |
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Answer» Correct Answer - B |
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| 715. |
Assertion: The isothermal curves intersect each other at a certain point. Reason: The isothermal changes takes place rapidly, so the isothermal curves have very little slope.A. If both assetion and reason are true and reaasons is the correct expanation of assetionB. If both assetion and reason are tur but reason is not the correct explanation of assetionC. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - D Isothermal cuves never intersect each other Isothermal changes take place slowly, as they have a very little slope. |
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| 716. |
Assertion : In an isothemal expansion the gas absorbs heat and does work while in an gas by the enviroment and heat is relaeased. Reason: In an isothermal process there is no change in internal energy of an ideal gas.A. If both assetion and reason are true and reaasons is the correct expanation of assetionB. If both assetion and reason are tur but reason is not the correct explanation of assetionC. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - A From first law of thermodynamics in isothermal expansion heat supplied to the gas equals the work done by the gas. |
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| 717. |
Which thermodynamic variable is defined by the first law of thermodynamics? |
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Answer» Internal energy. |
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| 718. |
In the question number 5, if the maiximum pressure is P then what is the pressur at the point 5? (in P-T diagram)A. `(2P)/(3)`B. `(4P)/(3)`C. `(3P)/(3)`D. None of these. |
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Answer» Correct Answer - C Given: `P_(1)=P_(2)=2P_(3)=P` Since `1rarr2 and 3rarr4` are isobaric porcesses `therefore` Heat received = `nC_(P)(T_(F)-T_(i))` Accrording to question, `C_(P)(T_(2)-T_(0))=C_(P)(T_(0)-T_(3))` For process `2rarr3` i.e isochoric process `(P_(2))/(t_(2))=(P_(5))/(t_(5))=(P_(3))/(T_(3))` `therefore (P)/(T_(2))=(P_(5))/(T_(0))=(P//2)/(T_(3))`(From eqn(i)) or `T_(2)=2T_(3)` and `P_(5)=(P)/(2)(T_(0))/(t_(3))` soliving eqn (ii) using eqn (iii) we get `P_(5)=(3)/(4)P` `therefore` pressure at point 5 is `(3P)/(4)` |
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| 719. |
Zeroth law of thermodynamics gives the concept ofA. internal energyB. heat contentC. pressureD. temperature |
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Answer» Correct Answer - D The zeroth law of thermodynamics leads to the concept of temperature. |
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| 720. |
For adiabatic process, wrong statement isA. dQ = 0B. dU + dWC. Q = constantD. Entropy is not constant |
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Answer» Correct Answer - D |
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| 721. |
Helium at `27^(@)C` has a volume of 8 litres. It is suddenly compressed to a volume of 1 litre. The temperature of the gas will be `[gamma=5//3]`A. `108^(@)`B. `9327^(@)C`C. `1200^(@)C`D. `927^(@)C` |
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Answer» Correct Answer - D |
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| 722. |
Which Thermodynamical variable is defined by the first law of thermodynamics? |
| Answer» Internal energy. | |
| 723. |
Which thermodynamical variable is defined by the Zeroth law of Thermodynamics? |
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Answer» Temperature. |
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| 724. |
Which one of the following is not a thermodynamical coordinate ?A. Gas constant(R )B. Pressure (P)C. Volume(V )D. Temperature (T) |
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Answer» Correct Answer - A Pressure (P) volume (V) and temperature (T) are the thermodynameic co- ordinates used to diescribe the state of the system whereas gas sconstant (R ) is auniversal gs constant whose value is 8.314 J `mol^(-1) K^(-1)` |
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| 725. |
The object under thermodynamical study is calledA. SystemB. UniverseC. SurroundingD. Boundary |
| Answer» Correct Answer - A | |
| 726. |
Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same velocity V. The mass of the gas in A is `m_A,` and that in B is `m_B`. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be `DeltaP and 1.5 DeltaP` respectively. ThenA. `4m_(A)=9m_(B)`B. `3m_(A)=3m_(B)`C. `3m_(A)=2m_(B)`D. `9m_(A)=4m_(B)` |
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Answer» Correct Answer - C For gas A `P_(1)=(Rt)/(M)(m_(A))/(V_(1))` `P_(2)=(RT)/(M)(m_(a))/(V_(2))` `DeltaP=P_(1)-P_(2)=(RT)/(M)(m_(A))[(1)/(V_(1))-(1)/(V_(2))]` Putting `V_(1)`=V and `V_(2)`=2V, we get `DeltaP=(RT)/(M)(m_(A))/(2V)` Form equation (i) and (ii) we get `3m_(A)=2m_(B)` |
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| 727. |
Can the work done during a cyclic process be zero? |
| Answer» Yes, work done in a cyclic process can be zero. When the processes reverse exactly under similar conditions retracting the samw P-V diagram. | |
| 728. |
What is the significance of area of closed curve on P-V diagrams? |
| Answer» The area gives us the work done in the process. | |
| 729. |
The molar specific heat of a gas at constant volume is `20` Joule `mol^(-1)K^(-1)`. The value of `gamma` for it will beA. `(11)/(10)`B. `(7)/(5)`C. `(5)/(3)`D. `(3)/(2)` |
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Answer» Correct Answer - B `C_(v)=(R)/((gamma-1))impliesgamma=1+(R)/(V_(v))` |
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| 730. |
The specific heat of air at constant pressure is `1.005 kJ//kg//K` and the specific heat of air at constant volume is `0.718 kJ//kg//K` . If the universal gas constant is `8.314 kJ//k` mole `K` find the molecular weight of air ?A. `28.97`B. `24.6`C. `22.8`D. `19.6` |
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Answer» Correct Answer - A `c_(p)-c_(v)=(R)/(M)impliesM=(R)/(c_(p)-c_(v))` |
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| 731. |
One mole of an ideal monoatomic gas at temperature `T_0` expands slowly according to the law `p/V` = constant. If the final temperature is `2T_0`, heat supplied to the gas isA. `2RT_(0)`B. `RT_(0)`C. `(3)/(2)RT_(0)`D. `(1)/(2)RT_(0)` |
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Answer» Correct Answer - A In a process `PV^(x)` = constant molar heat capacity is given by `C=(R )/(gamma-1)+(R )/(1-x)` As the process is `(P)/(V)` = constant, i.e `PV^(-1)`=constant , `threfore , x =-1` For an ideal monatiomic gas , `gamma=(5)/(3)` `C=(R )/((5)/(3)-1)+(R )/(1-(-1))=(3)/(2)R+(R)/(2)=2R` `DeltaQ = n C(DeltaT)=1(2R)(2R_(0)-T_(0))=2RT_(0)` |
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| 732. |
A sample of NaOH (s) was added to water in a calorimeter. The temperature was monitoredas the NaOH dissolved to give the data below . Determinethe heat released during the solution process.(Assume the solution specific heat is `4.18J.g^(-1)`.`K^(-1)`) A. `1.01xx10^(3)` joulesB. `2.66xx10^(3)`JoulesC. `1.01xx10^(4)`JoulesD. `1.11xx10^(4)`Joules |
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Answer» Correct Answer - d |
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| 733. |
Equilibrium constant for a reaction can be obtained by kinetics approach or by thermodynamics approch. While from kinetics approach at equililbrium, the rate of forward and backward will be same, from thermodynamics approach Gibbs free energy will be minimized at equilibrium. Using this information and following thermodynamics values, answer the question that follow: `DeltaG_(f)^(@)A(g)`=-200kcal/mole `DeltaG_(f)^(@)B(g)` =-320 kcal/mole `DeltaG_(f)^(@)C(g)`=-300kcal/mole `DeltaG_(f)^(@)D(l)`=-224.606 kcal/mole `DeltaG_(f)^(@)D(g)`= -226.9.9 kcal/ mole, All values at 500K Calculate equilibrium concentration of B(g) if A(g) at 10 bar, B(g) at 2 bar,C(g) at 20 bar is mixed with excess liquid D such that following equilbrium gets established at 500K: `A(g) B(g)iffC(g)+D(g)`A. 2MB. `(2)/(41.57)`C. `1M`D. `(1)/(41.57)` |
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Answer» Correct Answer - b |
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| 734. |
Equilibrium constant for a reaction can be obtained by kinetics approach or by thermodynamics approch. While from kinetics approach at equililbrium, the rate of forward and backward will be same, from thermodynamics approach Gibbs free energy will be minimized at equilibrium. Using this information and following thermodynamics values, answer the question that follow: `DeltaG_(f)^(@)A(g)`=-200kcal/mole `DeltaG_(f)^(@)B(g)` =-320 kcal/mole `DeltaG_(f)^(@)C(g)`=-300kcal/mole `DeltaG_(f)^(@)D(l)`=-224.606 kcal/mole `DeltaG_(f)^(@)D(g)`= -226.9.9 kcal/ mole, All values at 500K Calculate rate constant of the backward reaction for the following reaction at 500K: `A(g)+B(g)iffC(g)+D(l) " if " K_(f)=10"bar"^(-1) sec^(-1)`A. `10 "bar"^(-1) sec^(-1)`B. `0.1 "bar"^(-1) sec^(-1)`C. `0.1 sec^(-1)`D. `10 sec^(-1)` |
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Answer» Correct Answer - c |
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| 735. |
Heat capacity `(C_(p))` is an extensive property but specific heat (c) is intensive property. What will be the relation between `C_(p) " and " c` for 1 mole of water ? |
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Answer» For water, molar heat capacity, `C_(P) = 18 xx` Specific heat, C `C_(P) = 18 xx c` Specific heat `c = 4.18 Jg^(-1) K^(-1)` Heat capacity, `C_(P) = 18 xx 4.18 JK^(-1) mol^(-1)` `= 753 JK^(-1) mol^(-1)` |
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| 736. |
Heat capacity `(C_(p))` in an extensive property but specific heat (c ) is an intensive property . What will be the relation between `C_(p)` and c for 1 mol of water ? |
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Answer» For water, molar heat capacity `(C_(p)) = 18 xx `specific heat `:. C_(p)= 18 xx c` But specific heat of water , `c=4.18 J g^(-1) K^(-1)` `:. `Heat capacity , `C_(p) =18 xx 4.18 JK^(-1) mol^(-1) = 75.3 JK^(-1) mol^(-1)` |
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| 737. |
Separate out the following into extensive and intensive `:` Volume, Temperature, Pressure,Boiling point, Free energy |
| Answer» Volume and freeenergy are extensive , others are intensive | |
| 738. |
The enthalpy change involved in the oxidation of glucoseis `- 2880 kJ mol^(-1)` . Twenty five per cent of this energy is available for muscular work. If 100 kJ of muscular work is needed to walk one kilometer, what is the maximum distance that aperson will beableto work after eating 120 g of glucose ? |
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Answer» Energy available for muscular work`=( 25)/( 100) xx 2880 kJ mol^(-1) = 720 kJ mol^(-1)` Molar mass of glucose`= 180 g mol^(-1)` `:. `Energy availablefor muscular work from 120 g glucose `= ( 720 )/( 180) xx120 kJ = 480 kJ` In 100 kJ of muscular work, distance walked `= 1` km `:. ` In 480kJ of muscular work, distance walked `= ( 1)/( 100) xx 480 = 4.8 km` |
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| 739. |
The enthalpy change involved in the oxidation of glucose is `-2880 kJ mol^(-1)` 25% of this energy is available for muscular work. If 100 kJ of muscular work is neededto walk one kilometer, what is the maximum distance that a person will be able to walk after consuming 120 gm of glucose? |
| Answer» Correct Answer - 4.8 km | |
| 740. |
The enthalpy change involved in the oxidation of glucose is `-2880kJ mol^(-1)`. Twenty five per cent of this energy is available for muscular work . If `100kJ` of muscular work is needed to walk one kilometre, what is the maximum distance that a person will be able to walk after eating `120g` of glucose ? |
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Answer» Molar mass of glucose `(C_(6)H_(12)O_(6))` `=(6 xx 12 +12 xx1xx6 xx 16) g mol^(-1)` `= 180 g mol^(-1)` `C_(6)H_(12)O_(6)(s) rarr 6CO_(2)(g) +6H_(2)O(l) DeltaH =- 2880 kJ mol^(-1)` Enthalpy consumed in muscular work `=(2880kJ) ((25)/(100))` `=720 kJ` Amount of glucose in `120g = ((120g)/(180 g mol^(-1)))` `= 0.6667 mol` Enthalpy avaiable for muscular work from `120g` of glucose `=(720 kJ mol^(-1)) (0.667 mol) = 480 kJ` Distance to which a person can move `= ((1km)/(100 kJ)) (480 kJ) = 4.80 km` |
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| 741. |
`DeltaH` and `DeltaS are +ve`. Under what conditions, process will be spontaneous? |
| Answer» `DeltaH = +ve, DeltaS = +ve`. The reaction will be spontaneous if `DeltaG =- ve, i.e., T DeltaS gt DeltaH`. It will be spontaneous at high temperature. | |
| 742. |
Preduct that anhyrous `AICI_(3)` is covalent form the data given below, ionisation enegry for `AI = 51.37 kJ mol^(-1), Delta_(hyd) H` for `AI^(3+) =- 4665 kJ "mole"^(-1), Delta_(hyd)H` for `CI^(Theta) =- 381 kJ mol^(-1))` |
| Answer» The total hydration enegry of `AI^(3+)` and `3CI^(Theta)` ions is `DeltaH_(hyd) ={-4665 +3(-381)} kJ mol^(-1) = - 5808 kJ mol^(-1)` The enegry released is more than that required for the ionisation of `AI` to `AI^(3+)` (which is `-4665 kJ mol^(-1))` causing ionisation of `AICI_(3)` solution. Thus, the compound `AICI_(3)` becomes ionic in aqueous solution. | |
| 743. |
A thermodynamic process of one mole ideal monoatomic gas is shown in figure. The efficiency of cyclic process `ABCA` will beA. `25%`B. `12.5%`C. `50%`D. `(100)/(13)%` |
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Answer» Correct Answer - D |
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| 744. |
A thermodynamic process of one mole ideal monoatomic gas is shown in figure. The efficiency of cyclic process `ABCA` will beA. 0.25B. 0.125C. 0.5D. 0.077 |
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Answer» Correct Answer - D `W=(1)/(2)P_(0)V_(0), T_(A)=T_(0),T_(B)=2T_(0),T_(C)=4T_(0)` Heat supplied =`Q_(AB)+Q_(BC)=C_(V)T_(0)+C_(p)2T_(0)` `=(13)/(2)RT_(0)=(13)/(2)P_(0)V_(0)` `therefore` Efficiency of the cyclic process `((1)/(2) P_(0)V_(0))/((13)/(2)P_(0)V_(0))xx100=(1)/(13)xx100=7.7%` |
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| 745. |
Givecn that : `Zn+1//2O_(2)rarr ZnO+84000` cal ……………1 `Hg+1//2O_(2)rarr HgO+21700` cal …………2 The heat of reaction `(Delta H)` for, `Zn+HgO rarr ZnO+Hg` is :-A. 105700 calB. 62300 calC. `-105700` calD. `-62300` cal |
| Answer» Correct Answer - D | |
| 746. |
Which of the following represents the gravitational work ? (a) Qv (b) F. x (c) mgh (d) -P∆V |
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Answer» Answer: (c) mgh |
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| 747. |
Which of the following facts cannot be answered by the first law of thermodynamics?A. During a spontaneous process, energy of the universe is constantB. During a nonspontaneous process, energy of the universe is constantC. Transformation take place spontaneously in one direction but not in the other.D. All of these |
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Answer» Correct Answer - C First law fails to develop any criterion for the spontaneity of a process because the energy of the universe is constant in both natural and unnatural directions. |
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| 748. |
According to the equation `C_(6)H_(6(l))+7/2 O_(2(g)) rarr 6CO_(2(g))+3H_(2)O_((l)), DeltaH=-xkJ` The energy evolved when 3.9 gm of benzene is burnt in air is - 163.2 kJ heat of combustion of benzene isA. 32.46 kJB. 16.32 kJC. 326.4 kJD. `-3264 kJ` |
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Answer» Correct Answer - D `3.9 gm rarr 163.2 kJ` `78 gm rarr ?` |
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| 749. |
Consider the reacting at `300K` `C_(6)H_(6)(l)+(15)/(2)O_(2)(l)rarr 6CO_(2)(g)+3H_(2)O(l),DeltaH=-3271`kJ What is `DeltaU` for the combustion of `1.5` mole of benzene at `27^(@)C` ?A. `-3267.25 kJ`B. `-4900.88 kJ`C. `-4906.5 kJ`D. `-3274.75 kJ` |
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Answer» Correct Answer - B For combustion of 1 mole of benzene `" "Deltan_(g)=-1.5` `" "DeltaH=DeltaU+Deltan_(g)RT` `implies -3271 = DeltaU-(1.5 xx 8.314 xx 300)/(1000)` `implies" "DeltaU=-3267.25 " kJ"` For 1.5 mole of combustion of benzene `DeltaU=-3267.25 xx1.5` `=-4900.88 " kJ"` |
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| 750. |
In an exothermic reaction heat is evolved and system loses heat to the surrounding. For such systemA. `q_(P)` will be negativeB. `Delta_(r)H` will be negativeC. `q_(p)` will be positiveD. `Delta_(r)H` will be positive |
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Answer» Correct Answer - A::B For an exothermic reaction, `q_(p) = -ve, Delta, H = -ve` |
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