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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.
| 2751. |
For the adiabatic expansion of an ideal gas:A. `PV^(gamma)=` constantB. `TV^(gamma-1)=` constantC. `T^(gamma)P^(1-gamma)=` constantD. None of these |
| Answer» Correct Answer - A::B::C | |
| 2752. |
State the first law of thermodynamics.A. `DeltaU = Deltaq - W`B. `q = DeltaU - W`C. `qd +dW = 0`D. `DeltaU q +W` |
| Answer» The first law of thermodynamics states that `DeltaU = q +w`, i.e. the internal enegry of system is equal to the sum of heat and work. | |
| 2753. |
In which reaction(s), `DeltaS` in negative?A. `H_(2)O(l)toH_(2)O(s)`B. `3O_(2)(g)to2O_(3)(g)`C. `H_(2)O(l)toH_(2)O(g)`D. `N_(2)(g)+3H_(2)(g)to2NH_(3)(g)` |
| Answer» Correct Answer - A::B::D | |
| 2754. |
An ideal gas undergoing expansion in vacuum shows:A. `DeltaU=-0`B. `W=0`C. `q=0`D. All of these |
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Answer» Correct Answer - D Ideal gas does not show intermolecular forces of attractions. |
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| 2755. |
In which reaction(s), `DeltaS` in negative?A. `H_(2)O(l)rarr H_(2)O(s)`B. `3O_(2)(g)rarr 2O_(3)(g)`C. `H_(2)O(l)rarrH_(2)O(g)`D. `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` |
| Answer» In all these conversion, the degree of randomness decreases. | |
| 2756. |
Human body is an example of:A. Open system B. Closed systemC. Isolated systemD. None of these |
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Answer» Correct Answer - A Human body can exchange energy and matter with surroundings. This represents open system. |
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| 2757. |
Human body is an example of:A. Open systemB. Closed systemC. Isolated systemD. None of these |
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Answer» Correct Answer - A Human body can exchange energy and matter with surroundings. This represents open system. |
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| 2758. |
Define Heat capacity. |
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Answer» The heat capacity for one mole of the substance is the quantity of heat needed fo raise the temperature of one mole by one degree Celsius. |
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| 2759. |
Define specific heat. |
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Answer» Specific heat /specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or one Kelvin). |
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| 2760. |
One method to produce hydrogen on an industrial on an industrial scale is the reaction of methane with overheated water vapour at 1100 K to form hydrogen and carbon monoxide . The reaction is known as steam reforming The `K_(p)` of the reforming reaction at 1100K is 28.6 When 1 mol of methane and 1 Kmol fo water are reacted at 1100 K , calculate the percentage conversion of methane at equilibrium at a total pressure of 1.6 bar. In another experiment 1.0 Kmol of `CH_(4)` and 1.0 mol of `H_(2)O` are taken in a sealed vessel at 400 K and 1.6 bar. The temperature is raised to 1100K. |
| Answer» Correct Answer - Conversion (methene) = 75% | |
| 2761. |
Calculate the equilibrium pressure (in Pascal) for the conversion of graphite to diamong at `25^(@)C.` The densities of graphite and diamond may be takes to be `2.20` and `3.40` g//cc respectively independent of pressure. (Express your answer in scientific notation x `xx10gamma` and write the value of y.) [Given : `DeltaG_(298)^(@)` `(C_("graphite") rarr C _("diamond"))=2900J//mol]` |
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Answer» Correct Answer - 9 |
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| 2762. |
Indicator diagram is a graph between :(A) Pressure & temperature. (B) Pressure & volume. (C) volume & temperature. (D) none of the above pairs. |
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Answer» Answer is (B) Pressure & volume. |
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| 2763. |
The first law of thermodynamics is based on the first law of conservation of :(A) energy (B) mass (C) momentum (D) none of the above |
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Answer» Answer is (A) energy |
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| 2764. |
Adiabatic is a graph between: (A) V & T (B) T & P (C) P&V (D) PV &T |
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Answer» Answer is (C) P & V |
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| 2765. |
Consider the reaction, `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`, carried out at constant temperature and pressure. If `Delta H` and `Delta U` are enthalpy change and internal energy change respectively, which of the following expressions is true ?A. `Delta H=0`B. `Delta H=Delta U`C. `Delta H lt Delta U`D. `Delta H gt Delta U` |
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Answer» Correct Answer - C `N_(2)+3H_(2)rarr2NH_(3)" " Deltan=2-4= -2` `Delta H=DeltaU+Delta nRT=Delta U-2RT" " :. Delta H lt Delta U` |
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| 2766. |
What is the value of `Delta s^(@)` for the reaction below? `Fe_(2)O_(3)(s)+3CO(g)to2Fe(s)+3CO_(2)(g)` A. `-44.0JxxK^(-1)`B. `-11.8JxxK^(-1)`C. `15.5JxxK^(-1)`D. `42.8JxxK^(-1)` |
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Answer» Correct Answer - c |
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| 2767. |
A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion ( in the intermediate stages of expansion/ compression the states of gases are not defined). The work done can be calculated using `dw=-P_("ext")dV` while in case of reversible process the work done can be calculated using dw=-PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process, since `P=(nRT)/(V)`,so, `w=intdw=-int_(V_(i))^(V_(f))(nRT)/(V).dV=-nRT " "In((V_(f))/(V_(i)))` Since,dw=PdV, so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram. If four identical samples of an ideal gas initially at similar state `(P_(0),V_(0),T_(0))` are allowed to expand to double their volumes by four different processes. (P)By isothermal irreversible process (Q) By reversible process having equation `P^(2)V=` constant (R) By reversible adiabatic process (S)By irreversible adiabatic expansion against constant external pressure. Then in the graph shown the final state is represented by four different points then, the correct match can be :A. `1-P,2-Q,3-R,4-S`B. `1-Q,2-P,3-S,4-R`C. `2-R,3-Q,4-P,1-S`D. 3-Q,1-P,2-S,4-R` |
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Answer» Correct Answer - b |
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| 2768. |
When a system is taken from state `A` to state `B` along path `ACB` as shown in figure below, `80 J` of heat flows into the system and the system does `30 J` of work. If `E_(D)-E_(A)= -40 J`, the heat absorbed in the processes `AD` and `DB` are respectivelyA. `q_(AD)=30J` and `q_(DB)= - 90 J`B. `q_(AD)=-60J` and `q_(DB)= 30 J`C. `q_(AD)=30J` and `q_(DB)= 90 J`D. `q_(AD)=-30J` and `q_(DB)= 90 J` |
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Answer» Correct Answer - D In `ADB` process, `DB` process is isochoric so `w_(DB)=0` So, `Delta E_(AD)=q_(AD)+w_(AD)` `-40=q_(AD)+(-10), q_(AD)= -30 J` Now, `q_(AB)=q_(AD)+q_(DB), 60 =-30+ q_(DB)` `q_(DB)=90 J` |
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| 2769. |
A system has internal energy equal to `U_(1), 450J` of heat is taken out of it and `600J` of work is done on it. The final enegry of the system will beA. `(E_(1)+150)`B. `(E_(1)+1050)`C. `(E_(1)-150)`D. None of these |
| Answer» Correct Answer - A | |
| 2770. |
In which of the processes the internal energy of the system remains constant. (A) Adiabatic. (B) Isochoric. (C) Isobaric. (D) Isothermal |
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Answer» Answer is (D) Isothermal |
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| 2771. |
Both q & w are __________ function & q + w is a ___________ function :-A. State, StateB. State, pathC. Path, stateD. Path, path |
| Answer» Correct Answer - C | |
| 2772. |
Which statement is true for reversible process :-A. It takes place in single stepB. Driving force is much greater than opposing forceC. Work obtain is minimumD. None |
| Answer» Correct Answer - D | |
| 2773. |
The temperature of an ideal gas increases in an:A. Adiabatic compressionB. Adiabatic expansionC. Isothermal expansionD. Isothermal compression |
| Answer» Correct Answer - A | |
| 2774. |
`q=-w` is not true for :-A. Isothermal processB. Adiabatic processC. Cyclic processD. 1 and 3 both |
| Answer» Correct Answer - B | |
| 2775. |
The work done by a weightless piston in causing an expansing `Delta V` (at constant temperature), when the opposing pressure P is variable, is given by :A. `W = - int p Delta V`B. W = 0C. `W = -P Delta V`D. None |
| Answer» Correct Answer - A | |
| 2776. |
Temperature and heat are not :-A. Extensive propertiesB. Intensive propertiesC. Intensive and extensive properties respectivelyD. Extensive and intensive properties respectively |
| Answer» Correct Answer - D | |
| 2777. |
The work done by 100 calorie of heat in isothermal expansion of ideal gas is :-A. 418.4 JB. 4.184 JC. 41.84 JD. None |
| Answer» Correct Answer - A | |
| 2778. |
What is the change in internal energy of an gas sample over one complete cycle ? (A) Positive (B) Negative (C) Zero. (D) depends upon the nature of path |
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Answer» Answer is (C) Zero. After the complete cycle, there is no change in temperature. The internal energy will not change as it depends upon temperature. |
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| 2779. |
How are the following related ? ( Give mathematical relation ) (i) Free energy change and electrical work . (ii) Free energy change, enthalpy change and entropy change. |
| Answer» (i) Electrical work done `(nFE) = - Delta G ` (ii) `DeltaG = DeltaH - T DeltaS`. | |
| 2780. |
Blowing air with open mouth is an example ofA. Isochoric processB. Isobaric processC. Isothermal processD. Adiabatic process |
| Answer» Correct Answer - B | |
| 2781. |
What is thermodynamics? |
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Answer» Thermodynamics is the branch of physics that deals with interconversion of heat and other forms of energy. |
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| 2782. |
Change in internal energy in an isothermal process for ideal gas isA. ZeroB. (+) veC. (-) veD. Cannot be predicted |
| Answer» Correct Answer - A | |
| 2783. |
Explain whyThe climate of a harbour town is more temperate than that of a town in a desert at the same latitude. |
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Answer» The humidity in harbour town is generally much greater than humidity in a desert. Since humidity is a measure of water vapor content in the atmosphere and the specific heat of water vapor is very high (≃ 1.86 kJ kg-1 k-1 at 300k) the temperature fluctuations in harbour towns are generally lower than those in desert regions. Hence, the climate in harbour towns is more temperate than that of a town in a desert at the same latitude. |
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| 2784. |
A thermodynamic system goes from states (i) `P_(1)`V to `2P_(1)` , V (ii) P , V to P , 2 V . Then work done in the two cases isA. Zero zeroB. Zero, `PV_(1)`C. `PV_(1)`,ZeroD. `PV_(1),P_(1)V_(1)` |
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Answer» Correct Answer - B |
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| 2785. |
The internal energy of an ideal gas depends uponA. specific volumeB. pressureC. TemeratureD. Density |
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Answer» Correct Answer - C |
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| 2786. |
In changing the state of thermodynamics from A to B state, the heat required is Q and the work done by the system is W. The change in its internal energy isA. Q = WB. Q - WC. QD. `(Q-W)/(2)` |
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Answer» Correct Answer - B |
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| 2787. |
In thermodynamic process, 200 Joules of heat is given to a gas and 100 Joules of work is also done on it. The change in internal energy of the gas isA. `100j`B. `300j`C. `419j`D. `24j` |
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Answer» Correct Answer - B |
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| 2788. |
If the amount of heat given to a system be 35 joules and the amount of work done on the system be 15 joules , then the change in the internal energy of the system isA. `-50` joulesB. `20`joulesC. `30`joulesD. `50`joules |
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Answer» Correct Answer - D |
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| 2789. |
Heat given to a system is 35 joules and work done by the system is 15 joules. The change in the internal energy of the system will beA. `-50j`B. 20jC. 30 jD. 50j |
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Answer» Correct Answer - B |
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| 2790. |
We add the same amount of heat to ten grams of each of the following substance at `20^(@)C`. Which of the samples show the lowest temperature change?A. `Al (s)`B. `C_(6) H_(6) (l)`C. `Hg(l)`D. `H_(2) O (l)` |
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Answer» Correct Answer - D The higher the specific heat for a substance, the more heat is required to raise a given mass of sample by a given temperature change, so the less its temperature changes by a given amount of heat. Water with the highest specific heat undergoes the smallest temperature change. The ranking from lowest to highest final temperature is `H_(2) O lt C_(6) H_(6) lt Al M lt Hg` |
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| 2791. |
`Delta U=0` forA. Cyclic process, Adiabatic processB. Isothermal, Adiabatic processC. Cyclic process, Isothermal processD. Isochoric process, Isothermal process |
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Answer» Correct Answer - C `DeltaU=0` for isothermal and cyclic processes |
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| 2792. |
Is it possible to convert internal energy into works? |
| Answer» Yes, For example, in explosion of a bomb, chemical energy (which is a form of internal energy) is converted into kinetic energy. | |
| 2793. |
For an ideal monoatomic gas, molar heat capacity at constant volume `(C_(v))` isA. `(2)/(3) R`B. `(3)/(2) R`C. `(5)/(2) R`D. `(2)/(5) R` |
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Answer» Correct Answer - B If a molecule of the ideal gas contains `n` atoms, then the number of degrees of freedom (different modes of translational, rotational, and and vibrational motion) is given by `3n` Three of these are assigned to the translational motion of the molecules, leaving `3n - 3` degrees of greedom for rotational and vibrational motion. For and ideal monoatomic gas, such as `He, 3 (1) - 3 = 0`, thus, the entire contribution to internal energy is from translational motion. Since each translation motion contributes `(1//2)RT` to the internal energy. we have for one mole of ideal monoatimic gas, `U = 3 ((1)/(2) RT) = (3)/(2) RT` `C_(V) = (Delta U)/(Delta T) = ((3)/(2) R Delta T)/(Delta T) = (3)/(2) R` Since `C_(P) - C_(V) = R` (for 1 mol), `C_(P) = R + C_(V) = R + (3)/(2) R = (5)/(2) R` |
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| 2794. |
An ideal gas is compressed at a constant temperature, will its internal energy increase or decrease? |
| Answer» NO, because internal energy of an ideal gas depends only on temperature of the gas. | |
| 2795. |
The work done when a gas is compressed by an average pressure if 0.50 atm so as to decrease its volume from `400 cm^(3)` to `200 cm^(3)`A. 10.13 JB. 20.13 JC. 30.13 JD. 40.13 J |
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Answer» Correct Answer - A `W=- P DeltaV` |
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| 2796. |
Assertion: The internal energy of an ideal gas does not change during an isothermal process. Reason: The decrease in volume of a gas is compensated by a corresponding increase in perssure, when its temp. is held constant.A. If both, Assertion and Reason are true and Reason is the correct explanation of the Asserrion.B. If both,Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - B In isothermal process, temperature remains constant. Therfore, inernal energy of an ideal gas does not change-as it is dependent only of temperature- neither on pressure nor on volume. So the assertion is true but the reason is not a correct explanation of the assertion. |
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| 2797. |
During a process, the internal enegry of the system increases by `240kJ` while the system performed `90 kJ` of work on its surroundings. How much heta was transferred between the system and the surroundings during this process. In which direction did the heat flow? |
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Answer» `DeltaU = q+w or 240 =q - 90 or q = 330 kJ` Heat flows from surroundings to the system. |
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| 2798. |
Is the efficiency of a heat engine more in hilly areas than in plains? |
| Answer» Yes, this is because in hilly areas, temp. of surroundings `(T_(2))` is lower than that of plains. Therefore, `(T_(2)//T_(1))` is smaller in hilly areas than in plains. `eta= (1-(T_(2))/(T_(1)))` becomes more in hilly areas than in plains. | |
| 2799. |
Discuss whether the following phenomena are reversible.(i) Water fall, (ii) Rusting of iron, (iii) Electrolysis |
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Answer» (i) Water fall- The falling of water is not a reversible process. During the fall of water, the major part of its potential energy is converted into kinetic energy of water. On striking the ground, a part of it is converted into heat and sound energy. It is not possible to convert the heat and sound produced along with the K.E. of water into potential energy so as to make the water rise back to the initial height .Therefor, waterfall is not reversible process. (ii) Rusting of iron- In the rusting of iron, the iron oxidized by the oxygen from the air. Since, it is a chemical change, it is not a reversible process. (iii) Electrolysis- It is a reversible process, provided the resistance of the electrolyte to the flow of current is zero.On reversing the direction of current, the direction of motion of ions is reversed. |
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| 2800. |
Give two examples of reversible processes. Discuss their reversibility. |
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Answer» (i) Melting- as well as vaporzation are reversible processes.On reversing the conditions under which they occur, the vapurs condense into a liquid and the liquid solidifies. (ii) All isothermal and adiabatic processes in which no extraneous loss of heat occurs, can be retraced by reversing the boundary conditions and hence they are reversible processes. |
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