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2601.

There are two statement of same gas initially under similar initial state. Gases of both the samples are expanded. `I^(st)` sample using reversible isothermal process and `II^(nd)` sample using reversible adiabatic process till final pressure of both the samples becomes half of initial pressure. Then:A. Final volume of `I^(st)` sample `lt` final volume of `II^(nd)` sampleB. Final volume of `II^(nd)` sample `gt` work by gas in `I^(st)` sampleC. Final volumes will be equalD. Information is insufficient

Answer» Correct Answer - B
2602.

Find the change in internal energy if heat supplied to the gaseous system is 100 J and work done by the system is 80 J.A. 5 JB. 30 JC. 10 JD. 20 J

Answer» Correct Answer - D
2603.

Value of `C_(p)` for monatomic gas is `(5)/(2)R`. Find `C_(V)`.

Answer» As `C_(p)-C_(v)=R`
`thereforeC_(v)=C_(p)-R`
`=(5)/(2)R-R`
`=(3)/(2)R`.
2604.

For a gaseous system find change in internal energy if the heat supplied to the system is 50 J and work done by the system is 16 J.

Answer» `DeltaQ=DeltaU+DeltaW`
`thereforeDeltaU=50-16`
`thereforeDeltaU=34J`
2605.

Find `(C_(p))/(C_(v))` for monatomic ideal gas.

Answer» `C_(p)=(5)/(2)R and C_(V)=(3)/(2)R` for monatomic gas, `(C_(P))/(C_(V))=(5)/(3)`
2606.

For a gaseous system, change in internal energy and work done on the system are respectively 17 J and 41 J. find heat supplied / evolved from the system.

Answer» `DeltaQ=17-41=-24J`, evolved by 1st law.
2607.

What does a heat engine consist of ? OR What are the elements (parts) of a typical heat engine?

Answer»

The following are the parts of a typical heat engine :

(1) Working substance : It can be 

1. a mixture of fuel vapour and air in a gasoline (petrol) engine or diesel engine 

2. steam in a steam engine. The working substance is called a system.

(2) Hot and cold reservoirs : The hot reservoir is a source of heat that supplies heat to the working substance at constant temperature TH . The cold reservoir, also called the sink, takes up the heat released by the working substance at constant temperature TC < TH.

(3) Cylinder and piston : The working substance is enclosed in a cylinder fitted with (ideally) a movable, massless, and frictionless piston. The walls of the cylinder are nonconducting, but the base is conducting. The piston is nonconducting. The piston is connected to a crankshaft so that the work done by the working substance (mechanical energy) can be transferred to the environment.

2608.

What are the two basic types of heat engines?

Answer»

(i) External combustion engine in which the working substance is heated externally as in a steam engine.

(ii) Internal combustion engine in which the working substance is heated internally as in a petrol engine or diesel engine. 

[Note : A steam engine was invented by Thomas New-comen (1663-1729), English engineer. The first practical steam engine was constructed in 1712. The modem steam engine was invented in 1790 by James Watt (1736-1819), British instrument maker and engineer. A hot-air type engine was developed by Robert Stirling (1790-1878), Scottish engineer and clergyman.

A four-stroke internal combustion engine was devised by Nikolaus August Otto (1832-1891), German engineer. A compression-ignition internal combustion engine was devised by Rudolph (Christian Karl) Diesel (1858 – 1913), German engineer.]

2609.

What is calories? Define on calorie.

Answer»

Calories is C.G.S. unit of heat. One calorie is the amount of heat required to raise the temperature of 1 g of water through 1°C.

2610.

State the basic steps involved in the working of a heat engine.

Answer»

1. The working substance absorbs heat (QH) from a hot reservoir at constant temperature. TH . It is an isothermal process QH is positive.

2. Part of the heat absorbed by the working substance is converted into work (W), i.e. mechanical energy. In this case, there is a change in the volume of the substance.

3. The remaining heat (|QC| = |QH| – W) is transferred to a cold reservoir at constant temperature TC < TH . QC is negative.

2611.

State the difference between a reversible process and an irreversible process. OR Distinguish between a reversible process and an irreversible process.

Answer»

A reversible process is a bidirectional process, i.e., its path in P-V plane is the same in either direction. In contrast, an irreversible process is a undirectional process, i.e., it can take place only in one direction.

A reversible process consists of a very large number of infinitesimally small steps so that the system is all the time in thermodynamic equilibrium with its environment. In contrast an irreversible process may occur so rapidly that it is never in thermodynamic equilibrium with its environment.

2612.

What are the conditions for thermodynamic equilibrium?

Answer»

(i) Temperature of every part of the system must be the same. 

(ii) There should not be net unbalanced forces on a part or whole of the system. 

(iii) There should be no changes due to chemical reactions.

2613.

State the two forms of the second law of thermodynamics.

Answer»

Second law of thermodynamics :

1. It is impossible to extract an amount of heat QH from a hot reservoir and use it all to do work W. Some amount of heat QC must be exhausted (given out) to a cold reservoir. This prohibits the possibility of a perfect heat engine. This statement is called the first form or the engine law or the engine statement or the Kelvin-Planck statement of the second law of thermodynamics.

2. It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this. This prohibits the possibility of a perfect refrigerator. This statement is called the second form or the Clausius statement of the second law of thermodynamics.

Notes : 

1. Max Planck (Karl Ernst Ludwig) (1858- 1947) German physicist, put forward quantum therory of radiation.

2. Rudolf Clausius (1822-88) German theoretical physicist, made significant contribution to thermodynamics.

2614.

State and explain the limitations of the first law of thermodynamics.

Answer»

1. The first law of thermodynamics is essentially the principle of conservation of energy as there is a close relation between work and energy. We find that there is a net transfer of energy (heat) from a body at higher temperature to a body at lower temperature. The net transfer of energy from a body at lower temperature to a body at higher temperature is not observed though consistent with the first law of thermodynamics. 

2. If two containers, one containing nitrogen and the other containing oxygen, are connected to allow diffusion, they mix readily. They never get separated into the respective containers though there is no violation of the first law of thermodynamics. 

3. It is not possible to design a heat engine with 100% efficiency, though there is no restriction imposed by the first law of thermodynamics. 

4. At room temperature, ice absorbs heat from the surrounding air and melts. The process in the opposite direction is not observed, though consistent with energy conservation. These examples suggest that there is some other law operative in nature that determines the direction of a process

2615.

Why is mercury preferred a thermos-metric fluid?

Answer»

(i) Mercury is a shining bright fluid so it is easily visible in the glass tube. 

(ii) It is good thermal conductor.

2616.

Milk is poured into a cup of tea and is mixed with a spoon. What happens to the work done to mix the tea and milk?

Answer»

The work done in mixing milk gets converted into heat energy. In principle, the temperature of the tea and milk will rise.

2617.

Can two isothermal curves intersect each other?

Answer» No, If they intersect, then at two different temperatures (of the isothermals), volume and pressure of gas will be same, which is not possible.
2618.

Why enthalpy changes whereas, internal energy does not change due to change in heat energy ?

Answer»

Mostly chemical reaction occurs under constant atmospheric pressure. In this situation, heat change occurs on the system is different from the change under constant volume.

Hence, heat energy change occurs under constant pressure but internal energy does not change due to this heat energy change but enthalpy changes.

2619.

What are the limitations of the first law of thermodynamics?

Answer»

The first law of  thermodynamic has the following limitations:

(a) The first law does not indicate the direction in which the change car occur- 

For examples, (i) When two bodies at different tempeatures are put in thermal contact with each other, heat flows the body at higher temperature to the body at lower temperature. We now know that heat cannot flow the body at lower temperature to the body at higher temperature,even if first law of thermodynamic is not violated.

(ii) When a moving car is stopped by applying breaks, work done against friction is converted into heat. When the car cools down, it does not start moving with the conversion of all its heat energy into mechanical work. 

(b) The law given no idea about the extent of change.

(i) Our observation and experience tell that there appears to be no restriction conversion of mechanical work into heat.But, there are severe restrictions on the reverse process, i.e., Conversion of heat energy into mechanical energy.

(ii) We know that heat is not converted into mechanical energy all by itself. An external agency called the heat engine is required for the purpose.

(iii) No heat engine can convert all the heat energy received from the source into mechanical energy. The first law of thermodynamics is silent about all this.

(c) The first law of thermodynamics given no information about the source of heat i.e., whether it is a hot or a cold body.

2620.

Can heat be added to a system without increasing its temperature ?

Answer»

Yes, for example, in melting process and boiling process.

2621.

In summer, when the valve of a bicycle tube is removed, the escaping air appears cold. Why?

Answer» This happens due to adiabatic expansion of the air of the tube of the bicycle.
2622.

On removing the valve, the air escaping from a cycle tube becomes cool. Why?

Answer»

Tube is a thermal insulator. The removal of value makes the pressurized air inside the tube come out of the tube suddenly.

There are the conditions suitable for adiabatic change which ensure decrease in temperature during expansion. Thus the air escaping from the cycle tube becomes cool.

2623.

In summer, when the valve of a bicycle tube is removed, the escaping air appears cold. why?

Answer»

This happens due to adiabatic expansion of the air of the tube of the bicycle.

2624.

State planck's statement for the second law of thermodynamics. Explain it.

Answer»

Planck's Statement-" It is impossible to construct a heat engine operating on a cycle, that will extract heat from a reservoir and perform an equivalent amount of work".

This statement implies that the working substance, working in a cycle, cannot convert 100% of heat extracted into work, i.e., some of  the heat extracted has to be rejected to the sink.

Statement of the second law of thermodynamics starts from "it is impossible." Thus the second law is negative in from. Hence, it cannot be proved directly. However, the law is accepted universally and has been applied to a wide variety of phenomena like chemical reactions, thermoelectricity, electric, cells, study of solutions, change of state etc.

2625.

What are the signs of convention adopted in the first law of thermodynamics ?

Answer»

The system of taking plus/minus sign with the three quantities (dQ. dW and dU) involved, is as follows:

(a) When heat is supplied to a system. dQ is taken as positive.When a gas is compressed, work is done on the gas, dW is taken as negative.

(b) When temperature of a gas increases its internal energy increases, dU is taken as positive. When temperature of a gas decreases, its internal energy decreases, dU is taken as negative.

(c) When a gas expands, work done by the gas dW is taken as positive. When a gas is compressed, work is done on the gas, dW is taken as negative.

2626.

State the characteristics of the first law of thermodynamics.

Answer»

The following facts are worth noting regarding the first law of thermodynamics.

(a) This law, which is basically the law of conservation of energy, applies to every process in nature.

(b) The law applies equally to all the three phases of matter, i.e., solid liquid and gas.

(c) Out of the three quantities, dQ, dW and dU, the first two are dependent on the path the system follows in going from initial to final state. However, dU does not depend on path. It depends only on initial and final states of the system.

(d) While applying the first law of thermodynamics, care must be taken that all the three quantities dQ, dU and dW must be expressed either in joule or in calorie.

(e) Just as Zeroth law of thermodynamics introduces the concept of temperature, the first law of thermodynamics introduces the concept of internal energy.

(f) From the first law of thermodynamics, we learn that it is impossible to get work from any machine, without giving an equivalent amount of energy to the machine.

(g) The first law of thermodynamics establishes the essential equivalence between work and heat, as according to this law, internal energy (and hence temperature) of a system can be increased either by supplying heat to it or by doing work on the system or both.

2627.

Which of the two, diamond or graphite, has more enthalpy?

Answer»

Graphite has greater enthalpy since it is loosely packed.

2628.

How non-spontaneous process be converted to spontaneous process?

Answer»

∆G = ∆H + T∆S; when the signs for ∆H and ∆S are both the same, then temperature determines that the process is spontaneous. When ∆H is positive (unfavourable) and ∆S is positive (favourable), high temperatures are needed so the favourable ∆S term dominates making the process spontaneous (∆G < 0). When ∆H is negative (favourable) and ∆S is negative (unfavourable), low temperatures are needed so the favourable ∆H term dominates making the process spontaneous (∆G < 0).

2629.

Internal energy change is state function but work is not a state function. Why?

Answer»

Internal energy is state function because it describes quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state. In contrast, work is process quantity or path function, because its value depends on the specific transition (or path) between two equilibrium states.

2630.

Differentiate between isothermal and adiabatic changes.

Answer»
Isothermal changesAdiabatic changes
1. Temperature (T) remains constant, i.e., ΔT = 0.Heat constant (Q) remains constant, i.e., ΔQ = 0.
2. System is thermally conducting to the surroundings.System is thermally insulate from the surroundings.
3. The changes occur slowly.The changes occur suddenly.
4. Internal energy (U) remains constant i.e.,ΔU = 0.Internal energy changes, U ≠ constant.ΔU≠ 0.
5. Specific heat becomes infinite.Specific heat becomes zero.
6. Equation of isothermal changes is pV = constant.Equation of adiabatic changes is pVt =constant.
7. Slope of isothermal curve, dp/dV = -(P/V)Slope of adiabatic curve, dp/dV = -Y(P/V)
8. Coefficient of isothermal elasticity, Ei = pCoefficient of adiabatic elasticity Ea = γp.

2631.

Differentiate between isothermal and adiabatic process?

Answer»

Process in which temperature remains constant throughout the process is called isothermal process. When such a process occurs, heat can flow from the system to the surrounding and vice versa in order to keep the temperature constant. In an isothermal process, dT= 0.

When a process is carried out in such a way that no heat can flow from the system to the surroundings or vice versa i.e., system is completely insulated from the surroundings. In such a system, temperature of the system always changes. For an adiabatic process dQ = 0

2632.

What is the relation between E and H?

Answer»

E = Internal energy, H = Enthalpy,
∆H = ∆E + P∆V or ∆H = ∆E + ∆ngRT
here ∆E = Change in internal energy
∆H = Enthalpy change,
∆ ng = Number of moles of gaseous products – Number of moles of geseous reactants Where R = Gas constant
T = Temperature

2633.

In which state a substance has maximum entropy?

Answer»

Entropy by definition is the degree of randomness in a system. Out of three states of matter: Solid, Liquid and Gas, the gaseous particles move freely and therefore, the degree of randomness is the highest. Solids have highly ordered arrangement and has low entropy. The entropy of liquids is more than solids because molecules of liquid shows less ordered arrangement than solids. The molecules of gas are free to move and not constrained to be adjacent to each other. Therefore, Sgas >> Sliquid > Ssolid. In gaseous state, entropy is maximum.

2634.

Define combined form of first and second law of thermodynamics.

Answer»

In thermodynamics, the combined law of thermodynamics, also called the Gibb’s fundamental equation, is a mathematical summation of the first law of thermodynamics and the second law of thermodynamics summed into a single concise mathematical statement as
dU - TdS + PdV ≤ 0
Where dU is a variation in internal energy.
T is temperature, dS is variation in entropy
P is pressure
dV is variation in volume of a simple working body in which there is neither flows of particles out of the body nor external forces, other than gravity, acting on the body. In theoretical structure, in addition to the obvious inclusion of the first two laws, the combined law incorporates the implications of the zeroth law, via temperature T and the third law, through its use of free energy as related to the calculation of chemical affinities near absolute zero.

2635.

During isothermal expansion of ideal gas(a) Internal energy increases(b) Enthalpy decreases(c) Enthalpy is unchanged(d) Enthalpy decreases and becomes zero

Answer»

(d) Enthalpy decreases and becomes zero

2636.

Is entropy of universe constant?

Answer»

No, the entropy of the universe always increases in the course of every spontaneous change. It is the second law of thermodynamics.

2637.

Internal energy is(a) Partially potential and partially kinetic(b) Totally kinetic(c) Totally potential(d) None of these

Answer»

(a) Partially potential and partially kinetic

2638.

In which state, entropy of a substance is maximum?(a) Solid(b) Liquid(c) Gas(d) Same in all states

Answer»

The answer is (c) Gas

2639.

What are the necessary conditions for adiabatic process?

Answer»

The necessary conditions for adiabatic process are:-
(a) There should not be any exchange of heat between the system and its surrounding. All walls of the container and the piston must be perfectly insulating.
(b) The system should be compressed or allowed to expand suddenly so that there is no time for the exchange of heat between the system and its surroundings.

2640.

Choose the correct relationA. `eta=(1-beta)/(beta)`B. `beta=(1-eta)/(eta)`C. `eta=(beta)/(1-beta)`D. `beta=(eta)/(1-eta)`

Answer» Correct Answer - B
2641.

What will be the efficiency of Carnot engine if its source is at 500 K and sink at 300 K?

Answer»

Efficiency, η = (T- T1)/T2
T2 = 500 K (temperature of the source)
T1 = 300 K (temperature of the sink)
Therefore,
η = (T2 – T2])/r2
= (500 – 300) / 500
= 200/500 .
= 2/5
= 0.4

2642.

The source of a Carnot engine is at 500 K but sink is at 300 K. The efficiency of engine will be :(a) 0.2(b) 0.4(c) 0.6(d) 0.3

Answer»

The answer is (b) 0.4

2643.

Assertion: The efficiency of a carnot engine cannot be `100%`. Reason: This is because sink of heat engine cannot be maintained at `0 K`.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.

Answer» Correct Answer - A
Here, both assertion and reason are true, and reason is correct explanation of assertion
As efficiency, `eta= 1-(T_(2))/(T_(1)), T_(2)!= 0, :. eta!=1`
2644.

Even Carnot engine cannot give `100%` efficiency because we cannotA. Prevent radiationB. Find ideal sourcesC. Reach absolute zero temperatureD. Eliminate friction

Answer» Correct Answer - C
2645.

A carnot engine whose sink is at 300 K has an efficiency of 50. by how much should the temperature of source be increased so as the efficiency becomes 70% ?A. 100 KB. 200 KC. 300 KD. 400 K

Answer» Correct Answer - D
2646.

A cannot heat engine whose sink is at 200 K, has an efficiency 30%. By how much the temperature of the source be increased to have its efficiency equal to 50%. Keeping sink temperature constant.

Answer»

From the relation,

η = 1 − \(\frac{T_2}{T_1}\)

or, \(\frac{T_2}{T_1}\) = 1 - \(\frac{30}{100}\) = \(\frac{7}{10}\)

or, T1 \(\frac{10T_2}{7}\) = \(\frac{10\times200}{7}\)= 285.71 K

New efficiency is now 50%,

η' = 1 − \(\frac{T_2}{T_1}\)

\(\frac{T_2}{T_1}\) = 1 - η'

\(\frac{T_2}{T_1}\) = 1 - \(\frac{50}{100}\) or \(\frac{1}{2}\)

2T2 = T1

or,  T1 = 2 × 200 K

T1 = 400 K

Now increase in temperature of source,

= 400 K – 285.71 K

= 114.3 K

2647.

Define system.

Answer»

System:- A portion of universe which is under investigation, e.g., portion of a test tube, where reaction is taking place, is called system.

2648.

Define isothermal and adiabatic systems.

Answer»

Isothermal and adiabatic systems:- A process is said to be isothermal if the temperature of the system remains fixed. The system is in thermal contact with the surroundings and heat energy can be exchanged between the two. It can flow from system to the surroundings in case of exothermic process and vice versa in case of endothermic process. For a isothermal process, dT = 0 where dT is the change in temperature.

In an adiabatic process, no exchange of heat between the system and the surroundings is possible i.e., the system is completely isolated or insulated from the surroundings. In this case, the change carried in the system is exothermic, its temperature rises and if it is endothermic, there is a fall in temperature.

2649.

Which of the following is / are exothermic and which are endothermic?(i) Ca(g) → Ca2+(g) + 2e- (ii) O(g) + e- → O- (g) (iii) N2-(g) + e- → N3-(g)

Answer»

(i) Endothermic (ionisation energy is required) 

(ii) Exothermic (first electron affinity - energy is released) 

(iii) Endothermic (higher electron affinities energies are required)

2650.

Calculate the heat of reaction of the following reaction : C6H12O6(s) + 6O2(g) → 6CO2(g)+ 6H2O(g); ΔH = ? C(graphite) + O2(g) → CO2(g); DH = -395.0 kJ … (1) H2(g) + O2(g) → H2O(l); DH = -269.5 kJ … (2) 6C(graphit.e) + 6H2(g) + O2(g) → C6H12O6(s); DH = -1169.8 kJ … (3)

Answer»

Multiplying equation (1) and (2) each by 6 reversing (3), we get, 

6C(graphite) + 6O2(g) → 6CO2(g); ΔH = -2370 kJ … (4) 

6H2(g) + 3O2(g) → 6H2O(l); ΔH = -1616.4 kJ … (5) 

C6H12O6(s) → 6C(graphite) + 6H2(g) + 3O2(g) ; ΔH = +1169 - 8 kJ …. (6) 

Adding (4), (5) and (6), 

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

AH(C6H12O6) = -2370.0 - 1616.4 + 1169,8 = -2816.6 kJ