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2551.

A system absorbs 50 kJ heat and does 20 kJ of work. What is the net change in the internal energy of the system ?A. Increase by 30 kJB. Decrease by 30 kJC. Increase by 70 kJD. Decrease by 70 kJ

Answer» Correct Answer - A
`DeltaU=q+(-w)=50+(-20)=30kJ`
2552.

Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume `u=U/V propT^4` and pressure `P=1/3(U/V)`. If the shell now undergoes an adiabatic expansion the relation between T and R is :A. `T prop e^(-R)`B. `T prop e^(-3R)`C. `T prop 1/R`D. `T prop 1/(R^(3))`

Answer» Correct Answer - C
Here, `P= 1/3(U/V) and u=U/V propT^(4)`
or `u=U/V=KT^(4)`
`:.P=1/3KT^(4), As PV=nRT, :. P=(nRT)/V`
or `(nRT)/V= KT^(4) or (nR)/V=KT^(3)`
`:. V prop T^(-3) or VT^(3)= constant`
As volume, `V= 4/3 piR^(3) :. 4/3piR^(3)T^(3)=constant`
or `R^(3)T^(3)= constant`
`RT= constant or T prop 1/R`
2553.

Consider the `DeltaG_(f)^(@)` and `DeltaH_(f)^(@)` (kJ`//`mol) for the following oxides. Which can be most easily decomposed to form the metal and oxygen gas?A. `ZnO(DeltaG^(@)=-318.4, DeltaH^(@)=-348.3)`B. `Cu_(2)O(DeltaG^(@)=-146.0, DeltaH^(@)=-168.8)`C. `HgO(DeltaG^(@)=-58.5, DeltaH^(@)=-90.8)`D. `PbO(DeltaG^(@)=-187.9, DeltaH^(@)=-217.3)`

Answer» Correct Answer - C
`DeltaG^(@)=-RTlnK, " K"uarr"stability of reactant"darr`
2554.

The enthalpy change for a reaction does not depend upon:A. the physical states of reactants and productsB. use of different reactants for the same productC. the number of intermediate reaction stepsD. the differences in initial or final temperature of involved substances

Answer» Correct Answer - C
The change in enthalpy does not depend upon number of intermediate steps.
2555.

In an adibatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic conditions from the following :A. `q = 0, Delta T !=0, W = 0`B. `q!=0, Delta T = 0, W = 0`C. `q = 0, Delta T = 0, W = 0`D. `q =0, Delta T lt 0, W != 0`

Answer» Correct Answer - C
Free expansion, `W = 0`
Adiabatic process, `q = 0`
`Delta U = q + W = 0`, this means that internal energy remains constant. Therefore, `Delta T = 0` In ideal gas there is no intermolecular attraction.
Hence, when such a gas expands under adiabatic conditions into a vacuum, no heat is absrobed or evolved since no external work is done to separate the molecules
2556.

Which of the following does not represent enthalpy change during phase transformation?A. Standard enthalpy of fusionB. Standard enthalpy of vaporisationC. Standard enthalpy of sublimationD. Standard enthalpy of formation

Answer» Correct Answer - D
There is no phase change in formation. A new product is formed during the reaction.
2557.

An ideal gas expands from `1xx10^(-3)m^(3)` to `1xx10^(-2)m^(3)` at `300K` againts a constant pressure of `1xx10^(5)Nm^(-2)` . The work done is :A. `-900 J`B. `-900 kJ`C. 270 kJD. `+900 kJ`

Answer» Correct Answer - A
2558.

Which of the following `P-V` diagrams best represents an isothermal process?A. B. C. D.

Answer» Correct Answer - B
In an isothermal process, `PV=K=constant`
`P=K/V or P prop1/V` so choice (b) is correct
2559.

Which of the following `P-V` diagrams best represents an isothermal process?A. B. C. D.

Answer» Correct Answer - B
2560.

If heat given to a system is 6 kcal and work done is 6 k J . Then change in internal energy isA. 19.1kjB. 12.5kjC. 25kjD. Zero

Answer» Correct Answer - A
2561.

The molar heat capacity of water at constant pressure, `C_(p)` is `"75 J K"^(-1)"mol"^(-1)`. When 10 kJ of heat is supplied to 1 kg water which is free to expand, the increase in temperature of water isA. 2.4 KB. 4.8 KC. 3.2 KD. 10 K

Answer» Correct Answer - A
`q_(m)=mcDeltaT`
`q_(m)="10 kJ or 10000 J, m = 1000 g"`
`c="75 J K"^(-1)"mol"^(-1)."No. of moles "=(1000)/(18)="55.5 mol"`
`DeltaT=(10000)/(55.5xx75)=2.4K`
2562.

The ratio of P to V at any instant is constant and is equal to 1. for a monoatomic ideal gas undergoing a process. What is the molar heat capacity of the gas?A. `(3R)/(2)`B. `(4R)/(2)`C. `(5R)/(2)`D. 0

Answer» Correct Answer - B
2563.

200 joules of heat was supplied to a system at constant volume. It resulted in the increase in temperature of the system from 298 to 323 K. What is the change in internal energy of the system ?A. 400 JB. 200 JC. 50 JD. 150 J

Answer» Correct Answer - B
At constant volume, `DeltaV=0`
`w=-PDeltaV=0" "DeltaU=q=200J`
2564.

What would be the work done on`//` by the system if the internal energy of the system falls by the 100 joules even when 200 joules of heat is supplied to it ?

Answer» Correct Answer - Work done by the system `= 300 J`
2565.

. When heat energy of 1500 Joules , is supplied to a gas at constant pressure `2.1xx10^(5)N//m^(2),` there was an increase in its volume equal to `2.5xx10^(-3)m^(3)`.The increase in internal energy of the gas in joules isA. 45B. 525C. 975D. Zero

Answer» Correct Answer - C
2566.

An ideal gas involved in a reversible adiabatic process follow the law `T alpha P^(1//3` then what will be the value of `C_(p)//C_(v)?`A. `1`B. `(4)/(3)`C. `1.5`D. `1.333`

Answer» Correct Answer - C
2567.

One mole of `H_(2) at 25^(@)C` undergo combustion in the presence of excess `O_(2)` in a closed rigid adibatic container: `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)," "DeltaH=-240kJ//"mole"` Choose the incorrect option :A. `q=0`B. `w=0`C. `deltaU=0`D. `deltaT=0`

Answer» Correct Answer - D
2568.

In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.A. `q=0, DeltaT ne 0, W=0`B. `q ne0, DeltaT=0, W=0`C. `q=0, DeltaT=0, W=0`D. `q=0, DeltaT lt 0, W ne 0`

Answer» Correct Answer - C
For free expansion (i.e. in vacuum), `P_("ext")=0`.
Thus, `W-P_("ex")DeltaV=0`
`therefore" "DeltaU=q+W=0` which is true for isothermal process where T is constant i.e., `DeltaT=0.`
2569.

Select the option in which all the parametres are intensive :A. pH, volume, Electrodes potential.B. Molar enthalpy, Heat capacity, Resistivity.C. Temperature,Molality, Electromotive force.D. Mass, Pressure, Molar entropy.

Answer» Correct Answer - C
2570.

What are Gram specific heat ?

Answer»

Gram specific heat : The heat required to raise the temperature of one gram mass of a body through 1ºC (or 1 K) is called gram specific heat of the material of the body.

specific heat, c = Q/mΔT

Units : Calorie/gmºC (Practical), J/kg K(S.I.) 

Dimension : [L2T–2θ–1

2571.

Difference between Thermal Capacity and Water Equivalent.

Answer»

(1) Thermal capacity : It is defined as the amount of heat required to raise the temperature of the whole body (mass, m) through 1ºC or 1 K. Thermal capacity = m c = µC = Q/ΔT

Dimension : [ML2T–2θ–1], 

Unit : call ºC (practical) Joule K (S.I.)

(2) Water Equivalent : Water equivalent of a body is defined as the mass of water which would absorb or evolve the same amount of heat as is done by the body in rising or falling through the same range of temperature. It is represented by W. 

If m = Mass of the body, c = Specific heat of body 

∴ Water equivalent (W) = mc gm 

2572.

Define heat and Temperature.

Answer»

The energy associated with configuration and random motion of the atoms and molecules with in a body is called heat. 

(1) Units : Joule (S.I.) and calorie (Practical unit) 

(2) The ratio of work done (W) to heat produced (Q) is constant. 

W/Q = J or W = JQ

J is called mechanical equivalent of heat and has value 4.2 J/cal. 

1 calorie = 4.186 Joule = 4.12 Joule 

(3) Heat is a path dependent and is taken to be positive if the system absorbs it and negative if releases it.

Temperature is defined as the degree of hotness or coldness of a body. Heat flows from higher temperature to lower temperature. 

Two bodies are said to be in thermal equilibrium when both the bodies are at the same temperature. Temperature α kinetic energy [As E = 3/2 RT]

2573.

What is meant by Scales of Temperature ?

Answer»

The Kelvin temperature scale is also known as thermodynamic scale. The S.I. unit of temperature is kelvin and is defined as (1/273.16) of the temperature of the triple point of water. The triple point of water is that point on a P–T diagram where the three phases of water, the solid, the liquid and the gas, can coexist in thermal equilibrium. 

To construct a scale of temperature, two fixed points are taken. First is the freezing point of water, it is called lower fixed point. The second is the boiling point of water, it is called upper fixed point.

2574.

Define Expansion of Gases.

Answer»

Gases have no definite shape, therefore gases have only volume expansion.

2575.

The energy change due to the reaction `2Na(s) +CI_(2) (g) rarr2Na CI(s)` is `-826kJ` The consumption of 1 mo1 of `CI_(2)` gas contracts the system by `22.4L` at 1 atm What is the enthalphy change of the reaction .

Answer» `W = P DeltaV = 1atm xx 22.4 L = 22.4 L atm`
`= 22.4 L atm xx 101.3J`
`= 2.27 xx 10^(3) = 2.27 kJ`
The work done on the system is to be subtracted form the energy change to get enthalpy change,
`DeltaH = DeltaU +P DeltaV`
`=- 826 kJ + (-2.27 kJ) =- 828.27 kJ`
2576.

Calculate the total work done by the gas in the process. (given ln2=0.693).

Answer»

Total work done by gas, WTotal​ = WAB ​+ WBC ​ + WCA

WAB​ = nRT ln 4V/V ​= 2nRT ln2 = 2PV ln2.

Also PA​VA ​= PB​VB​ (As AB is an isothermal process)

or, PB​ = PAVA/VB ​​= PV/4V ​= P/4​.

In the step BC, the pressure remains constant. Hence the work done is,

WBC​ = P/4​(V − 4V) = −3PV/4​.

In the step CA, the volume remains constant and so the work done is zero. The net work done by the gas in the cyclic process is

W = WAB ​ + WBC ​ + WCA

= 2PVln2 − 3PV/4 ​+ 0

Hence, the work done by the gas is 0.636 PV.

2577.

The heat of neutralization fo a stron acid by a strong base is a constantA. `-57.0 kJ mol^(-1)`B. `-13.7 kcal mol^(-1)`C. `-5.7 xx 10^(4) J mol^(-1)`D. `-13.7 xx 10^(3) cal mol^(-1)`

Answer» The heat of neutralisation for strong acid and base is equal to `-57.3 kJ mol^(-1) or -13.7 kcal mol^(-1)`
2578.

The enthalpy change for the process `C("graphite") rarr C(g)`is calledA. Heat of vaporisationB. Heat fo sublimationC. Heat of allotropic changeD. Heat of atomisation

Answer» `C("graphite") rarr C(g)`. The process is sublimation as well conversion of crystalline allotropic from (graphite) to amorphours form (gas carbon).
2579.

`1` mole of gas occupying `3` litre volume is expanded against a constant external pressure of `1` atm to a volume of `15` litre. The work done by the system is:A. `1.215xx10^(3) J`B. `12.15xx10^(3) J`C. `121.5xx10^(3) J`D. None of these

Answer» Correct Answer - A
`-w_(irr)=P_(ext)(V_(2)-V_(1))`
`=1xx(15-3)=12` litre atm
`=(12xx1.987xx4.184)/0.0821=1.215xx10^(3)J`
2580.

Work done in expansion of an ideal gas from `4` litre to `6` litre against a constant external pressure of `2.1 atm` was used to heat up `1` mole of water at `293 K`. If specific heat of water is `4.2 J g^(-1)K^(-1)`, what is the final temperature of water?

Answer» `W-2.1[6-4]=-2.1xx2`
`= -4.2 atmxxlit`
`= -4.2xx101.325J`
This work is used to heat up the water
specific heat of `H_(2)O=4.2(J)/("gram".K)`
Heat required for increasing temperature by `1^(@)C` of `1 "mole"=4.2xx18=75.6 J`
`4.2xx101.325=75.6[T-293]`
`5.63=T-293`
`T=298.63K`
2581.

The molar heat capacity , `C_(v)` of helium gas is `3//2R` and is independent of temperature. For hydrogen gas, `C_(v)` approaches `3//2R` at very low temperature, equal `5//2R` at moderate temperature and is higher than `5//2 R` at high temperatures. Give a reason for the temperature dependence of `C_(v)` in case of hydrogen, in not more than two or three sentences.

Answer» Correct Answer - Hydrogen is diatomic so at high temperature rational and vibrational motion also counts.
Helium `(He)` gas is monoatomic and it has three translational degree of freedom. Hence, contribution of each transiational degree of freeedom towards `C_(v)` being `R//2`, so the total contribution towards `C_(v)=3xxR//2`. Hydrogen molecule is diatomic. At low temperature, rotational and vibrational contribution for `H_(2)` are zero. So, `C_(v)` for `H_(2)` at low temperature continues to be `3R//2`. At moderate temperature, rotational contribution `(=2xxR//2)` also becomes dominant and hence total contribution towards `C_(v)=(3R)/(2)+R=(5R)/(2)`. At even high temperature, vibrational contribution `(=1xxR)` also becomes significant. Hence total contibution towards `C_(v)=(3R)/(2)+R+R=(7R)/(2)`
2582.

Phase transitions are ubiquitous in nature. We are all familiar with the different phase of water (vapour, liquid and ice) and with the change from one to another, the change of phase are called phase transitions. There are six ways a substance can change between these three phase, melting, freezing, evaporating, condensing sublimation and decomposition. At `1 atm` pressure vaporisation of `1` mole of water from liquid `(75^(@)C)` to vapour `(120^(@)C)`. `C_(v)(H_(2)O,l)=75 J "mole"^(-1)K^(-1), C_(p)(H_(2)O,g)=33.3J"mole"^(-1)K^(_1)` `Delta H_(vap)` at `100^(@)C=40.7KJ//"mole"` Calculate change in internal energy when Water vapour at `100^(@)C` to `120^(@)C` ?A. `666 J`B. `2998 J`C. `499.72 J`D. `166.28 J`

Answer» Correct Answer - C
2583.

Phase transitions are ubiquitous in nature. We are all familiar with the different phase of water (vapour, liquid and ice) and with the change from one to another, the change of phase are called phase transitions. There are six ways a substance can change between these three phase, melting, freezing, evaporating, condensing sublimation and decomposition. At `1 atm` pressure vaporisation of `1` mole of water from liquid `(75^(@)C)` to vapour `(120^(@)C)`. `C_(v)(H_(2)O,l)=75 J "mole"^(-1)K^(-1), C_(p)(H_(2)O,g)=33.3J"mole"^(-1)K^(_1)` `Delta H_(vap)` at `100^(@)C=40.7KJ//"mole"` Calculate change in internal energy when Water liquid at `100^(@)C` to vapour at `100^(@)C` ?A. `40700 J`B. `37598.878 J`C. `43801.1J`D. `31012.J`

Answer» Correct Answer - B
2584.

Phase transitions are ubiquitous in nature. We are all familiar with the different phase of water (vapour, liquid and ice) and with the change from one to another, the change of phase are called phase transitions. There are six ways a substance can change between these three phase, melting, freezing, evaporating, condensing sublimation and decomposition. At `1 atm` pressure vaporisation of `1` mole of water from liquid `(75^(@)C)` to vapour `(120^(@)C)`. `C_(v)(H_(2)O,l)=75 J "mole"^(-1)K^(-1), C_(p)(H_(2)O,g)=33.3J"mole"^(-1)K^(_1)` `Delta H_(vap)` at `100^(@)C=40.7KJ//"mole"` Calculate change in internal energy when Water liquid at `75^(@)C` to `100^(@)C` ?A. `1875 J`B. `13125 J`C. `-1875 J`D. `-13125 J`

Answer» Correct Answer - A
2585.

For a Van der Walls gas find : (a) the equation of the adiabatic curve in the variables `T, V` , the difference of the molar heat capacities `C_p - C_v` as a function of `T` and `V`.

Answer» (a) From the first law for an adiabatic
`dQ = dU + pd V = 0`
From the previous problem
`d U =((del U)/(del T))_V dT + ((del U)/(del V))_T dV = C_V dT + (a)/(V^2) dV`
So, `0 = C_V dT + (RT dV)/(V - b)`
This equation can be integrated if we assume that `C_V` and `b` are constanr then
`(R)/(C_V) (dV)/(V - b) + (dT)/(T) = 0` or, `1n T +(R)/(C_V) 1n (V - b) = constant`
or, `T(V - b)^(R//c_v) = constant`
(b) We use
`dU = C_V dT + (a)/(V^2) dV`
Now, `dQ = C_V dT + (RT)/(V - b)dV`
So along constant `p`,
`C_p = C_V + (RT)/(V - b) ((del V)/(del T))`
Thus `C_p - C_V = (RT)/(V - b)((del V)/(del T))_p`, But `p = (RT)/(V - b)- (a)/(V^2)`
On differentiating, `0 = (-(RT)/((V - b)^2) +(2 a)/(V^2))((del V)/(del T))_p + (R)/(V - b)`
or, `T((del V)/(del T))_p = (RT//V -b)/((RT)/((V - b)^2 )-(2 a)/(V^3)) = (V -b)/(1-(2a(V - b)^2)/(RT V^3))`
and `C_p - C_V = (R)/(1 -(2a(V - b)^2)/(RTV^3))`.
2586.

Can a system be heated and its temperature remains constant?

Answer»

∆Q = ∆U + W (from I law of thermodynamics)

or, ∆Q = nCv∆T + ∆W  (∵ ∆U = nCv∆T)

if system has constant temperature inspite of heat supplied then supplied then

∆T = 0 and ∆Q = ∆W. It means heat supplied (∆Q) to the system is used in doing work (∆W) against the surroundings.

2587.

Can mechanical energy be converted continuously and completely into heat? Is the reverse also possible?

Answer»

The mechanical energy can be completely and continuously converted into heat. Bu the reverse is not possible.

2588.

No real engine can have efficiency grater than that of a Carnot engine working between the same temperatures.Why?

Answer»

A Carnot engine is an ideal heat engine form the following points of view:

(i) There is absolutely no friction between the walls of cylinder and piston.

(ii) The working substance is an ideal gas.

In a real engine, these conditions cannot be fulfilled and hence no heat engine working between the same two temperatures can have efficiency greater than than of carnot engines.

2589.

Can we make heat to flow from a body at lower temperature to a body at higher temperature?

Answer»

Yes, it can be done with the help of an external agent.

2590.

On what factors, the efficiency of a Carrot engine depends?

Answer»

On the temperature of source of heat and the sink.

2591.

In a Carnot engine, temperature of sink is increased. What will happen to its efficiency?

Answer»

We know that, 

η = \(1-\frac{T_2}{T_1}\) 

On increasing the temperature of sink (T2), the efficiency of the Carnot engine will decrease.

2592.

Is Carnot engine a real engine?

Answer»

No, it is an ideal engine which practically not possible to construct.

2593.

Is coefficient of performance of refrigerator a constant quantity?

Answer»

No, the coefficient of performance of a refrigerator decreases as its inside temperature decreases.

2594.

Under what ideal condition the efficiency of a Carnot engine be 100%.

Answer»

If the temperature of sink is zero kelvin.

2595.

Is coefficient of performance of a refrigerator a constant quantity ? 

Answer»

No. As the inside temperature of the refrigerator decreases, its coefficient of performance decreases.

2596.

A system changes from state X to Y with a change in internal energy measuring to `"25 kJ mol"^(-1)`, by a reversible path and returns from Y to X by an irreversible path. What will be the net change in internal energy?A. 25 kJB. gt 25 kJC. lt 25 kJD. zero

Answer» Correct Answer - D
For a cyclic process, net change in internal energy is zero. Change in internal energy does not depend on the path by which final stage is reached.
2597.

if it is a carnot cycle and a heat engine working on this cycle takes 200 J heat from the source, find amount of heat converted into work.

Answer» Correct Answer - 50J
2598.

Find the efficiency of carnot engine whose source and sink are at `927^(@)C` and `27^(@)C`.

Answer» Correct Answer - By `neta=1-(T_(2))/(T_(1))`
0.75
2599.

A carnot engine has efficiency of 80%. If its sink is at `127^(@)C`, the find the temperature of source.

Answer» Correct Answer - By `neta=1-(T_(2))/(T_(1))`
`1727^(@)C`
2600.

Why is the work done in isobaric process not zero always?

Answer» Heat absorbed goes partly to increase the internal energy and partily to do work.