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1501.

Rank the enthalpy of fusion , sublimation and vaporization for water :A. sublimation =vaporization =fusionB. vaporization> sublimation> fusionC. fusion> sublimation > vaporizationD. fusion > vaporization >sublimation

Answer» Correct Answer - d
1502.

Define (i) Molar enthalpy of fusion (ii) Molar enthalpy of vaporization?

Answer»

(i) The enthalpy change that accompanies melting of one mole of a solid substance at its melting point is called molar enthalpy of fusion. 

(ii) Amount of heat required to vaporize one mole of a liquid at constant temperature and understand pressure (1 bar) is called molar enthalpy of vapourisaton.

1503.

Define (a) Enthalpy of atomization (b) Lattice enthalpy.

Answer»

Enthalpy of atomization: It is defined as the enthalpy change accompyning the breaking of one mole of a substance completely into its atoms in the gas phase. 

H2(g) → 2H(g); ΔaH° = 435.0kJ mol-1 

Lattice enthalpy: Lattice of an comp; the enthalpy change which occure when one mole of an ionic compound dissociates into its gaseous ionic state. 

NaCl(s) → Na+ (g) + Cl (g); ΔH = +788 ks / mol

1504.

Find the pressure of saturated tungsten vapour at a temperature `T = 2000 K` if a tungsten filament is known to lose a mass `mu = 1.2-10^-13 g//(s. cm^2)` from a unit area per unit time when evaporating into high vacuum at this temperature.

Answer» Here we must assume that `mu` is also the rate at which the tungsten filament loses mass when in an atmosphere of its own vapour at this temperature and that `eta` (of the previous problem) `~~ 1`. Then
`p = mu sqrt((2 pi RT)/(M)) = 0.9 n Pa`
from the previous problem where `p` = pressure of the saturated vapour.
1505.

Which of the following has highest heat of combustion out of the following and why?(a) C2H6 (b) C2H4 (c) C2H2 (d) CH4

Answer»

C2H6(g) will have highest heat of combustion because it has highest molecular weight and second highest calorific value. 

ΔHc = calorific value of kJ/g × mol. wt. = 52 × 30 = 1560 kJ/mol.

1506.

What fraction of water supercooled down to the temperature `t = -20 ^@C` under standard pressure turns into ice when the system passes into the equilibrium state ? At what temperature of the supercooled water does it turn into ice completely ?

Answer» When a portion of supercool water turns into ice some heat is liberated, which should heat it upto ice point. Neglecting the variation of specific heat of water, the fraction of water turning inot ice is clearly
`f = (c|t|)/(q) = 0.25`
where `c` = specific heat of water and `q` ~~ latent heat of fusion of ice, clealy `f = 1` at `t = - 80^@C`.
1507.

Find the internal pressure `p_i` of a liquid if its density `rho` and specific latent heat of vaporization `q` are known. The heat `q` is assumed to be equal to the work performed against the forces of the internal pressure, and the liquid obeys the Van der Waals equation. Calculate `p_i` in water.

Answer» The internal pressure being `(a)/(V^2)`, the work done in condensation is
`int_(V_l)^(V_s) (a)/(V^2) dV = (a)/(V_l) - (a)/(V_g) ~~ (a)/(V_l)`
This by assumption is `Mq, M` being the molecular weight and `V_l, V_g` being the molae volumes of the liquid and gas.
Thus `p_l = (a)/(V_l^2) = (Mq)/(V_l) = rho q`
where `rho` is the density of the liquid . For water `p_i ~~ 3.3 xx 10^13 atm`.
1508.

What fraction of the volume of a vessel must liquid ether occupy at room temperature in order to pass into crictical state when critical temperature is reached ? Ether has `T_(cr) = 467 K, p_(cr) = 35.5 atm p M = 74 g//mol`.

Answer» The vessel is such that either vapour or liquid of mass `m` occupies it at critical point. Then its volume will be
`v_(Cr) = (m)/(M) V_(Mcr) = (3)/(8) (RT_(Cr))/(P_(Cr))(m)/(M)`
The corresponding volume in liquid phase at room temperature is
`V = (m)/(rho)`
Where `rho` = density of liquid ether at room temperature. Thus
`eta = (V)/(V_(Cr)) = (8 Mp_(Cr))/(3 RT_(Cr) rho) ~~ 0.254`
using the given data (and `rho = 720 gm` per litre).
1509.

Calculate the Van der Waals constant for carbon dioxide it its critical temperature `T_(cr) = 304 K` and critical pressure `p_(cr) = 73 atm`.

Answer» `P_(Cr)/(RT_(Cr)) = (a//27 b^2)/(8a//27 b) = (1)/(8 b)`
Thus `b = R (T_(Cr))/(8 p_(Cr)) = (0.082 xx 304)/(73.8) = 0.043 litre//mol`
and `((RT_(Cr))^2)/(P_(Cr)) = (64 a)/(27)` or `a = (27)/(64)(RT_(Cr))^2//P_(Cr) = 3.59 (atm.litre^2)/((mol)^2)`.
1510.

Calculate the temperature and density of carbon dioxide in critical state, assuming the gas to be a Van der Waals one.

Answer» `T_(Cr) = (8)/(27) (a)/(bR) = (8)/(27) xx (3.62)/(.043 xx .082) ~~ 304 K`
`rho_(Cr)= (M)/(3 b) = (44)/(3 xx 43) gm// c.c = 0.34 gm//c.c`.
1511.

Write the Van der Waals equation via the reduced parameters `pi, v` and `tau`, having taken the corresponding critical values for the units of pressure, volume, and temperature. Using the equation obtained, find how many times the gas temperature exceeds its critical temperature if the gas pressure is `12` times as high as critical pressure, and the volume of gas is equal to half the critical volume.

Answer» `(p + (a)/(V_M^2))(V_m - b) = RT`
or `(p+(a)/(V_m^2))/(P_(Cr))xx (V_m - b)/(V_(M_Cr)) = (8)/(3) (T)/(T_(Cr))`
or `(pi + (a)/(P_(Cr) V_m^2)) xx (v - (b)/(V_(MCr))) = (8)/(3) tau`,
where `pi = (p)/(p_(Cr)), v = (V_m)/(V_(MCr)), tau = (T)/(T_(Cr))`
or `(pi + (27 b^2)/(V_M^2))(v - (1)/(3)) = (8)/(3) tau`, or `(pi + (3)/(v^2))(v - (1)/(3)) = (8)/(3) tau`
when `pi = 12` and `v = (1)/(2), tau = (3)/(8) xx 24 xx (1)/(6) = (3)/(2)`.
1512.

Knowing the Van der Waals constant, find : (a) the maximum volume which water of mass `m = 1.00 kg` can occupy in liquid state , (b) the maximum pressure of the saturated water vapour.

Answer» (a) The ciritical Volume `V_(MCr)` is the maximum volume in the liquid phase and the minimum volume in the gaseous. Thus
`V_(max) = (1000)/(18) xx 3 xx .030 litre ~~ 5 litre`
(b) The critical pressure is the maximum possible pressure in the vapour phase in equilibrium with liquid phase. Thus
`p_(max) = (a)/(27 b^2) = (5.47)/(27 xx .03 xx .03) = 225 atmosphere`.
1513.

Differentiate between Cyclic and Non-cyclic Process.

Answer»

A cyclic process consists of a series of changes which return the system back to its initial state. 

In non-cyclic process the series of changes involved do not return the system back to its initial state. 

(1) In cyclic process change in internal energy is zero and temperature of system remains constant. 

(2) Heat supplied is equal to the work done by the system. 

(3) For cyclic process P–V graph is a closed curve and area enclosed by the closed path represents the work done. 

If the cycle is clockwise work done is positive and if the cycle is anticlockwise work done is negative. 

1514.

Define Irreversible Process. Give an examples.

Answer»

Irreversible process : Any process which is not reversible exactly is an irreversible process. 

Examples of irreversible processes are : 

(i) Sudden expansion or contraction 

(ii) Heat transfer between bodies

1515.

First law of thermodynamics does not forbid flow of heat from lower teperature to higher temperature. Comment.

Answer» First law of thermodynamics tells only about the conversion of mechanical energy into heat energy and vice-versa. As such it does not put any conditions as to why heat cannot flow from lower temperature to higher temperature.
1516.

Assertion : First law of thermodynamics does not forbid flow of heat from lower temperature to higher temperature. Reason : Heat supplied to a system always equal to the increase in its internal energy.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.

Answer» Correct Answer - B
First law of thermodynamics tells only about the conversion of mechanical energy into the heat energy and vice versa.
1517.

What are two essential features of Carnot’s ideal heat engine ?

Answer»

(i) Source and sink have infinite heat capacities. 

(ii) Each process of the engine’s cycle is fully reversible.

1518.

List the Properties of thermal radiation.

Answer»

Properties of thermal radiation :

(1) Thermal radiations are also called infra-red radiations. 

(2) Medium is not required for the propagation of radiations. 

(3) Every body whose temperature is above zero Kelvin emits thermal radiation. 

(4) Their speed is equal to that of light. 

(5) They follow laws of reflection refraction, interference diffraction and polarisation.

1519.

What is the specific heat of a gas in an isothermal process ?

Answer»

The specific heat of a gas in an isothermal process is Infinite. 

1520.

Why spark is produced when two substances are struck hard against each other ?

Answer»

Work is converted into heat.

1521.

Calculate the heat of neutralisation from the following data: `200mL` of `1M HCI` is mixed with `400mL` of `0.5M NaOH`. The temperature rise in calorimeter was found to be `4.4^(@)C`. Water equivalent of calorimeter is `12g` and specific heat is `1cal mL^(-1) degree^(-1)` for solution.

Answer» Given total solution `= 200 +400 = 600mL`
`200mL` of `1M HCI` neutralise `400mL` of `0.5M NaOH mEq = N xx V`
`:. mM`or `mEq` of acid and base `= 200` (For monovalent neture)
i.e., `200mEq` of `HCI` react with `200mEq` of `NaOH` to produc heat `=DeltaH`
`1000mEq of HCI + 1000 mEq of NaOH` will given heat `=5 xx DeltaH` i.e., heat of neutralisation.
Now, `DeltaH`, i.e., heat produced during neutralisation of `200mEq` of acid and base
`=` Heat taken up by calorimeter `+` solution
`= m_(1)s_(1)DeltaT +m_(2)s_(2)DeltaT`
`=12 xx 4.4 xx 600 xx1xx4.4 = 2682.8 cal`
Heat of neutralisation `=- 5xx 2692.8 cal=-13.464 kcal.`
1522.

A thermodynamic quantity is thatA. Which is used in thermochemistry.B. Which obeys all laws of thermodynamic.C. Quantity whose value depends only upon the state of the system.D. Quantity which is used in measuring thermal changes.

Answer» Quantity whose value depends only upon the state of the system.
1523.

Thermodynamic equilibrium involvesA. Chemical equilibriumB. Thermal equilibriumC. Mechanical equilibriumD. All the the three

Answer» Correct Answer - All the the three
1524.

Identify intensive property from the followingA. VolumeB. MassC. EnthalpyD. Temperature

Answer» Temperature is independent of the mass of substance.
1525.

At a particlular temperature `H^(o+) (aq) +OH^(Theta)(aq) rarr H_(2)O(l),DeltaH =- 57.1 kJ` The approximate heat evolved when `400mL` of `0.2M H_(2)SO_(4)` is mixed with `600mL` of `0.1M KOH` solution will beA. `3.426 kJ`B. `13.7kJ`C. `5.2 kJ`D. `55 kJ`

Answer» Number of equivalent of `H_(2)SO_(4)` taken
`= (0.2 xx2xx 400)/(1000) = 0.16`
Number of equivalent of `KOH` added `=(600xx0.1)/(1000) = 0.06`
Number of equivalents of acids and bases which neutralized each other `= 0.06`
`:.` Heat evolved `= 0.06 xx 57.1 kJ = 3.426 kJ`
1526.

Whenever an acid is neutralised by a base, the net reaction is `H^(o+)(aq)+overset(Theta)OH(aq)rarr H_(2)O(l),DeltaH =- 57.1 kJ` Calculated the heat evolved for the following experiments? a. `0.50 mol` of `HCI` solution is neutralised by `0.50mol` of `NaoH` solution. b. `0.50mol` of `HNO_(3)` solution is mixed with `0.30mol` of `KOH` solution. c. `100mL` of `0.2M HCI` is mixed with `100mL` fo `0.3M NaOH` solution. d. `400mL` of `0.2M H_(2)SO_(4)` is mixed with `600mL` of `0.1M KOH` solution.

Answer» According to the reaction
`H^(o+)(aq)+overset(Theta)OH(aq) rarr H_(2)O(l),DeltaH =- 57.1 kJ`
when `1mol` of `H^(o+)` ions and `1mol` of `overset(Theta)OH` ions are neutralised, `1mol` of water is formed and `57.1 kJ` of enegry is released.
a. `0.50 mol HCI -= 0.50 mol H^(o+)` ions
`0.50mol NaOH -= 0.50 mol overset(Theta)OH` ions
On mixing, `0.50 mol` of water is formed.
Heat evolved for the formation of `0.50mol` of water `=57.1 xx 0.5 = 28.55 kJ`
b. `0.50 mol HNO_(3)-= 0.50 mol H^(o+)` ions
`0.30 mol KOH -= 0.30 overset(Theta)OH` ions
i.e., `0.30 mol of H^(o+)` ions reacts with `0.30 mol` of `overset(Theta)OH` ions to form `0.30 mol` of water molecules
Heat evolved in the formation of `0.3 mol` of water `=57.1 xx 0.3 = 17.13kJ`
c. `100 mL` of `0.02 M HCI` will given
`((0.2)/(1000xx100)) = 0.02 mol of H^(o+)` ions and `100mL` of `0.3M NaOH` will given
`((0.3)/(1000)xx100) = 0.03 mol of overset(Theta)OH` ions i.e, `0.02mol` of `H^(o+)` ions reacts with `0.02` mole of `overset(Theta)OH` ions to form `0.02 mol` of water molecules.
Heat evolved in the formation of `0.02`mole of water `=0.02 xx 57.1 = 1.142 kJ`
d. `400mL of 0.2M H_(2)SO_(5)` will give
`((2x0.2)/(1000)xx400) = 0.16 mol of H^(o+)` ions and `600mL` of `0.1M KOH` will given
`((0.1)/(1000)xx600) = 0.60 mol of overset(Theta)OH` ions i.e, `0.06 mol of H^(o+)` ions reacts with `0.06 mol of overset(Theta)OH` ions to form `0.06 mol` of water molecules. Heat evolved in the formation of `0.06mol` of water `=0.06 xx 57.1 = 3.426 kJ`
1527.

The heat content of the products is more than that of the reactants in an…………….reaction.

Answer» Correct Answer - Endothermic
Endothermic, Since `H_(P) gt H_(R)` or `H_(R) = Delta H = +ve`
1528.

Work done in expansion of anideal gas from 4 litre against a constant external preesure 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 (in K) ?

Answer» Correct Answer - 299
1529.

Work done in expansion of an idela gas from `4L` to `6L` against a constant external pressure of `2.5 atm` was used to heat up `1mol` of water at `293K`. If specific heat of water is `4.184J g^(-1)K^(-1)`, what is the final temperature of water.

Answer» Since work is done against constant pressure and is done in one step, therefore, it is irreversible
`:. W =- P_(ex) DeltaV`
`DeltaV = (6 -4) = 2L`
`W =- 2.5 xx 2=- 5L atm`
`(1L atm = 101.262J) =- 5 xx 101.262J =- 506.31J` or
`= (5 xx 1.987)/(0.0821) cal = (5 xx 1.987 xx 4.184)/(0.0821) = - 506.31J`
Now this work is done in heating `1mol` of `H_(2)O`.
`w = n xx` Specific heat `xx DeltaT`
`506.31 = 1 xx 4.184 xx 18 xx DeltaT`
Specific heat `= 4.184J g^(-1) = 4.184 xx 18J mol^(-1)]`
`DeltaT = 6.723`
Final temperature =` T_(1) + DeltaT = 293 + 6.723 = 299.723K`
1530.

Which is (are) state properties? (P) enthalpy (Q) heat (C ) VolumeA. P onlyB. Q onlyC. P and R onlyD. Q and R only

Answer» Correct Answer - C
1531.

What will be `DeltaG` for the reaction at `25^(@)C` when partial pressures of reactants `H_(2), CO_(2),H_(2)O and CO` are 10, 20, 0.02 and 0.01 atm respectively. (Given : `G_(H_(2)O)^(@)=-"228.58 kJ, "G_(CO)^(@)=-"137,15 kJ, "G_(CO)^(@)=-"394.37 kJ"`.)A. `+5.61kJ`B. `-5.61 kJ`C. `7.09 kJ`D. `-8.13kJ`

Answer» Correct Answer - B
`H_(2(g))+CO_(2(g))hArr H_(2)O_((g))+CO_((g))`
`DeltaG^(@)" for reaction "=[G_(H_(2)O)^(@)+G_(CO)^(@)]-[G_(H_(2))^(@)+G_(CO_(2))^(@)]`
`[-228.58-137.15]-[0-394.37]=28.64kJ`
Also, `DeltaG=DeltaG^(@)+2.303 RT log Q`
`=28.64 +2.303xx8.314 xx10^(-3)xx298log.(0.02xx0.01)/(10xx20)`
`Delta=-5.61kJ`
Thus, reaction will occur in forward direction.
1532.

How much work is done by the gas when `1.00g` of sodium azide, `NaN_(3)` (`M=65.01 gxxmol^(-1)`), decomposes in a container of changeable volume (e.g. and airbag in a car) against a constant pressure of 1.00 atm at 298K? `2NaN_(3)(s)to2Na(s)+3N_(2)(g)`A. `+114 J`B. `+57.2J`C. `-114J`D. `-57.2J`

Answer» Correct Answer - D
1533.

How does an air conditioner differ from a refrigerator? Define the coefficient of performance of an air conditioner and express it in terms of heat current.

Answer»

The working of an air conditioner is exactly similar to that of a refrigerator, but the volume of the chamber/room cooled by an air conditioner is far greater than that in a refrigerator. The evaporator coils of an air conditioner are inside the room, and the condenser outside the room. A fan inside the air conditioner circulates cool air in the room.

The coefficient of performance, K, of an air conditioner is defined as K = \(|\frac{Q_C}W|\), where QC is the heat absorbed and W is the work done. The time rate of heat removed is the heat current, H = \(\frac{|Q_C|}t\), where t is the time in which heat |QC|, is removed.

\(\therefore\) K = \(|\frac{Q_C}W|\) = \(\frac{|Q_C|/t}{|W|/t}\) = \(\frac HP\).

where H = |QC|/t is the heat current and P( = |W|/t) is the time rate of doing work, i.e., power.

1534.

Explain the energy flow in a refrigerator and define the coefficient of performance of a refrigerator.

Answer»

In a refrigerator, QC is the heat absorbed by the working substance (refrigerant) at a lower temperature TC , W is the work done on the working substance, and QH is the heat rejected at a higher temperature TH . The absorption of heat is from the contents of the refrigerator and rejection of heat is to the atmosphere. Here, QC is positive and W and QH are negative. In one cycle, the total change in the internal energy of the working substance is zero.

∴ QH + QC = W ∴ QH = W – QC 

∴ -QH = QC – W 

Now, QH < 0, W < 0 and QC > 0 

∴ |QH| = |QC| + |W| 

The coefficient of performance (CoP), K, or quality factor, or Q value of a refrigerator is defined as

K = \(\frac{|Q_C|}{|W|}\) = \(\frac{|Q_C|}{|Q_C|-|Q_H|}\)

[Note: K does not have unit and dimensions or its dimensions are [M°L°T°.]

1535.

– ∆G is the net work done by the system except ……(a) electrical work (b) expansion work (c) chemical work (d) photo chemical work

Answer»

(b) expansion work

1536.

Write a note about mechanical work. Give its unit.

Answer»

1. Mechanical work is defined as force multiplied by the displacement through which the force acts.

2. Whenever a force (F) acts on an object and the object undergoes a displacement (x) in the direction of the force, then the mechanical work is said to be done.

3. Mathematically w = F . x 

w = F. x = N.m 

w = Joule

1537.

Define electrical work. Give its unit.

Answer»
  • When a charged body moves from one potential region to another electrical work is done.
  • If the electrical work done is QV. Where V is the potential difference and Q is the quantity of electricity.

w = QV 

w = Coulomb . Volts 

w = Joule

1538.

Write a note about the sign convention of heat.

Answer»
  • The symbol of heat is q. 
  • if the heat flows into the system from the surroundings, the energy of a system increases. Hence it is taken to be positive (+ q). 
  • if heat flows out of the system into the surroundings. energy of the system decreases. Hence it is taken to be negative (- q).
1539.

Explain about gravitational work. Give its unit.

Answer»

1. When an object is raised to a certain height against the gravitational field, gravitational work is done on the object.

2. For example, if an object of mass ‘m’ is raised through a height ‘h’ against acceleration due to gravity ‘g’ , then the gravitational work carried out is ‘mgh’.

w = m.g.h 

w = Kg .ms-2.m 

w = Kg m2 s-2 

w = Joule

1540.

What is meant by open system? Give example.

Answer»
  • A system which can exchange both matter and energy with its surroundings is called an open system.
  • Hot water contained in an open beaker is an example for open system.
  • In this system, both water vapour and heat is transferred to the surroundings through the imaginary boundary.
  • All living things are open systems because they continuously exchange matter and energy with the surroundings.
1541.

What is meant by work? Give its unit.

Answer»
  • Work is defined as the force (F) multiplied by the displacement (x).
  • – w = F. x
  • Minus (-) sign indicates the work done by the system
  • Unit of work: The SI unit of work is Joule (J).
1542.

Explain a closed system with an example.

Answer»
  • A system which can exchange only energy but not matter with its surroundings is called a closed system.
  • Here the boundary is sealed but not insulated. Hot water contained in a closed beaker is an example for a closed system.
  • In this system heat is transferred to the surroundings but no water vapour can escape from this system.
  • A gas contained in a cylinder fitted with a piston constitutes a closed system.
1543.

What is meant by isolated system? Give example.

Answer»
  • A system which can exchange neither matter nor energy with its surroundings is called an isolated system.
  • Here boundary is scaled and insulated.
  • Hot water contained in a thermos flask, is an example for an isolated system
1544.

Define 1. System 2. Surroundings

Answer»

1. System: A system is defined as any portion of matter under thermodynamic consideration. which is separated from the rest of the universe by real or imaginary boundaries. 

For example, water taken in a beaker, balloon filled with air, seed, plant, flower and bird.

2. Surroundings: Everything in the universe that is not the part of system and can interact with system is called as surroundings.

1545.

A gas can expand from 1oo ml to 250 ml under a constant pressure of 2 atm. The work done by the gas is …(a) -30.39 J (b) 25 J (c) 5 kJ (d) 16 J

Answer»

(a) -30.39 J

∆V = expansion in volume = 100 to 250 

∆V = V2 – V1 = 250 – 100 

∆V = 150 ml = 0.15 litre 

Work done = ? 

Pressure = 2 atm 

w = -P∆V 

= -2 x 0.15 litre x 101.3 JL-1 atm-1 

= -30.39 J.

1546.

An ideal gas expands in volume from 105 x 10-3 m3 to 1 x 10-2 m3 at 300K against a constant pressure at 105 x 105 Nm5 . The work done is …(a) -900 J(b) 900 kJ (c) 270 kJ(d) -900 kJ

Answer»

Answer: (a) -900 J

1547.

What are the limitations of the thermodynamics ?

Answer»
  • Thermodynamics suggests feasibility of reaction but fails to suggest rate of reaction. It is concerned only with the initial and the final states of the system. It is not concerned with the path by which the change occurs.
  • It does not reveal the mechanism of a process.
1548.

What are the scope of thermodynamics?

Answer»

The scope of thermodynamics:

  • To derive feasibility of a given process. 
  • It also helps in predicting how far a physical (or) chemical change can proceed. until the equilibrium conditions are established.
1549.

Identify the state quantity among the following (a) q (b) q – w (c) q + w (d) q / w

Answer»

Answer: (b) q – w

1550.

In general, for an exothermic reaction to be spontaneous (a) temperature should be high (b) temperature should be zero (c) temperature should be low (d) temperature has no effect

Answer»

(c) temperature should be low