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4051.

The molar enthalpy of vaporization of benzene at its boiling point `(353 K)` is `29.7 KJ//"mole"`. For how long minute) would a `11.4` Volt source need to supply a `0. 5A` current in order to vaporise `7.8 g` of the sample at its boiling point ?

Answer» Correct Answer - 9
4052.

It the enthalpy of vaporization of benzene is `308/ kJ mol^(-1)` at boiling point `(80^(@)C)`, calculate the entroy `(j mol^(-1) K^(-1))` in changing it from liquid to vapor.A. 308B. 0.873C. 0.308D. 873

Answer» Correct Answer - D
According to thermodynamics,
`Delta_(vap) S = (Delta_(vap) H)/(T_(b))`
`= (308 xx 10^(3) J mol^(-1))/((273 + 80) K)`
`= 873 K^(-1) mol^(-1)`
Note that `T_(b)` is the boiling point of benzene.
4053.

Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below `2Zn(s)+ O_(2)(g) to 2 ZnO_(s): DeltaH=-693.8 kJ mol^(-1)`A. The enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one omle of oxygen by 693.8 kJ.B. The enthalpy of two moles of ZnO is more than the total enthapy of two moles of Zn and one moles of oxygen by 693.8 kJ.C. `693.8 kJ mol^(-1)` energy is evolved in the reaction .D. `693.8 kJ mol^(-1)` energy is absorbed in the reaction .

Answer» Correct Answer - a.c
4054.

Which of the following are correct regarding standerd molar entroy at 298K ?A. `S_(O3)^(@)gtS_(O2)^(@)`B. `S_(C_(2)H_(6))^(@) gt S_(CH_(4))^(@) `C. `S_(H^(+(aq)))^(@)gt0`D. `S_(Na(s))^(@)gt0`

Answer» Correct Answer - a.b,d
4055.

`4Li(s)+O_(2)(g)to2Li_(2)O(s)` At `25^(@)C,DeltaH` for this reaction is -598.8 kilo joules per mole of Li should be reacted with excess `O_(2)(g)` in order to relaase 150 Kj?A. 0.874gB. 1..74gC. 3.15gD. 7.01g

Answer» Correct Answer - c
4056.

Find the adiabatic exponent `gamma` for a mixture consisting of `v_1` moles of a monatomic gas and `v_2` moles of gas of rigid diatomic molecules.

Answer» Obviously
`(1)/(R ) C_V = (3)/(2) gamma_1 + (5)/(2) gamma_2`
(Since a monoatomic gas has `C_V = (3)/(2) R` and a diotomic gas has `C_V = (5)/(2) R`. [The diatomic molecule is rigid so no vibration])
`(1)/(R) C_p = (3)/(2) gamma_1 + (5)/(2) gamma_2 + gamma_1 + gamma_2`
Hence `gamma = (C_p)/(C_V) = (5 gamma_1 + 7 gamma_2)/(3 gamma_1 + 5 gamma_2)`.
4057.

Find the change in internal energy (in joule) of a gas when it absorbs `40 calories` of heat and performs work equal to `16 J`.

Answer» Correct Answer - 8
Here, `dU=? dQ=40cal.= 40xx4.2= 168 J`
`dW= 160J`
From `dQ= dU+dW`
`dU= dQ-dW= 168-160= 8 J`
4058.

Efficiency of the engine in the above questions isA. `30%`B. `10%`C. `20%`D. `80%`

Answer» Correct Answer - C
`eta= 1-(T_(2))/(T_(1))= 1-(400)/(500)= 1/5= 20%`
4059.

The pressure -volume work for an ideal gas canbe calculated by using the expression `w= -int_(V_(l))^(V_(f))p_(ex)dV`.The work can also be calculated from the `pV` - plotby using the area under the curve within the specified limit. When anideal gas is compressed (a)reversibly or (b) irreversibly from volume`V_(i)` to `V_(f)`, choose the correct option.A. w ( reversible)=w ( irreversible)B. w( reversible) `lt` w ( irreversible)C. w ( reversible) `gt` w ( irreversible)D. w(reversible)`=`w ( irreversible) `+p_(ex) DeltaV`

Answer» Correct Answer - b
Look at Fig.page. Area under the curvein Figure. (a) Which is for irreversible compression is more than the area under tha curve in Figure. Which is for reversible compression. Hence, for compression `w( rev) lt w ( "irrev")`.
4060.

Two similar reactions are given below `:` `H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(g), DeltaH = = DeltaH_(1)` `H_(2)(g)+(1)/(2) O_(2)(g) rarr H_(2)O(l), DeltaH= DeltaH_(2)` In terms of magnitude of `DeltaH`,A. `DeltaH_(1) gt DeltaH_(2)`B. `DeltaH_(1) lt DeltaH_(2)`C. `DeltaH_(1) = DeltaH _(2)`D. None of these is correct

Answer» Correct Answer - b
`DeltaH_(2) gt DeltaH_(1)` because when `H_(2)O(g)` consdenseto`H_(2)O(l)`, heat of condensation is evolved.Hence, `DeltaH_(1) ltDeltaH_(2)`.
4061.

`50 L` of a certain liquid is confined in a piston system at the external pressure `100 atm`. This pressure is suddenly released and liquid is expanded against the constant atmospheric pressure, volume of the liquid increases by `1 L` and the final pressure on the liquid is `10 atm`. Find the workdone.A. `1 L. atm`B. `5 L.atm`C. `5000 L. atm`D. `50 L.atm`

Answer» Correct Answer - A
4062.

In an adiabatic process ,no transfer of heat takes place between system and surroundings.Choose the correct option for free expansion of and ideal gas under adiabatic condition from the following `:`A. `q=0 , Delta cancel(=) 0,w=0`B. `q cancel(=) 0, DeltaT =0,w=0`C. `q =0,DeltaT=0, w=0`D. `q=0,DeltaT=0,w cancel (=) 0`

Answer» Correct Answer - c
For free expansion , `P_(ext) = 0,w = - P _(ext) DeltaV=0`
For adiabatic process , `q=0`
`:. DeltaU =q+w= 0`. This means that internal energy remains constant which is so at constant temperature. Hence `DeltaT =0`
4063.

One mole of an ideal monoatomic gas at temperature T and volume 1 L expands to 2 L against a constant external pressure of one atm under adibatic conditions, then final temperature of gas will be :A. `T+(2)/(3xx0.0821)`B. `T-(2)/(3xx0.0821)`C. `(T)/(2^(5//3-1)`D. `(T)/(2^(5//3+1)`

Answer» Correct Answer - B
4064.

Enthalpy change of `CH_(4)+(1)/(2) O_(2) rarr CH_(3) OH` is negative . If enthalpy of combustion of`CH_(4)` and`CH_(3)OH` are x and y respectively, then which relation is correct ?A. `x gt y`B. `x lty`C. `x=y`D. `x gey`

Answer» Correct Answer - B
(i) `CH_(4) + 2O_(2)rarr CO_(2)+ 2H_(2)O , DeltaH= x`
(ii) `CH_(3) OH + (3)/(2) O_(2) rarr CO_(2)+ 2H_(2)O, DeltaH =y`
(i) - (ii) gives `DeltaH ` for the given reaction, i.e., `DeltaH = x-y `.As it is -ve , `x lty`
4065.

A real gas is subjected to an adibatic process causing in a change of state from (3 bar, 50 L, 500 K) to (5 bar, 40 L, 600 K) against a constant pressure of 4 bar. The magnitude of enthalpy change for the process is :A. 4000 JB. 5000 JC. 9000 JD. 1000 J

Answer» Correct Answer - C
4066.

Which quantity out of `Delta_(r)G` and`Delta_(r)G^(c-)` will be zero at equilibrium ?

Answer» `Delta_(r)G= Delta_(r)G^(@) + RT ln K`. When reaction attains equilibrium `Delta_(r) G=0` so that`Delta_(r)G^(@)=- RT` ln K . Thus `Delta_(r)G^(@) =0` only when `K+1` . For all other values of `K, Delta_(r)G^(@) ` will be non-zero.
4067.

Which quantity out of `Delta_(r)G " and " Delta_(r) G^(Θ)` will be zero at equilibrium ?

Answer» Gibbs energy for a reaction in which all reactants and products are in standard state. `Delta_(r)G^(Θ)` is related to the equilibrium constant of the reaction as follows
`Delta_(r)G = Delta_(r)G^(Θ) + RT ln K`
At equilibrium, `0 = Delta_(r) G^(@) + RT ln K " " ( :. Delta_(r) G = 0)`
or `Delta_(r) G^(Θ) = - RT ln K`
`Delta_(r) G^(Θ) = 0` When `K = 1`
For all other values of `K, Delta_(r)G^(Θ)` will be non-zero.
4068.

An ideal gas is subjected to two different reversible expansion process, `PV^(3)=K_(1)` and `PV^(5//2)=K_(2)` from same initial state to same final volume. Which of the follwing statements is correct regarding the two process :A. The work done by the gas in second process is more as compared to first process.B. The magnitude of work done can be compared only by knowing value of `K_(1) "and" K_(2)`.C. Heat capacity of the gas for the two process is sameD. Final state for the two process will be same

Answer» Correct Answer - A
4069.

Consider the following nine phase transformations. `H_(2)O(s,1atm,273 K)rarr H_(2)O(l,1 atm, 3273 K)` `H_(2)O(s,1atm,300K)rarr H_(2)O(l,1 atm, 300 K`) `H_(2)O(s,1 atm,200 K)rarrH_(2)O(l,1 atm, 200 K`) `H_(2)O (s,0.5 atm, 273 K)rarrH_(2)O`(l,0.5 atm 273 K) `H_(2)O(s,2 atm, 273 K)rarrH_(2)O (l,2 atm,273 K)` `H_(2)O(l, 1 atm, 400 K)rarrC_(6)H_(6)`(g),1atm, 353K) `C_(6)H_(6)(l,1 atm, 400 K)rarrC_(6)H_(6)(g,1atm,400 K`) `C_(6)H_(6)(l,1atm,300 K)rarrC_(6)H_(6)`(g,1atm,300K) `C_(6)H_(6)(l,2atm,327 K)rarrC_(6)H_(6)`(g,2atm,323 K) Given : `T_(nbp)`of `C_(6)H_(6)(l)=353 K` Now a four digit number abcd is fined as follows: Digit a It is number of phase transformations for which `DeltaS_(total)=0` Digit b It is number of phase transformations for which `DeltaS_(total)lt0` Digit c It is number of phase transfprmation for which `DeltaS_(total)gt0` Digit d It is number of phase transformation for which `DeltaHgt0` Determine the number abcd [Note:If a =1, b=2,c=3 and d=4, then answeris 1234]

Answer» Correct Answer - 2439
4070.

Discuss the factors which influence the value of enthalpy change in chemical reactions.

Answer»

The enthalpy change depends upon the following factors:

(i) Chemical composition of the reactants:- The amount of heat released or absorbed depends upon the strength of the chemical bonds of the products relative to those of the reactants.

(ii) Amount of the reactants:- The value of heat change in a reaction depends upon the amounts of the reactants. The value of ΔH is expressed in such away that it indicates the amount of material as well as quantity of heat.

(iii) Conditions of a reaction:- The amount of heat change in a reaction also depends upon the physical state of the reactants as well as products. Therefore, it is essential to indicate the physical state of every reactant and product of a reaction.

4071.

Explain enthalpy changes during phase transformations.

Answer»

Phase changes involve energy changes. Ice, for example, requires heat for melting. Normally this melting takes place at constant pressure (atmospheric pressure) and during phase change, temperature remains constant (at 273 K).

H2O (s) → H2O (l);

ΔfusHΘ = 6.00 kJ mol-1

Here Δfus HΘ is enthalpy of fusion in standard state. If water freezes, then process is reversed and an equal amount of heat is given off to the surroundings, that is

ΔfreezeHΘ = -Δfus HΘ

This enthalpy is called enthalpy of fusion or molar enthalpy of fusion.

Δfus HΘ = HΘliquid - HΘsolid

Melting of a solid is endothermic, so all enthalpies of fusion are positive.

Amount of heat required to vaporize one mole of a liquid at constant pressure and temperature is called its enthalpy of vaporization or molar enthalpy of vaporization. Here constant temperature is its boiling point, TP.

In case of water vaporization

H2O (l) → H2O (g); ΔvapHΘ = + 40.79 kJ mol-1

Sublimation is a direct conversion of a solid into its vapour. Solid CO2 or 'dry ice' sublimes at 195 K with ΔsubHΘ = 25.2 kJ mol-1, napthalene sublimes slowly and for this ΔsubHΘ = 73.0 kJ mol-1.

Because enthalpy is a state property, the enthalpy of sublimation can be expressed as

ΔsubHΘ = ΔfusHΘ + ΔfusHΘ.

4072.

Explain extensive properties.

Answer»

Extensive properties:- The property whose value depends upon the quantity of matter contained in any system is called extensive property. Thus, an extensive property depends upon the size of the system. The properties such as mass, volume, area, energy, heat capacity etc, are extensive properties.

4073.

Explain intensive properties.

Answer»

Intensive properties:- The property whose value depends upon the nature of the substance, and in independent of its amount in the system is called an intensive property. Thus, an intensive property does not depend upon the size of the system. The properties such as density, temperature, pressure, concentration, viscosity, refractive index, surface tension, etc. are some intensive properties.

4074.

Write the meaning of each of the terms in the equation ΔE = q + W.

Answer»

The equation ΔE = q + W is a mathematical statement of the first law of thermodynamics. In this equation ΔE is the increase in the internal energy of the system, q is the heat absorbed by the system and W is the work done on the system. Delta (Δ) is not used before q and W, because these depend on path and there is no initial and final heat or work on the system. The concept of heat or work comes into picture only during the change. If there is no change, there is no heat or work. But the system has energy.

4075.

Two moles of an ideal monoatomic gas at `27^(@)C`occupies a volume of V . If the gas is expanded adiabatically to the volume , 2V then the work done by the gas will be `[lambda=5//3,R=8.31j//molK]`A. `-2767.23j`B. `2767.23j`C. `2500j`D. `-2500j`

Answer» Correct Answer - B
4076.

It has been found that 221.4 J is needed to heat 30 g of ethanol from 15°C to 18°C. Calculate molar heat capacity of ethanol.

Answer»

Molar heat capacity, cm = Specific heat capacity x molar mass

therefore, cm (ethanol) = 2.46 Jg-1°C-1 x 46 g mol-1

= 113.2 J mol-1°C-1

The molar heat capacity of ethanol is 113.2 J mol-1°C-1

(or 113.2 J mol-1 K-1).

4077.

1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in figure. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.

Answer» The given diagram represent that the prcess is carried out in infinite steps, hence it is isothermal reversible expansion of the ideal gas from pressure 2.0 atm to 1.0 atm 298 K.
`W = - 2.303 nRT log. (p_(1))/(p_(2))`
`W = - 2.303 xx 1 mol xx 8.314 JK^(-1) xx 298 K log 2 " " ( :. (p_(1))/(p_(2)) = (2)/(1))`
`W = - 2.303 xx 1 xx 8.314 xx 298 xx 0.3010 J`
`W = - 1717.46 J`
4078.

Two rigid adiabatic vessel A and B which initially ,contain two gases at different temperature are connected by pipe line with value of negligible volume .The vessel A contain 2 moles Ne gas `(C_(p.m)=(5)/(2)R)` at 300 K, vessel B contain 3 moles of `SO_(2) gas (C_(p.m)=4 R)` at 400 K . The volume of the A and B vessel is 4 and 6 litre repectively . The final total pressue (in atm ) when valve is opened and 12 kcal heat supplied throught it to vessels . [Use :` R =2 cal //"mole" K "and" R=0.08 L . atm//"mole"` K as per desire ] A. `3.5 atm`B. `7 atm `C. `35 atm`D. `70 atm`

Answer» Correct Answer - C
4079.

A system contains 1 mole of a monoatomic ideal gas . Now 1 mole of a diatomic non-reacting ideal gas is added into the system at constant volume and temperature. Due to addition diatomic gas ,Choose the incorrect statement , regarding the new system :A. Enthalpy of system will increaseB. Adiabatic coefficient `(lambda=C_(p)//C_(V))` of the system will decreaseC. Internal energy of the system remains constant .D. Pressure energy system will increase .

Answer» Correct Answer - C
4080.

In the statement ΔE = q + W. Why is ΔE regarded as a state function while q and W are individually not state functions?

Answer»

ΔE is regarded as state function because it depends upon initial and final states of the system are not on the path e.g., CO2 obtained from various methods has some value of ΔE.

Heat (g) and work (w) are not the state functions because they are path dependent.

4081.

"Neither the enthalpy change nor the entropy change alone can be used to explain the spontaneity of the reaction". Explain

Answer»

Reaction are feasible in the direction of decrease in energy. Thus the reactions which proceed with a decrease in enthalpy (ΔHnegative) should be spontaneous. This is found to be so in many cases. However, there are some spontaneous reactions should process with an increase in entropy. Once again there are many spontaneous reactions which proceed with a decrease in entropy (ΔSnegative).

Thus, it appears which explains the spontaneity of any reaction is called free energy change ΔG. This is related to ΔH and ΔS by the reaction.

ΔG = ΔH - TΔS

The reactions which proceed with negative ΔG are spontaneous.

4082.

Enthalpy diagram for a particular reaction is given in figure. Is it possible to decide spontaneity of a reaction from given diagram. Explain.

Answer» No, enthalpy is one of the contributing factors in deciding spontaneity but it is not the only factor. Another contributory factor, entropy factor has also to be taken into consideration
4083.

State a chamical reaction in which `DeltaH` and `DeltaU` are equal .

Answer» `H_(2)(g) +I_(2) (g) rarr 2HI(g)`
4084.

A Carnot engine works between `120^(@)C` and `30^(@)C`. Calculate the efficiency. If the power produced by the engine is `400W`, calculate the heat abosorbed from the source.

Answer» Efficiency, `eta =(T_(2)-T_(1))/(T_(2))`
Here `T_(2) = 273 +120 = 393K`
`T_(1) = 273 +30 = 303 K`
`:. eta = (393-303)/(393) = 0.229 = 22.9%`
Again, `eta = (Q_(2)-Q_(1))/(Q_(2)) = (W)/(Q_(2))`
Thus, heat absorbed form the source.
`Q_(2) = (W)/(eta) = (400)/(0.229) = 1747W`
4085.

The volume of `1kg` of oxygen gas at `NPT` is `0.7m^(3)` . Specific heat of oxygen at constant volume is `653 Jkg^(-1)k^(-1)` . The specific heat of oxygen at constant pressure in `Jkg^(-1)k^(-1)` is (atmospheric pressure `10^(5)`Pa)A. `713`B. `813`C. `913`D. `1013`

Answer» Correct Answer - C
`PdW=RdT=(C_(P)-C_(V))dT`
4086.

A resistor of resistance 200 Ω carries a current of 2 A for 10 minutes. Assuming that almost all the heat produced in the resistor is transferred to water (mass = 5 kg, specific heat capacity = 1 kcal/kg), and the work done by the water against the external pressure during the expansion of water can be ignored, find the rise in the temperature of the water. (J = 4186 J/cal)

Answer»

Data : I = 2 A, R = 200 Ω, t = 10 min = 10 × 60

s = 600 s, M = 5 kg, S = (1 kcal/kg) (4186 J/kcal) 

= 4186 J/kg

Q = ∆ U + W = MS ∆T + W ≃ MS ∆T ignoring W..

Also, Q = I2 Rt 

∴ I2 RT = MS∆T

∴ The rise in the temperature of water = ∆T = \(\frac{I^2Rt}{MS}\)

\(\frac{(2)^2(200)(600)}{(5)(4186)}\)°C = 22.93°C

4087.

In a thermodynamics process, 300 joule of heat is supplied to a gas and 200 joule of work is done by the gas. What is the change in internal energy of the system?

Answer» There will be a net increase in internal energy of the system by `300-200= 100 jo u l e`
4088.

A gas expands with temperature according to the relation `V = kT^(2//3)`. What is the work done when the temperature changes by `30^(@)C`?A. `10R`B. `20R`C. `30R`D. `40R`

Answer» Correct Answer - B
4089.

A gas expands with temperature according to the relation `V = kT^(2//3)`. What is the work done when the temperature changes by `30^(@)C`?A. 10 RB. 20 RC. 30 RD. 40 R

Answer» Correct Answer - B
4090.

A gas expands with temperature according to the relation `V=KT^(2/3)`.Work done when the temperature changes by 60K is.A. 10RB. 30RC. 40RD. 20R

Answer» Correct Answer - C
Here
dW =PdV=`(RT)/(V)dV`
Given `V=KT^(2//3)` therefore `dV=K(2)/(3)(dT)/(T)`
From (i) `W=overset(T_(2))underset(T_(1))int RT (dV)/(V)=overset(T_(2))underset(T_(1))int RT(2)/(3) (dT)/(T)`
`W=(2)/(3)R(T_(2)-T_(1))=(2)/(3)Rxx60 =40R`
4091.

50g of oxygen at NTP is compressed adiabatically to a pressure of 5 atmosphere. The work done on the gas, if `gamma =1.4 and R = 8.31 J mol^(-1)K^(-1)` isA. `-5173 J`B. `1131 J`C. `-1364 J`D. `5673 J`

Answer» Correct Answer - A
Here, `T_(1)=273 K, P_(1)=1 atm, P_(2)=5atm`
no of moles of oxygen `=(50)/(32)`
`rArr T+2=T_(1)(P_(1)/(P_(2)))^(gamma-1)/(gamma) = 3273((5)/(1))^(14-1)/(1.4)`
`=273xx(5)^(2//7)=273xx1.58=432.37`
`W=(nR)/(gamma-1)[(t_(1)-t_(2))]=(50)/(32)xx(8.31)/(1.4-1)(273-432.37)`
=-5173 J
4092.

A perfect gas contained in a cylinder is kept in vacuum. If the cylinder suddenly bursts, then the temperature of the gasA. Remains constantB. Become zeroC. IncreaseD. Decrease

Answer» Correct Answer - A
4093.

What is the network done in a reversible process?

Answer»

Zero. Because, the heat absorbed and change in internal energy are zero.

4094.

State the relation between molar specific heat at constant pressure and that at constant volume for an ideal gas.

Answer»

Relation between molar specific heat at constant pressure and that at constant volume for an ideal gas is,

Cp – Cv = R,

where Cp is the molar specific heat at constant pressure, Cv is the molar specific heat at constant volume and R is the Universal gas constant.

4095.

What is the network done by an ideal gas in an Isochoric process?

Answer»

Zero. Because as the heat absorbed by the gas is entirely used to raise its internal energy.

4096.

Define thermal equilibrium.

Answer»

A system is said to be in thermal equilibrium if the macroscopic variables that characterise the system to not change whit time.

4097.

The temperature of the system decreases in the process ofA. Free expansionB. Isothermal expansionC. Isothermal compressionD. Adiabatic expansion

Answer» Correct Answer - D
4098.

State zeroth law of thermodynamics.

Answer»

Zeroth law of thermodynamics states that – two systems in thermal equilibrium with a third system separately by are in thermal equilibrium with each other.

4099.

Two bodies A and B are said to be in thermal equilibrium with each other if they have sameA. pressureB. volumeC. temperatureD. Area

Answer» Correct Answer - C
4100.

What is an Isotherm?

Answer»

An isotherm is a P – V curve for a fixed temperature.