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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.
| 3101. |
In a refrigerator, heat from inside at `270K` is transferred to a room at `300K`. How many calories of heat will be delivered to the room for each joule of electrical energy consumed ideally? |
| Answer» Correct Answer - `2.38 cal`. | |
| 3102. |
For a reversible reaction `A hArr B`. Find `(log_(10)K)/(10)` at `2727^(@)`C temperature Given `Delta_(r)H^(0)` = - 54.07 kJ `mol^(-1)` `Delta_(r)S^(0)` = 10 `JK^(-1)` R = 8.314 `JK^(-1) mol^(-1)` |
| Answer» Correct Answer - 1 | |
| 3103. |
The rapid depletion of fossil fuels has inspired extensive research in the area of alternative and renewale energy sources. Of these, hydroden is the most Contemplated fuel of the future . Howevercost effective production and hazard free storage are major issuses is using `H_(2)` (Note : use the data in table-1 given at the end of partA, whenever necessary.) A cylinder contains hydrogen at a pressure of 80Moa at `25^(@)C` , Assuming ideal behaviour , Calculate the density of hydrogen in the cylinder in `Kg//m^(3)` . |
| Answer» Correct Answer - `p=64.58 kg m_(-3)` | |
| 3104. |
The rapid depletion of fossil fuels has inspired extensive research in the area of alternative and renewale energy sources. Of these, hydroden is the most Contemplated fuel of the future . Howevercost effective production and hazard free storage are major issuses is using `H_(2)` (Note : use the data in table-1 given at the end of partA, whenever necessary.) A cylinder contains hydrogen at a pressure of 80Moa at `25^(@)C` , Assuming ideal behaviour , Calculate the density of hydrogen in the cylinder in `Kg//m^(3)` . |
| Answer» Correct Answer - `p=64.58 kg m_(-3)` | |
| 3105. |
The rapid depletion of fossil fuels has inspired extensive research in the area of alternative and renewale energy sources. Of these, hydroden is the most Contemplated fuel of the future . Howevercost effective production and hazard free storage are major issuses is using `H_(2)` (Note : use the data in table-1 given at the end of partA, whenever necessary.) If the miximum theoretical work (calculate in 4.3 (i))0 is used to run an electric motor of 1 watt, under standard potential conditons. (i) for how many months the motor will be run? (ii) What is the value of the current produced by this motor? (Assume 30 days in all months.) Table-1: `{:(,H_(2)(g),O_(2)(g),H_(2)O(I)),(S_(298)^(@) K//"JMolK"^(-1),131,205,70),(,H_(2)O(l),CO_(2)(g),),(DeltaH_(f)^(@)//"KMol"^(-1), -266 KJ "mol"^(-1),-394 KJ "mol"^(-1),):}` |
| Answer» Correct Answer - `(i) 46.3` months (ii) `I=0.813 A` | |
| 3106. |
When 12.0 g of carbon reacted with limited quantity of oxygen, 57.5 kcal of heat was produced, calculate the number of moles of CO produced `(Delta_(f)H(CO_(2))=-94.5 cal, Delta_(f)H(CO)= -21.41kcal.`A. 0.5 molB. 0.46 molC. 0.64 molD. 0.74 mol |
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Answer» Correct Answer - A `12g C=1` mole, suppose CO produced `=x` moles. Then `CO_(2)` produced `=(1-x)` mole `:. X(-21.41)+(1-x)(-94.05)= -57.5` This on solving gives `x=0.54` mole. |
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| 3107. |
The densities of graphite and diamond at `298K` are `2.25` and `3.31gcm^(-3)` , respectively. If the standard free energy difference `(DeltaG^(0))` is equal to `1895Jmol^(-1)` , the pressure at which graphite will be transformed into diamond at `298K` isA. `9.92 xx 10^(8) Pa`B. `9.92 xx 10^(7) Pa`C. `9.92 xx 10^(6) Pa`D. `9.92 xx 10^(5) Pa` |
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Answer» Correct Answer - D Since we are dealing with standard conditions, the pressure must be 1 bar `(10(5) Pa)` and the temperature must be specified, usually `298 K`. |
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| 3108. |
Two moles of an ideal gas initially at `27^(@)C` and one atmospheric pressure are compressed isothermally and reversibly till the final pressure of the gas is 10 atm. Calculate q, w AND `Delta U` for the process. |
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Answer» Here, `n = 2` moles , `T = 27^(@) C=300 K, P _(1)= 1 atm,P _(2) =10 atm` `w= 2.303n RT log. (P_(2))/(P_(1)) = 2.303 xx 2 mol xx 8.314 JK^(-1) mol^(-1) xx 300 K xx log. (10)/(1) = 11488 J ` For isothermal compressionof ideal gas, `Delta U = 0` Further, `Delta U = q+w :. q = -w = -11488J` |
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| 3109. |
A sample of hydrogen of mass `6g` is allowed to expand isothermally at `27^(@)C` till its volume is doubled. (a) How may moles of hydrogen are there? (b) What is final temperature of `H_(2)`? (c ) Caculate work done during expansion. |
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Answer» (a) Total mass of hydrogen `=6g`. As molar mass of hyderogen is 2 gram, therefore, number of moles of hydrogen `=6//2=3`. (b) As expansion is isothermal, temperature remains constant at `27^(@)C`. (c ) `W=2.3026 nRT"log"_(10)(V_(2))/(V_(1))` `=2.3026xx3xx8.31xx(27+273)log_(10)2` `=2.3026xx3xx8.31xx300xx0.3010` `=5183.5J` |
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| 3110. |
`2A(s)hArrB(g)+2C(g)+3D(g)` Total pressure developed in closed container by decomposition of A at equibrium is 12 atm at `727^(@)C`. Calculate `DeltaG^(@)` (in kcal), of the reaction at `727^(@)C`. (`R=2 cal//"mole"-K,In 2=0.7, In =1.1`) Round off your answer to integer (without sign). |
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Answer» Correct Answer - 18 |
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| 3111. |
A person of mass 60 kg wants to lose 5kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt of expending 7000 k cal, how many times must he go up and down to reduce his weight by 5 kg ? |
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Answer» Given, `" "` height of the stairs = h= 10 m Energy produced by burning 1 kg of fat = 7000 kcal `therefore` Energy produced by burning 5 kg of fat = `5xx7000` = 35000 kcal `" "=35xx10^(6)` cal Energy utilised in going up and down one time ` " "=mgh+(1)/(2)mgh= (3)/(2)mgh` `" "(3)/(2)xx60xx10xx10` `" "=9000 J=(9000)/(4.2)=(3000)/(1.4) ` cal `therefore` Number of times, the person has to go up and down the stairs `" "=(35xx10^(6))/((3000//1.4))=(3.5xx1.4xx10^(6))/(3000)` `" "16.3xx10^(3)` times |
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| 3112. |
Graph for one mole gas is given below the process which occurs in going from, `B rarr C` isA. IsothermalB. AdiabaticC. IsobaricD. Isochoric |
| Answer» Correct Answer - C | |
| 3113. |
Work is the mode of transference of energy. If the system involves gaseous substance and there is difference of pressure between system and surroundings, such a work is referred to as pressure - volume work `(W_(PV)= -P_(ext)DeltaV)`. It has been observed that reversible work done by the system is the maximum obtainable work. `w_(rev) gt w_(ir r)` The works of isothermal and adiabatic processes are different from each other. for isothermal reversible proces, `W_("isothermal reversible")=2.303 nRT log_(10) (V_(2)/V_(1))` `W_("adiabatic reversible")=C_(V) (T_(1)-T_(2))` Calculate work done when 1 mole of an ideal gas is expanded reversibly from 20 L to 40 L at a constant temperature of 300 K.A. `w_(1) gt w_(2) gt w_(3_ gt w_(4)`B. `w_(3) gt w_(2) gt w_(1) gt w_(4)`C. `w_(3) gt w_(2) gt w_(4) gt w_(1)`D. `w_(3) gt w_(1) gt w_(2) gt w_(4)` |
| Answer» Correct Answer - D | |
| 3114. |
`5` moles of a liqiud `L` are converted into its vapour at its boiling point `( 273^(@)C)` and at a pressure of `1 "atm"`. If the value of latent heat of vapourisation of liquid `L` is `273 L atm//"mole"`, then which of the following statements is`//`are correct. Assume volume of liquid to be negligible and valpour of the liquid to behave ideally.A. Work done by the system in the above process is `224 L` atm.B. The enthalpy change `(Delta H)` for the above process is `1365 L` atm (with respect to magnitude only)C. The internal energy of the system increases in the above process.D. The value of `Delta U` for the above process is `1589 L` atm. |
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Answer» Correct Answer - A::B::C `1xxV_(t)=5xxRxx546=224 L` `W= -P_(ext)(Delta V)= -1 "atm"(224 L)= -224 L "atm"` `:.` work done by system `=224 L "atm"` Enthalpy change `(DeltaE)=q=273.xx5 =1365 K atm` `Delta U= q+W=1365-224=1141 L "atm"` |
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| 3115. |
Statement -1: For every chmical reaction at equilibrium , standard Gidds energy of reaction is zero Statement-2: At constant temperature and pressure , chemical reactions are spontaneious in the direction of decreasing gibbs energy.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is falseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - D At equilibrium `DeltaG` (Gibbs energy) =0 but `DeltaG^(@)` (standard Gibbs energy) `ne 0` As `DeltaG` (Gibbs energy) is more negative reaction will be more spontaneous. |
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| 3116. |
For the process `H_(2)O(l) (1 "bar", 373 K) rarr H_(2)O(g) (1"bar", 373 K)` the correct set of thermodynamic parameters isA. `DeltaG=0, DeltaS =+ve`B. `DeltaG=0, DeltaS =-ve`C. `DeltaG=+ve, DeltaS =0`D. `DeltaG=-ve, DeltaS =+ve` |
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Answer» Correct Answer - A `H_(2)O(l,1"bar", 373 K) rarr H_(2)O(g. 1"bar",373K)` `DeltaS gt 0," "DeltaH gt 0, " " DeltaG=0` |
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| 3117. |
Match the thermodynamic processes given under column I with the expressions given under column II. |
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Answer» Correct Answer - `A to r,t; B to p,q,s, ;C to p,q,s;D to p,q,s,t;` |
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| 3118. |
For the process `H_(2)O(l) (1 "bar", 373 K) rarr H_(2)O(g) (1"bar", 373 K)` the correct set of thermodynamic parameters isA. `DeltaG = 0, DeltaS=+`B. `DeltaG = 0, DeltaS=-`C. `DeltaG =+" " DeltaS=0`D. `DeltaG =-" " DeltaS=+` |
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Answer» Correct Answer - a |
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| 3119. |
The standard enthalpy of formation (`DeltaH_(f)^(@))` at `398` K fir methane , `CH_(4(g))` is `748 KJ mol^(-1)`. The additional information required to determine the average energy for `C-H` bond formation would be :A. the dissociation energy of `H_(2)` and enthalpy fo sublimation of carbonB. latent heat of vapourisation of methaneC. the first four ionization energies of carbon and electron gain enthalpy of hydrogenD. the dissociation energy of hydrogen molecule , `H_(2)` |
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Answer» Correct Answer - A `C=2H_(2) rarr CH_(4), DeltaH^(@) = -74.8 KJ mol^(-1)` In order to calculate average energy for C-H bond formation we should know the following data. `C("graphite") rarr C(g), DeltaH, `=enthalpy of sublimation of carbon `H_(2)(g) rarr 2H(g) , DeltaH^(@)` bond dissociation energy of `H_(2)` |
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| 3120. |
A rigid and insulated tank of `3m^(3)` volume is divided into two compartments. One compartment of volume of `2m^(3)` contains an ideal gas at `0.8314` Mpa and 400 K while the second compartment of volume of `1m^(3)` contains the same gas at `8.314` Mpa and 500 K. If the partition between the two compartments is rptured, the final temperature of the gas is :A. 420 KB. 450 KC. 480 KD. None of these |
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Answer» Correct Answer - C Mole of the gas in the first compartment `n_(1)=(P_(1)V_(1))/(RT_(1))=-(0.8314xx10^(6)xx2)/(8.314xx400)=500` Similarly, `n_(2)=2000` The tank is rigid and insulated hence w = 0 and q = 0 therefore `DeltaU=0` Let `T_(f)` and `P_(f)` denote the final temperature and pressure respectively `DeltaU=n_(1)C_(V,m)[T_(f)-T_(1)]+n_(2)C_(V,m)[T_(f)-T_(2)] =0` `500(T_(f)-400)+2000(T_(f)-500)=0` `T_(f)=480K` |
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| 3121. |
A container of volume `2m^(3)` is divided into two equal compartments, one of which contains an ideal gas at 400 K. The other compartment is vacuum. The whole system is thermally isolated from its surroundings. The partition is removed and the surrounding. The partition is removed and the gas expands to occupy the whole volume of the container.Its temperature now would beA. `400 K`B. `250 K`C. `200 K`D. `100 K` |
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Answer» Correct Answer - A `dU=0, dT=0` |
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| 3122. |
A container of volume 2L is seperated into equal compartments. In one compartment, one mole of an ideal monoatomic gas is filled at 1 bar pressure and the other compartment is completely evacuted. A pihole is made in the seperator so gas expands to occupy full 2 L and heat is supplied to gas so that finally pressure of gas equals 1 bar. Then : A. `Delta U=Delta H= 150 J`B. `Delta H= 250 J`C. `Delta U= 100 J`D. `Delta U=Delta H=0` |
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Answer» Correct Answer - B Final temperature of gas must be double of the intial temperature `Delta H= Delta U+Delta(PV)=nC_(v)DeltaT+Delta(PV) DeltaT=T_(f)-T_(i)=2T_(i)` `=(nR)/((gamma-1))T_(i)+ Delta (PV)=(3)/(2)xx100+100=250J` |
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| 3123. |
A container of volume 2L is seperated into equal compartments. In one compartment, one mole of an ideal monoatomic gas is filled at 1 bar pressure and the other compartment is completely evacuted. A pihole is made in the seperator so gas expands to occupy full 2 L and heat is supplied to gas so that finally pressure of gas equals 1 bar. Then : A. `DeltaE=DeltaH=150 J`B. `DeltaH=250 J`C. `DeltaE=100 J`D. `DeltaE=DeltaH=0` |
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Answer» Correct Answer - D |
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| 3124. |
If `DeltaG = -177`K cal for `" "(1)2Fe(s)+(3)/(2)O_(2)(g) rarr Fe_(2)O_(3)(s)` and `DeltaG =- 19` K cal for `" "(2)4Fe_(2)O_(3)(s)+ Fe(s) rarr 3Fe_(3)O_(4)(s)` What is the Gibbs free energy of formation fo `Fe_(3)O_(4)(s)`?A. `+229.6` Kcal/molB. `-242.3` Kcal/molC. `-727` Kcal/molD. `-229.6` Kcal/mol |
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Answer» Correct Answer - B `DeltaG^(@)` for `3Fe(s) + 2O_(2)(g) rarr Fe_(3)O_(4)(s)` can be obtained by taking `[(2) + 4 xx (1)] xx (1)/(3)` Hence , we get `DeltaG_(f) =[-19 + 4 xx (-177)] xx (1)/(3)=-242.3 K " cal for " 1 mol Fe_(3)O_(4)` |
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| 3125. |
Calculate the free energy change in dissolving one mole of sodium chloride at `25^(@)C` . Lattic energy `= + 777.8 kJ mol^(-1)`. Hydration energy of `NaCl = - 774.1 kJ mol^(-1)` and `Delta S ` at `25^(@)C = 0.0 43kJ K^(-1) mol^(-1)` |
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Answer» Correct Answer - `DeltaG = - 9.1 kJ mol^(_1)` `DeltaH = `Lattice energy `+` Hydration energy `= 777.8-774.1kJ mol^(-1) = 3.7 kJ mol^(-1)` `DeltaG = DeltaH - T DeltaS= 3.7 kJ mol^(-1) -298 K ( 0.043kJ K^(-1) mol^(-1)) = ( 3.7 - 12.8) kJ mol^(-1) = - 9.1 kJ mol^(-1)` |
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| 3126. |
The free energy changes for the two reactions given below are `:` (i) `SO_(2)(g) + Cl_(2)(g) rarr SO_(2)Cl_(2)(g) , DeltaG = - 2270 cal` (ii) `S( ` rhom) `+ O_(2)(g) + Cl_(2) (g) rarr SO_(2)Cl_(2)(g) , Delta G = - 74060 cal` Find the `Delta G` for the reaction `:` S ( rhom ) `+ O_(2)(g) rarr SO_(2)(g)` |
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Answer» Correct Answer - `-71790 cal` Eqn. (ii) - Eqn. (i) gives the required result. |
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| 3127. |
Calculate the equilibrium constant, K for the following reaction at 400 K ? `2NOCl(g) hArr 2NO(g) + Cl_(2)(g)` Given that ` Delta_(r) H^(@) = 80.0 kJ mol^(_1)` and `Delta_(r)S^(@) =120 J K^(-1) mol^(-1)` at 400 K. |
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Answer» `Delta_(r) G^(@) = Delta_(r)H^(@) - TDelta_(r)S^(@) = 80000J - 400 xx 120 J = 32000 J` Now, `Delta_(r) G^(@) = -2.303 RT log K ` `:. 32000 = - 2.303 xx 8.314 xx 400 xx logK ` or ` log K = - 4.1782 = bar(5).8218` `:. K = ` Antilog `bar(5).8218 = 6.634 xx 10^(-5)` |
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| 3128. |
The absolute temperature (Kelvin scale ) T is related to the temperature TC on the Celsius scale by TC = T - 273.15.Why do we have 273.15 in this relation and not 273.16 ? |
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Answer» 273.16 K is the triple point of water. In the relation, θ = T - 273.15,273.15 is the temperature on the absolute scale corresponding to 0°C. |
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| 3129. |
One mole of an ideal monoatomic gas is heated at a constant pressure of one atmosphere from`0^(@)` to `100^(@)C.` Then the change in the internal energy isA. `6.56`joulesB. `8.32xx10^(2)`joulesC. `12.48xx10^(2)`joulesD. `20.80joules` |
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Answer» Correct Answer - C |
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| 3130. |
If the ratio of specific heat of a gas of constant pressure to that at constant volume is `gamma`, the change in internal energy of the mass of gas, when the volume changes from `V` to `2V` at constant pressure `p` isA. `R(lambda-1)`B. pVC. `PV(lambda-1)`D. `lambdapV(lambda-1)` |
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Answer» Correct Answer - C |
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| 3131. |
The latent heat of vaporisation of water is 2240 J/gm. If the work done in the process of expansion of 1 g of water is 168 J, then increase in internal energy isA. 2408 jB. 2240 jC. 2072jD. 1904j |
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Answer» Correct Answer - C |
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| 3132. |
If a system undergoes contraction of volume then the work done by the system will beA. It introduces the concept of the internal energyB. It introduces the concept of the entropyC. It is not applicable to any cyclic processD. None of the above |
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Answer» Correct Answer - C |
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| 3133. |
An ideal gas is taken through a cyclic thermodynamic process through for steps. The amount of heat involved in four steps are `Q_(1)=6000J,Q_(2)=-5000JQ_(3) andQ_(4)=400J` respectively. If efficincy of cycle is `10x%` then find value of x? |
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Answer» Correct Answer - B The amount of heat supplied `=10,000J` The work done `=2000j%` efficincy `(W)/(Q)xx100=(2000)/(10000)xx100=20` |
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| 3134. |
An ideal diatomic gas under goes a process in which its internal energy chnges with volume as given `U=cV^(2//5)` where c is constant. Find the ratio of molar heat capacity to universal gas constant `R` ? |
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Answer» From given relation we can write `T prop V^(2//5)`…….(1) ideal gas equation `implies PV=nRT` ……(2) form (1),(2) `PV prop V^(2//5)` (or) `PV^(2//5)`=const ………..(3) Comparing equation (3) with polytropic equation `PV^(x)=` const, we have `X=(3)/(5)` `:. C=(R)/(lambda-1)+(R)/(1-x)=5R :. (C)/(R)=5` |
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| 3135. |
The volume of an ideal gas is `V` at pressure `P`. On increasing the pressure by `DeltaP`, change in volume of gas is `DeltaV_(1)`, under isothermal conditions and `DeltaV_(2)` under adiabatic conditions. Which is more and why? |
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Answer» Under isothermal conditions, `K_(i)=(DeltaP)/(DeltaV_(1)//V)=P`, and under adiabatic conditions, `K_(a)= (DeltaP)/(DeltaV_(2)//V)=gammaP` Dividing, we get, `gamma=(DeltaV_(1))/(DeltaV_(2))` As `gammagt1, :. DeltaV_(1)gtDeltaV_(2)` |
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| 3136. |
A vessel containing 5 litres of a gas at 0.8 m pressure is connected to an evacuated vessel of volume 3 litres. The resultant pressure inside with be (assuming whole system to be isolated)A. 4/5 mB. 0.5 mC. 2.0 mD. 3/4 m |
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Answer» Correct Answer - B |
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| 3137. |
For an isothermal expansion of a perfect gas, the value of `(DeltaP)/(P)` isA. `-lambda^(1//2)(DeltaV)/(V)`B. `(DeltaV)/(V)`C. `-lambda(DeltaV)/(V)`D. `-lambda^(2)(DeltaV)/(V)` |
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Answer» Correct Answer - B |
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| 3138. |
An air bubble of diameter `d=4mum` is located in water at a depth `h=5.0m` considering standard atmospheric pressure at 1 atm, find the pressure in the air-bubble? |
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Answer» The pressure has terms due to hydrotatic pressure and capillarity and they add `p = p_0 + rho gh + (4 alpha)/(d)` =`(1 + (5 xx 9.8 xx 10^3)/(10^5)+ (4 xx .73 xx 10^-3)/(4 xx 10^-6) xx 10^-5) "atoms" = 2.22` atom. |
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| 3139. |
On a hot summer day we want to cool our room by opening the refrigerator door annd closing all the windows and doors. Will the process work ? |
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Answer» No. whena refrigerator is working in a closed room with its door closed, it is rejecting heat from inside to the air in the room. So temperature of room increases gradually. When the door of refrigerator is kept open, heat rejected by the refrigerator to the room will be more than the heat taken by the refrigetator from the room (by an amount equal to work done by the compressor). Therefore, temp. of room will increase at a slower rate compared to the first case. |
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| 3140. |
Assertion : It is not possible for a system, unaided by an external agency to transfer heat from a body at lower temperature to another body at higher temperature. Reason : According to Clausius statement, “ No process is possible whose sole result is the transfer of heat from a cooled object to a hotter objectA. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of t he assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 3141. |
Assertion : A room can be cooled by opening the door of a refrigerator in a closed room. Reason : Heat flows from lower temperature (refrigerator) to higher temperature (room).A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of t he assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - D |
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| 3142. |
A swimmer coming out from a pool is covered a film of water weighing about 18g. How much heat must be supplied to evaporate this water at `298K` ? Calculate the internal energy of vaporisation at`100^(@)C`. `Delta _(vap) H^(@)` for water at 373K `= 40.66 kJ mol^(_1)`. |
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Answer» The process of evaporation is `: 18 H_(2) O(l ) rarr 18 g H_(2)O(g) ` No. of moles of 18 g `H_(2) O = (18g )/(18 g mol^(-1))= 1 mol` `Delta n_(g) = 1-0 =1 mol` `:. Delta _(vap) U^(@) = V_(vap) H^(@) - Delta n_(g) RT = 40.66 k J mol^(-1) - (1 mol ) (8.314 xx 10^(-3) k J K^(-1) mol^(-1) ) ( 298 K) ` `= 40.66 k J mol^(-1) - 3.10 kJ mol^(-1) = 37. 56 kJ mol^(-1)` |
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| 3143. |
A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaporisation at 298K. `Delta_("vap")H^(Ө)` for water at 298K =44.01 kJ `"mol"^(-1)` |
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Answer» We can represent the process of evaporation as `underset(1"mol")(H_(2)O)(1)overset("vaporisation")rarrunderset(1"mol")(H_(2)O)(g)` No. of moles in 18 g`H_(2)O(1)` is `=(18g)/(18g"mol"^(-1))=1"mol"` Heat supplied to evaporate18g water at 298 K `" " nxxDelta_("vap")H^(Ө)` `" " =(1"mol")xx(44.01 kJ "mol"^(-1))` `" " =44.01 kJ` (assuming steam behaving as an ideal gas). `Delta_("vap")U=Delta_("vap")H^(Ө)-pDeltaV=Delta_("vap")H^(Ө)-Deltan_(g)RT` `Delta_("vap")H^(v)-Deltan_(g)RT=44.01kJ` `-(1)(8.314kJ^(-1)"mol"^(-1))(298K)(10^(-3)kJ J^(-1))` `Delta_("vap")U^(v)=44.01 kJ-2.48kJ` `" "41.53 kJ` |
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| 3144. |
A swimmer coming out from a pool is covered with a film of water weighing about 18 g. how much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaperization at `100^(@)C`. `Delta_(vap)H^(Theta)` for water at 373 K = 40.66 kJ `mol^(-1)` |
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Answer» The process of evaporation can be represented as `underset("1 mol")underset("18 g")(H_(2)O(I)) overset("Vaporisation") rarr underset("1 mol") underset("18 g")(H_(2)O(g)), Delta_(Vap)H^(Theta)=40.66 kJ" "mol^(-1)` Assumming steam behaving as an ideal gas `{:(Delta_(Vap)U^(Theta),=,Delta_(Vap)H^(Theta)-p Delta V),(,=,Delta_(Vap)H^(Theta)-Delta n_(g)RT),("Here" Delta n_(g),=,1-0 = "1 mol"),(Delta_(Vap)U^(Theta),=,"40.66 kJ mol"^(-1)-("1 mol")(8.314 xx 10^(-3)"kJ mol"^(-1))("373 K)),(,=,40.66 kJ mol^(-1)-3.10),(,=,37.56 kJ mol^(-1)):}` |
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| 3145. |
A gas is compressed at a constant pressure of `50N//m^(2)` from a volume `10m^(3)` to a volume of `4m^(3)`. 100J of heat is added to the gas then its internal energy isA. Increased by 400 JB. Increased by 200 JC. Increased by 100 JD. Decreased by 200j |
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Answer» Correct Answer - A |
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| 3146. |
Find the possible efficiency of a cycle working between `27^(@)C` and `127^(@)C`. |
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Answer» We cannot find efficiency fo a cycle until we have complete information about all processes involved but we can find maximum efficiency between given temperature which corresponds Carnot cycle. `implieseta_(max)=1-(T_(L))/(T_(H))=1-(300)/(400)=(1)/(4)` `implieseta_(max)=25%` Hence `eta` of the cycle `le25%`. |
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| 3147. |
The work done by a system is `10J`, when `40J` heat is supplied to it. Calculate the increase in the internal enegry of system. |
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Answer» `W =- 10 J, q = 40 J` From the first law, `DeltaU = q +w = 40 - 10 = 30J` |
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| 3148. |
A gas undergoes a change of state during which 100J of heat is supplied to it and it does 20J of work. The system is brough back to its original state through a process during which 20 J of heat is released by the gas. What is the work done by the gas in the second process?A. `60J`B. `40J`C. `80J`D. `20J` |
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Answer» Correct Answer - A |
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| 3149. |
A cyclic process involves three processes: A to B, B to C, C to A, heat exchanges during processes are: `Q_(AB)=+40J` `Q_(BC)=+20J` `Q_(CA)=-10J` Find net work done and efficiency. |
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Answer» `sumQ=40+20-10=50J`. `sumQ_(+ve)=40+20=60J` For cyclic process `sumW=sumQ=50J` [As `DeltaU=0`] `implies eta=(sumQ)/(sumQ_(+ve))=(50)/(60)=(5)/(6)` |
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| 3150. |
We consider a thermodynamic system. If `DeltaU` represents the increase in its internal energy and W the work done by the system, which of the following statements is true?A. `DeltaU=-W` in an adiabatic processB. `DeltaU=W` in an isothermal processC. `DeltaU=-w` in an isothermal processD. `DeltaU=W`in an isothermal process |
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Answer» Correct Answer - A |
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