4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 24 InterviewSolution

1151.	Distinguish between isothermal and adiabatic processes.
Answer» A thermodynamic process in which temperature remains constant is called isothermal process. The change in internal energy for this process is zero. A thermodynamic process in which there is no transfer of heat from system to surroundings is called adiabatic process. The change in internal energy in this process is nonzero.

Discussion

1152.	1 kg of water initially at a temperature 27°C is heated by a heater of power 1KW. If the lid is opened, heat is lost at a constant rate of 200J/S. Find the time required for water to attain a temperature of 80°C with the lid open. (Specific heat of water = 4.2kJ/ kg/k)
Answer» Amount of water = 1 kg Net heat supplied = (1000 – 200) = 800 W Change in temperature = ∆ T = (353 – 300) = 53K ∴ ∆ T = 53K Time required, t = H/R where H is the required energy and R is the rate of heat transfer H = mC ∆T = 1 × 4.2 × 10³× 53 = 2.226 × 10⁵ J R = 8 × 10² J/s ∴ t = \(\frac{2.226 \times 10^5}{8\times 10^2}\) = 2.783 × 10² ∴ t ≈ 279 = 4 min 39 secs.

1152.

1 kg of water initially at a temperature 27°C is heated by a heater of power 1KW. If the lid is opened, heat is lost at a constant rate of 200J/S. Find the time required for water to attain a temperature of 80°C with the lid open. (Specific heat of water = 4.2kJ/ kg/k)

Answer»

Amount of water = 1 kg

Net heat supplied = (1000 – 200) = 800 W

Change in temperature = ∆ T

= (353 – 300) = 53K

∴ ∆ T = 53K

Time required, t = H/R

where H is the required energy and R is the rate of heat transfer

H = mC ∆T

= 1 × 4.2 × 10³× 53

= 2.226 × 10⁵ J

R = 8 × 10² J/s

∴ t = \(\frac{2.226 \times 10^5}{8\times 10^2}\)

= 2.783 × 10²

∴ t ≈ 279 = 4 min 39 secs.

Discussion

1153.	Statement-1: The internal energy of a system does not change when there is not change in temperature Statement-2: The internal energy of a gaseous system is state function.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
Answer» Correct Answer - D

Discussion

1154.	Statement-1: For a gaseous system `C_(p)` is always greater than `C_(v)` Statement-2: Work done by a gas at constant volume is zero.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
Answer» Correct Answer - A

Discussion

1155.	Statement-1: A refrigerator transfers heat from lower temperature to higher temperature. Statement-2: Heat can be transferred from lower temperature to higher temperature by itself according to first law of thermodynamics.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
Answer» Correct Answer - B

Discussion

1156.	Obtain an expression for the coefficient of performance of a refrigerator.
Answer» A refrigerator transfers heat from a body at lower temperature to a body at higher temperature by doing work on it. If Q₂ is the heat absorbed from body at temperature T₂(sink) and Q₁is the heat liberated by the refrigerator to a body at temperature T₂ (source) then work done by there refrigerator, W = Q₁– Q₂ ∴ The coefficient of performance, α = \(\frac{Q_2}{W}\) = \(\frac{Q_2}{Q_1-Q_2}\) \(=\frac{T_2}{T_1-T_2}.\)

1156.

Obtain an expression for the coefficient of performance of a refrigerator.

Answer»

A refrigerator transfers heat from a body at lower temperature to a body at higher temperature by doing work on it. If Q₂ is the heat absorbed from body at temperature T₂(sink) and Q₁is the heat liberated by the refrigerator to a body at temperature T₂ (source) then work done by there refrigerator,

W = Q₁– Q₂

∴ The coefficient of performance,

α = \(\frac{Q_2}{W}\)

= \(\frac{Q_2}{Q_1-Q_2}\)

\(=\frac{T_2}{T_1-T_2}.\)

Discussion

1157.	In a heat engine, the temperature of the source and sink are 500 K and 375 K. If the engine consumes `25xx10^5J` per cycle, find(a) the efficiency of the engine, (b) work done per cycle, and (c) heat rejected to the sink per cycle.A. `6.25xx10^(5) J`B. `3xx10^(5) J`C. `2.19xx10^(5) J`D. `4xx10^(4) J`
Answer» Correct Answer - A Here `T_(1)=500 K, T_(2)=375 K,Q_(1)=25xx106(5) J` `therefore eta =1-(T_(2))/(T_(1))=1-(375)/(500)=0.25 = 25%` `W=etaQ_(1)=0.25xx25xx2510^(5)=6.25xx10^(5)J`

Discussion

1158.	Two chunks of metal with heat capacities `C_(1)` and `C_(2)`, are interconnected by a rod length `l` and cross-sectional area `S` and fairly low heat conductivity `K`. The whole system is thermally insulated from the environment. At a moment `t = 0` the temperature difference betwene the two chunks of metal equals `(DeltaT)_(0)`. Assuming the heat capacity of the rod to be negligible, find the temperature difference between the chucks as a function of time.
Answer» Suppose the chunks have temperatures `T_1, T_2` at time `t` and `T_1 - dT_1, T_2 + dT_2` at time `dt + t`. Then `C_1 dT_1 = C_2 dT_2 = (kS)/(l)(T_1 - T_2)dt` Thus `d Delta T = -(kS)/(l) ((1)/(C_1) + (1)/(C_2)) Delta T dt ` where `Delta T = T_1 - T_2` Hence `Delta T = (Delta T)_0 e^(-t//tau)` where `(1)/(tau) = (ks)/(l) ((1)/(C_1) + (1)/(C_2))`.

Discussion

1159.	Find the temperature distribution in a substance palced between two parallel plates kept at temperatures `T_1` and `T_2`. The plate separation is equal to `l`, the heat conductivity coefficient of the substance `x overline prop V overline T`.
Answer» `Q = k(del T)/(del x) = -A sqrt(T) (del T)/(del x)` =`-(2)/(3) A (del T^(3//2))/dx, (A = constant)` =`(2)/(3) A((T_1^(3//2) - T_2^(3//2)))/(l)` Thus `T^(3//2) = constant - (x)/(l) (T_1^(3//2) - T_2^(3//2))` or using `T = T_1` at `x = 0` `T^(3//2) T_1^(3//2)+(x)/(l) (T_2^(3//2) - T_1^(3//2))` or `((T)/(T_1))^(3//2) = 1 + (x)/(l) (((T_2)/(T))^(3//2) - 1)` `T = T_1 [1 + (x)/(l) {((T_2)/(T_1))^(3//2) - 1}]^(2//3)`.

Discussion

1160.	The space between two large horizontal plates if filled with helium. The plate separation equal `l = 50 mm`. The lower plates is kept at a temperature `T_1 = 290 K`, the upper, at `T_2 = 330 K`. Find the heat flow density if the gas pressure is close to standard.
Answer» `k = (1)/(3) sqrt((8 RT)/(pi M)) (1)/(sqrt(2) pi d^2 n) mn (R(i)/(2))/(M) = (R^(3//2) i T^(3//2))/(3 pi^(3//2) d^2 sqrt(M) N_A)` Then from the previous problem `q = (2 i R^(3//2) (T_2^(3//2) - T_1^(3//2)))/(9 pi^(3//2) d^2 sqrt(M) N_A l), i = 3 here`.

Discussion

1161.	The space between two large parallel plates separated by a distance `l = 5.0 mm` is filled with helium under a pressure `p = 1.0 Pa`. One plate is kept at a temperature `t_1 = 17 ^@C` and the other, at a temperature `t_2 = 37 ^@C`. Find the mean free path of helium atoms and the heat flow density.
Answer» At this pressure and average temperature `= 27^@C = 300 K = T = ((T_1 + T_2))/(2)` `lamda = (1 kT)/(sqrt(2) pi d^2 p) = 2330 xx 10^-5 m = 23.3 nmm gt gt 5.0 mm = l` The gas is ultrathin and we write `lamda = (1)/(2) l` here Then `q = k(dT)/(dx) = k(T_2 - T_1)/(l)` where `k = (1)/(3) lt v gt xx (1)/(2) l xx (MP)/(RT) xx (R)/(gamma -1) xx (1)/(M) = (plt v gt)/(6 T(gamma -1))l` and `q = (plt v gt)/(6T(gamma -1))(T_2 - T_1)` where `lt v gt = sqrt((8 RT)/(M pi))`. We have used `T_2 - T_1 lt lt (T_2 + T_1)/(2)` here.

Discussion

1162.	The given P-V diagram expansion of a gs which one of the following statement is true? A. A is isothermal and B is adiabatic processB. A is adiabatic and B is isothermal processC. Both are isothemal processD. Both are adiabatic process
Answer» Correct Answer - A As slop of curve B is more than slope of aA therefore curve A is showing isothermal process and curve B is showing adiabatic process.

Discussion

1163.	The possibility of increase in the temperature of gas without adding heat to it happens inA. Adiabatic exapansionB. isothermal expansionC. adiabatic compressionD. isothermal compression
Answer» Correct Answer - C In adiabatic process `dQ=0 rarrdU +dW=0 or dU=-dW` In the process of compression work is done on the gas therefore dW is negative .Hence dU is positive i.e internal energy of gas increases and threefore temperature of the gas also increses.

Discussion

1164.	In the figure given two processes `A and B` are shown by which a thermodynamic system goes from initial to final state F. if `DeltaQ_(A)` and `DeltaQ_(B)` are respectively the heats supplied to the systems then A. `DeltaQ_(A)=DeltaQ_(B)`B. `DeltaQ_(A)leDeltaQ_(B)`C. `DeltaQ_(A)ltDeltaQ_(B)`D. `DeltaQ_(A)gtDeltaQ_(B)`
Answer» Correct Answer - D

Discussion

1165.	Two mole of an ideal monatomic gas are confined within a cylinder by a mass less spring loaded with a frictionless piston of negligible mass and crossectional area `4xx10^(-3)m^(2)`. The gas is heated by a heater for some time. During this time the gas expands and does `50J` of work in moving the piston through a distance of `0.01`m. The temperature of gas increases by `50`k. The force constant of spring isA. `189.6 N m ^(-1)`B. `18.96 N m ^(-1)`C. `1896 N m ^(-1)`D. `2896 N m ^(-1)`
Answer» Correct Answer - C When the gas is heated it expands `&` pushes the piston by x. If k is force constanat of spring and `A` is area of crossection of the piston. It `P_(0)` is atmospheric pressure then at equilibrium of piston the pressure of the gas on the piston `P=P_(0)+(kx)/(A)` The increas in the volume of the gas by samll movement of x of piston is `dv=Adx` `W=int_(0)^(X)Pdv=P_(0)Ax+(1)/(2)kx^(2)` Putting `A=4xx10^(-3)m^(2),x=0.1m, W=50 J`. `P_(0)=1.013xx10^(5) Nm^(-2)` in above equation `:. k=1896 Nm^(-1)`

Discussion

1166.	Consider `P-V` diagram for an ideal gas shown in figure. Out of following diagrams(figure). Which represents the `T-P` diagram? A. (iv)B. (ii)C. (iii)D. (iv)
Answer» Correct Answer - C In (figure) `T` is constant and `P_(1) gt P_(2)`. This situation is represented by curve (iii) in (figure). 3 in which `P_(1) gt P_(2)` and straight line graph is prallel to pressure axis indicating constant temperature. Choice (c ) is correct.

Discussion

1167.	Can two isothermal curves intersect each other?
Answer» No, if they do so, then at two different temperature (or the isothermals), volume and pressure of gas will same, which is not possible.

Discussion

1168.	Can whole of work be converted into heat?
Answer» Yes, through friction.

Discussion

1169.	In case of compression, isothermal curve lies…..the adiabatic curve. Fill in the blanksA. aboveB. belowC. sometimes above and other time belowD. cannot say
Answer» Correct Answer - B `PV= "constant" , PV^(gamma)= "constant"`

Discussion

1170.	Does the internal energy of an ideal gas change in an isothermal process? In an adiabatic process.
Answer» In an isothermal process, T = constant, dT = 0. Hence, dU = 0. In an adiabatic process, Q = constant, dQ = 0. dU = - dW ≠ 0.

1170.

Does the internal energy of an ideal gas change in an isothermal process? In an adiabatic process.

Answer»

In an isothermal process, T = constant, dT = 0.

Hence, dU = 0.

In an adiabatic process, Q = constant, dQ = 0.

dU = - dW ≠ 0.

Discussion

1171.	Calculate the work done (in J) when 4.5 g of `H_(2)O_(2)` reacts against a pressure of 1.0 atm at `25^(@)C` `2H_(2)O_(2)(l)rarr O_(2)(g) +2H_(2)O(l)`A. `-1.63 xx 10^(2)`B. `4.5 xx 10^(2)`C. `3.2 xx 10^(2)`D. `-6.1 xx 10^(2)`
Answer» Correct Answer - A `w=-Deltan_(g)RT=(-4.5 )/(34xx2)xx8.314 xx298`, `w=-163.9 J`

Discussion

1172.	Consider the following data : `Delta_(f)H^(@)(N_(2)H_(4),l)=50kJ//mol,Delta_(f)H^(@)(NH_(3),g)=-46kJ//mol` `B.E(N-H)=393" kJ//mol and B.E."(H-H)=436kJ//mol` `Delta_("vap")H(N_(2)H_(4),l)=18kJ//mol` The N-N bond energy in `N_(2)H_(4)` is :A. 226 kJ/molB. 154 kJ/molC. 190 kJ/molD. None of these
Answer» Correct Answer - C `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)rarrNH_(3)(g)` Let B.E. of `N-=N` is x `-46 = (x)/(2)+(3)/(2)xx436 - 3xx393 implies x = 958` `N_(2)H_(4)(l)rarrN_(2)(g)+2H_(2)(g),` `Delta_(r )H=-50 " kJ"//"mol"` `Delta_(r)H=[(Delta_("vap")H(N_(2)H_(4),l)),(+4xxB.E.(N-H)),(+B.E.(N-N))]` `-((B.E.(N-=N)),(+2B.E.(H-H)))` `-50 =(18+4xx393+y)-(958 + 2xx436)` `-50=(1590+y)-(1830)` `B.E. (N-N) or y = 190 kJ/mol

Discussion

1173.	To an expanding gas, no external energy is suppiled. Will the gas do any work? If yes, what will be the source of energy?
Answer» Yes, the gas will do work in expansion. Energy spent comes from internal energy of the gas.

Discussion

1174.	Does the internal energy of an ideal gas change in an isothermal process? In an adiabatic process?
Answer» In an isothermal process , `T= const`. `dT=0 :. dU=0`. In an adiabatic process, `Q= constant` `dQ=0 :. dU= -dW!=0`.

Discussion

1175.	Can we increase the temperature of gas without supplying heat to it?
Answer» Yes, the temperature of gas can be by compressing the gas under Adiabatic condition.

Discussion

1176.	4 millimoles of an ideal monoatomic gas is taken through a cyclic process ABCDA, is shown in the figure, AB and CD are two isothermal curves having temperature 300 K and 600 K respectively. The total work done by the gas during the cyclic process is `(xRlog_(e)2)/(5)`. find the value of x .
Answer» Correct Answer - 6

Discussion

1177.	First law of thermnodynamics is given byA. dQ = Du + pdVB. dQ = dU x PdVC. dQ = (dU+dV)PD. dQ = pdV +dV
Answer» Correct Answer - A

Discussion

1178.	Statement-1: If the pressure and temperature of a gas sample are doubled then the volume of the gas will remain unchanged. Statement-2: Molar specific heat for an adiabatic process is zero. Statement-3: Heat energy is a path function.A. FTFB. TFTC. TFFD. FTT
Answer» Correct Answer - D

Discussion

1179.	Assertion : A hollow metallic closed container maintained at a uniform temperature cab act as a source of black body radiation. Reason : All metals act as a black body.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 - C Here, Assertion is true but Reason is false because the metals with highly polised surface cannot act as black bodies.

Discussion

1180.	Aseertion: Thermodynamics process in nature are irreversible. Reason: Dissipactive effects cannot be eliminated.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
Answer» Correct Answer - A

Discussion

1181.	Aseertion: Thermodynamics process in nature are irreversible. Reason: Dissipactive effects cannot be eliminated.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 the correct explanxtion of Assertion because the thermodynamical process taking place in nature are irreversible as they involve some loss of energy in their working.

Discussion

1182.	Aseertion: Thermodynamics process in nature are irreversible. Reason: Dissipactive effects cannot be eliminated.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 - A The thermodynamic process is irreversible as there always occurs a oss of energy due to energy spent in working against the dissipativbe force, which is not recovered back. Otjher irresversible process alspo occurs in nature such as friction where extra work is required to cancel out the effect of friction.

Discussion

1183.	The molar heat capacity `(C_(P))` of `CD_(2)O` is 10 cals at 1000 K. The change in entropy associated with cooling of 32 g of `CD_(2)O` vapour form 1000 K to 100 K at constant pressure will be : (D= deuterium, at mass =2u)A. `23.03" cal deg"^(-1)`B. `-23.03" cal deg"^(-1)`C. `2.303" cal deg"^(-1)`D. `-2.303" cal deg"^(-1)`
Answer» Correct Answer - B `DeltaS=2.303 nC_(p)"log" T_(2)/T_(1)`

Discussion

1184.	A very large swimming pool filled with water of temperture equal to `20^(@)`C is heated by a resistor with a heating power of 500 W for 20 minutes. Assuming the water in the pool is not in any contact with anything besides the resistor, determine the following quanties. Is the change of entropy of the resistor positive , negative or zero? `(i) DeltaS_("rev") gt 0 " " (ii) DeltaS_("rev")=0 " "(iii) DeltaS_("rev") lt 0`
Answer» Correct Answer - `DeltaS_("rev") =0`

Discussion

1185.	A very large swimming pool filled with water of temperture equal to `20^(@)`C is heated by a resistor with a heating power of 500 W for 20 minutes. Assuming the water in the pool is not in any contact with anything besides the resistor, determine the following quanties. Is the change of entropy of the water positive, negative or zero? `(i) DeltaS_("pool") gt 0 " " (ii) DeltaS_("pool")=0" "(iii) DeltaS_("pool") lt 0`
Answer» Correct Answer - `DeltaS_("pool") gt 0`

Discussion

1186.	A very large swimming pool filled with water of temperture equal to `20^(@)`C is heated by a resistor with a heating power of 500 W for 20 minutes. Assuming the water in the pool is not in any contact with anything besides the resistor, determine the following quanties. Is the process reversible ? (Y/N)
Answer» Correct Answer - The answer is No(N).

Discussion

1187.	A very large swimming pool filled with water of temperture equal to `20^(@)`C is heated by a resistor with a heating power of 500 W for 20 minutes. Assuming the water in the pool is not in any contact with anything besides the resistor, determine the following quanties. Is the change of entropy of the system positive , negative or zero? `(i) DeltaS_("total") gt 0 " " (ii) DeltaS_("total")=0 " " (iii) DeltaS_("total") lt 0`
Answer» Correct Answer - `DeltaS_("total") lt 0`

Discussion

1188.	A very large swimming pool filled with water of temperture equal to `20^(@)`C is heated by a resistor with a heating power of 500 W for 20 minutes. Assuming the water in the pool is not in any contact with anything besides the resistor, determine the following quanties. The heat delivered to the water.
Answer» Correct Answer - `Q=500 W xx20"min" xx60=600 KJ`

Discussion

1189.	Since 1891 lighting lamps have been manufactured in the Netherlands. The improvement today in comparison to the first lamp is enormous , especially with the intorduction of the gas discharge lamps. The life- time has increased by orders of magnitude . the colour is also an important aspect. Rare earth metal compunds like `CeBr_(3)` are now included to reach a colour temperature of 6000K in the lamp the compounds are ionic solids at room temperature , and upon heating they sublime partially to give a vapour of neutral metal halide molecules . To achieve a high vapour pressure, the sublimation enthalpy should be as low as possible. Conclusion in relation to the previous answers. was adding CsBr a good idea ? Mark the correct answer. `implies` Adding CsBr is counterproductive. `implies` Adding CsBr has no influence `implies` Adding CsBr is advantageous `implies` From these data no clear answer can be given.
Answer» Correct Answer - The third answer is correctg : Adding CsBr is advantageous.

Discussion

1190.	Since 1891 lighting lamps have been manufactured in the Netherlands. The improvement today in comparison to the first lamp is enormous , especially with the intorduction of the gas discharge lamps. The life- time has increased by orders of magnitude . the colour is also an important aspect. Rare earth metal compunds like `CeBr_(3)` are now included to reach a colour temperature of 6000K in the lamp the compounds are ionic solids at room temperature , and upon heating they sublime partially to give a vapour of neutral metal halide molecules . To achieve a high vapour pressure, the sublimation enthalpy should be as low as possible. Give the reaction equations of the thermochemical cycle (Law of Hess) for this process in which some steps involve `CeBr_(4)^(-)` ions monmelecular ions and /or neutral molecules in the phase. Step 1: `overset(+H_(1))rarr" "+` Step 2 : `+" "overset(+H_(2)) rarr` Step 3 : `+" "overset(+H_(3))rarr` Step 4: `+" "overset(+H_(4))rarr` Total : `(CsCeBr_(4))_("lattice") overset(+H_("total"))rarr` `(CeBr_(3))_("molecules") + (CsBr)_("molecules")`
Answer» Correct Answer - Step 1: (CsCeBr_(4))_("lsttice" overset(+H_(1)) rarr Cs_(-) + CeBr_(4)^(-)` step 2: `CeBr_(4)^(-) overset(+H_(2))rarr Ce^(2+) + 4Br^(-)` Step 3: `Ce^(3+) + 3Br^(-) overset(+H_(3))rarr (CeBr_(3))_("molecules")` Steps 4 : `Cs^(-) z+ Br^(-) overset(+ H_(4)) rarr (CsBr)_("molecules")` _________ `(CsCeBr_(4))_("lattice" overset(+total) rarr (CeBr_(3))_("molecules") + (CaBr)_("molecules")`

Discussion

1191.	One mole of `Cl_(2(g))` which may be assumed to obey the ideal gas law, initially at 300 K and `1.01325 xx 10^(7)` Pa, is expanded against a constant exteranl pressure of `1.01325 xx 10^(5)` Pa to a final pressure of `1.01325 xx 10^(6)` Pa . As a result of the expansion , the gas cooled to a temperature of 239 K (which is the normal boiling point of `Cl_(2)`) and 0.100 mol of `Cl_(2)` condensed. The enthalpy of vaporization of `Cl_(2(l))` is 20.42 `KJ "mol"^(-1)` at the normal volume is `C_(v)=28.66 JK^(-1) "mol"^(-1)` and the density of `Cl_(2(l))` is 1.56 g `cm^(-1)` (at 239 K) . Assume that the molar heat capacity at constant pressure for `Cl_(2(g))` is `C_(p)=C_(v)+R.` `(1 atm = 101325 xx 10^(5) Pa , R=8.314510 Jk^(-1) "mol"^(-1) = 0.0820584 L atm K^(-1)"mol"^(-1))` (i) Either draw a complete molecular orbital energy diagram or write the complete electronic configuration of `Cl_(2)` Predict the bond order of `Cl_(2)` and thus whether this molecules will be diamagnetic , ferromagnetic ,or paramagnetic. (ii) For the changes decribed above, calculate the change in the internal energy `(DeltaE)` and the change in the entropy `(DeltaS_("sys"))` of the system.
Answer» Correct Answer - (i) Electronic configuration fo a Ci atom: `1s^(2)2s^(2)2px^(2)2py^(2)2pz^(2)3s^(2)3px^(2)3py^(2)3pz^(1)` Significant atomic orbitals (AO) =1(K) + 4(L) + 4(M) =9AO Number of electrons in these Aos:17 Number of molecular orbitals (MO) equals number of AOs: Thus `2 xx [1(K) + 4(L) + 4(M) ]=18` MOs are present in a `Cl_(2)` molecule In the information of `Cl_(2): 2 xx 17 = 34` electrons to go into the 18 MOs. MO decription of `Cl_(2)` : `1sigma^(2)1sigma^()2sigma^(2)2sigma^()3sigma^(2)1pi^(4)3sigma^(2)4sigma^(2)4sigma^(2)5sigma^(2)2pi^(4)2pi^(4)` or `(KK)(LL)(sigma3s)^(2)(sigma^()3s)^(2)(sigma3p)^(2)(pi3p)^(4)(pi^()3p)^(4)` `(sigma1s)^(2)(sigma^()1s)^(2)(sigma2s)^(2)(sigma^()2s)^(2)(sigma2p_(z))^(2)(p2p_(y))^(2)(p2p_(x))^(2)(p^()2p_(x))^(2)(p^()2p_(y))^(2)(p^()2p_(z))^(2)` `(sigma3s)^(2)(sigma^()3s)^(2)(sigma3p_(z))^(2)(p3p_(x))^(2)(p3p_(y))^(2)(p^()3p_(x))^(2)(p^()3p_(y))^(2)(p^()3p_(z))^(0)` Or `(KK)(LL) (sigma3s)^(2)(sigma3p_(z))^(2)(pi3p_(z))^(2)(pi3p_(x))^(2)(pi3p_(y))^(2)(pi3p_(x))^(2)(pi^()3p_(x))^(2)(sigma^()3p_(y))^(2)(sigma^()2p_(z))^(0)` ``assumption :- bond formation is along the z-axis (equivalent formulae for x or y axes are accepted ) Bond order is given by `(n-n^())//2:` `n=18; n^(**) = 16` (18-16)/2 =1 (1 `sigma` bond , no `pi` bond) The `Cl_(2)` molecules has a bond order of 1. The `Cl_(2)` molecules is diamagnetic since there are no unpaired electrons. (ii) Summary of the changes involved: `Cl_(2)(g) 1"mol" 300K underset("cooled"1.013 xx 10^(7) "Pa" (100 atm))overset(DeltaE_(1))rarr Cl_(2)(g) 1" mol" 239 K underset(1.013 xx 10^(5) "Pa" 239 K (1 atm)) overset(DeltaE_(2))rarr Cl_(2)(I) 0.1 "mol"` The total process is an expanison plus an isobaric change of phase (gas to liquid) and since the internal energy (E) is a function of state, the total change in the internal energy is `DeltaE = DeltaE_(1) + DeltaE_(2)`. Process 1: `DeltaE_(1) = Int nC_(v)dT = 1 xx 28.66 xx 239 - 300=-1748.3 J` Note a) `DeltaE` for a perfect gas is a function only of T b) `C_(v)` is constant c) "-" sign means a loss of energy due to the work needed for expansion of 1 mole of gas. process 2: For convenience , the data were manipulated in atm, equivalent procedure in Pa will require the appropriate conversion factor. From an energetic pint fo view, the liquid formation Process 2 can be split into two separate steps: `dot" "` the vaporization heat loss (decreased internal energy,-) from the system into surrounding (since the process takes place at constant pressure, the heat is equal to the change in the enthalpy) `dot " "` the work done by the surrounding in compressing the system to a smaller volume (increased internal energy, +). Volume of gas which condensed is `V=nRT//P = (0.1 xx 0.0820584 xx 239)//1 = 1.96 dm^(3)` Volume of liquid `Cl_(2): (0.1 xx 2 xx 35.454)//1.56 = 4.54 cm^(3)` `DeltaE_(2)=DeltaH_(2)-int P_("ext") DeltaV(phase change)=DeltaH_(2)-P_("ext") (V_(1)-V_(g))` But `V_(1)` is approximately 0 and can be neglected (ca. 4.5` cm^(3)` liquid volume vs. ca. 17.6 `dm^(3)`; ca 0.03% error) `DeltaE_(2) =(0.1)(-DeltaH_("vap")) + P_("ext")Vg` `=0.1 xx (.^(-)C 20420)+ (1 xx 1.96 L ) xx 101.325 J dm^(-3) atm^(-1) = -2042.0 + 198.5 =-1843.5` `DeltaE=DeltaE_(1) + DeltaE_(2) =-1748.3 + -1843.5)=-3591.8` Entropy S is a function of two variables of state. Since in Process 1 the known variables are T and `P_(1)` expression of S is chosen as S(T,P). `Delta S_("sys") = DeltaS_(1) + Deltas_(2)` and `overset(-)C_(p)=overset(-)C_(v) + R =28.66 + 8.314 = 36.97 J K^(-1) "mol"^(-1)` `DeltaS_(1) nC_(p) "In" (T_(2))/(T_(1)) - nR "In" (P_(2))/(P_(1)) = 1.0 xx 36.97 "in" (239)/(300) - 8.314 "in" (1)/(100) ==-8.40 + 38.29` `=29.89 J K^(-1)` For the phase transition (constant temperature) , by definition `DeltaS_(2) = Q//T` Since the pressure is constant in this case ,` Qp//T = Qp//T = DeltaH//T` `DeltaS_(2) = (DeltaH_(2))/(T) = (0.1 xx (-20420))/(239) =-8.54 JK^(-1)` `DeltaS_("sys") =29389 - 8.54 = 21.35 JHK^(-1)`

Discussion

1192.	Define entropy.
Answer» Entropy:- It is thermodynamics property which measures randomness or disorders of a system. The more is disorder or randomness, higher will be entropy, e.g., Solid < Liquid < Gas. Entropy of a system is state function, i.e., it depends upon initial and final states of the system. When the state of a system changes, the entropy also changes, ΔS = \(\frac{q_{Res}}{T}\) = \(\frac{ΔH_{rev}}{T}\) Where q is the heat supplied isothermally. T is absolute temperature. For a irreversible spontaneous process ΔS = +ve. For a reversible change at equilibrium ΔS = 0. Entropy increases with increase in temperature, decreases with decrease in and temperature.

1192.

Define entropy.

Answer»

Entropy:- It is thermodynamics property which measures randomness or disorders of a system. The more is disorder or randomness, higher will be entropy, e.g., Solid < Liquid < Gas.

Entropy of a system is state function, i.e., it depends upon initial and final states of the system.

When the state of a system changes, the entropy also changes,

ΔS = \(\frac{q_{Res}}{T}\) = \(\frac{ΔH_{rev}}{T}\)

Where q is the heat supplied isothermally. T is absolute temperature.

For a irreversible spontaneous process ΔS = +ve. For a reversible change at equilibrium ΔS = 0. Entropy increases with increase in temperature, decreases with decrease in and temperature.

Discussion

1193.	Define change in internal energy.
Answer» Change in internal energy:- Change in internal energy is the difference in the internal energies of the products and the reactants involved in the reaction. Thus ΔE = E_(products) - E_(Reactants) = E_ρ - E_r The value of ΔE is positive if E_ρ > E_r and is negative in case. E_ρ < E_r.

1193.

Define change in internal energy.

Answer»

Change in internal energy:- Change in internal energy is the difference in the internal energies of the products and the reactants involved in the reaction. Thus

ΔE = E_(products) - E_(Reactants)

= E_ρ - E_r

The value of ΔE is positive if E_ρ > E_r and is negative in case. E_ρ < E_r.

Discussion

1194.	When a system is taken from A to C through path ABC, 10 J of heat flows to the system and 4 J of work is done by the system. How much heat flows into the system in path ADC, if the work done by the system is 3 J ?
Answer» Correct Answer - 9

Discussion

1195.	A carnot engine operates between temperature `600 K` and `300 K`. It absorbs `100 J` from the source. Calculate the heat transferred to the sink.
Answer» Using: `eta_("Carnot") =1 - (T_(1))/(T_(2)) =1- (300)/(600) = (1)/(2) = 50%` Also, `eta_("carnot") = ("Work done")/("Heat abosorbed") = ("Work done")/(100)` `rArr` Work done `=100 xx (1)/(2) = 50 J` and the heat lost to the surrounding `=100 - 50 = 50 J`

Discussion

1196.	An ideal gas has a specific heat at constant pressure `C_(p)=5/2R`. The gas is kept in a closed vessel of volume `0.0083m^(3)`, at temperature of `300 K` and pressure `1.6xx10^(6)N//M^(2)`. An amount of `2.49xx10^(4) J` of energy is supplied to the gas. calculate the final temperature and pressure of the gas.
Answer» Correct Answer - `T=675 K,P=3.58xx10^(6) N//M^(2)`

Discussion

1197.	An idealgas has `C_(p) =( 5)/(2) R`. The gas is kept is a closed container of volume `0.0083m^(3)` at a temperature of 300 K and a pressure of `1.6 xx 10^(6) Nm^(-2)`. An amount of energy equal to `2.49 xx10^(4) J` is supplied to the gas. Calculated the final temperature of the gas.
Answer» No. of moles of the gas `(n) = (PV)/(RT) = ( 1.6 xx 10^(6) xx 0.0083)/( 8.3 xx300) =5.33` Given `C_(p) = ( 5)/(2) R` `:. C_(v) = C_(p) -R = ( 5)/(2) R-R= (3)/(2) R=(3)/(2) xx 8.3 = 12.45 JK ^(-1) mol^(-1)` Heat supplied at constant volume `= n xx C_(v) = DeltaT` `:.2.49 xx 10^(4) = 5.33 xx 12.45 xx Delta T ` or `DeltaT = 375K` `:. `Final temperature `= 300 + 375 K = 675K`

Discussion

1198.	Given an idealgas is expanded adiabatically and irreversibley form volume `V_(1)` to `V_(2))` , then which one of the following is correct ?A. `DeltaS` ( system ) `= 0 & DeltaS` ( surroundings ) `= + ve`B. `DeltaS` ( system ) `= +ve & DeltaS` ( surroundings ) `= 0`C. `DeltaS` ( system ) `= 0 & DeltaS` ( surroundings ) `= 0`D. `DeltaS` ( system ) `= +ve & DeltaS` ( surroundings ) `= -ve`
Answer» Correct Answer - a For adiabatic process,`q=0` and hence `DeltaS` ( system ) `= 0` . Irrevrsible process is accompaniedby increase of entropy, therefore,`DeltaS` ( surrounding) `= + ve`.

Discussion

1199.	In the following questions a statement of Assertion (A) followed by a statement of Reason (R ) is given. Choose the correct option out of the choices given below. Assertion (A) `:` Spontaneous process is an irrerversible process and may be reversed by some external agency. Reason (R ) `:` Decrease in enthalpy isa contributory factor for spontaneity.
Answer» Correct Answer - B Correct explanation. Spontaneous processes are accompaniedby decrease in energy and increasein randomness.

Discussion

1200.	If `1 gram` of oxygen at `760 mm` pressure and `0^(@)C` has its volume double in an adiabatic change , calculate the change in internal energy. Take `R=2 cal. Mol e^(-1)K^(-1)= 4.2J cal^(-1) and gamma=1.4`.
Answer» Correct Answer - `-43.37J` In an adiabatic change, `dQ=0` From `dU+dW=dQ=0` `dU= -dW= -(r(T_(2)-T_(1)))/(1-gamma)` For `T_(2), use T_(2)=T_(1)((V_(1))/(V_(2)))^((gamma-1))` and `r=R/M`.

Discussion

Explore topic-wise InterviewSolutions in .

Distinguish between isothermal and adiabatic processes.

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Does the internal energy of an ideal gas change in an isothermal process? In an adiabatic process?

Can we increase the temperature of gas without supplying heat to it?

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Define entropy.

Define change in internal energy.

When a system is taken from A to C through path ABC, 10 J of heat flows to the system and 4 J of work is done by the system. How much heat flows into the system in path ADC, if the work done by the system is 3 J ?

A carnot engine operates between temperature `600 K` and `300 K`. It absorbs `100 J` from the source. Calculate the heat transferred to the sink.

An idealgas has `C_(p) =( 5)/(2) R`. The gas is kept is a closed container of volume `0.0083m^(3)` at a temperature of 300 K and a pressure of `1.6 xx 10^(6) Nm^(-2)`. An amount of energy equal to `2.49 xx10^(4) J` is supplied to the gas. Calculated the final temperature of the gas.

If `1 gram` of oxygen at `760 mm` pressure and `0^(@)C` has its volume double in an adiabatic change , calculate the change in internal energy. Take `R=2 cal. Mol e^(-1)K^(-1)= 4.2J cal^(-1) and gamma=1.4`.