4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 64 InterviewSolution

3151.	Calculate the standard internal energy change for the following reactions at `25^(@)C`: `2H_(2)O_(2)(l) rarr 2H_(2)O(l) +O_(2)(g)` `Delta_(f)H^(Theta) at 25^(@)C` for `H_(2)O_(2)(l) =- 188 kJ mol^(-1) H_(2)O(l) =- 286 kJ mol^(-1)`
Answer» `DeltaH^(Theta) = sum DeltaH^(Theta) ("product") - DeltaH^(Theta) ("reactants")` `=2 (-280) +0 -2(-188)` `=- 572 + 376 =- 196 kJ` `Delta n(g) = 1 - 0 = 1` `DeltaH^(Theta) = DeltaU^(Theta) + Deltan(g) RT :. DeltaU^(Theta) = DeltaH^(Theta) - DeltanRT` `=- 196 -(1xx 8.314 xx 10^(-3) xx 298)` `=198.4775 kJ`

Discussion

3152.	Ice-water mass ratio is maintntained as `1:1` in a given system containing water in equilibrium with ice at constant pressue . If `C_(p)` (ice) =`C_(p)` (water) =4.18 J `mol^(-1)K^(-1)` molar heat capacity of such a system is :A. ZeroB. infinityC. `4.182 JK^(-1) mol^(-1)`D. `75.48 JK^(-1) mol^(-1)`
Answer» Correct Answer - B

Discussion

3153.	The enthalpy of reaction for the reaction `:` `2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l) ` is `Delta_(r)H^(c-) =- 572 k J mol^(-1)` What will be standard enthalpy of formation of `H_(2)O(l) ` ?
Answer» Given`: 2H_(2)(g) + O_(2)(g) rarr2H_(2)O(l), Delta_(r)H^(@)= - 572kJ mol^(-1)` Enthalpy of formation is the enthalpy change of the reaction when 1 mole of the compound is formed from its elements, i.e., we aim at `H_(2)(g)+ (1)/(2) O_(2)(g) rarr H_(2)O(l), Delta_(r)H^(@) = ?` This can be obtained by dividing the given equation by 2. Hence, `Delta_(2)H^(@) ( H_(2)O) = ( - 572kJ mol^(-1))/( 2) =-286kJ mol^(-1)`

Discussion

3154.	Using the data provided, calculate the bond energy `( kJ mol^(-1)) ` of a `C-=C` bond in `C_(2)H_(2)`. That eneergy is ( take the bond energy ofC-H bond as350 kJ `mol^(-1))``2C(s) + H_(2)(g) rarr C_(2)H_(2)(g)" " DeltaH = 225 kJ mol^(-1)` `2C(s) rarr 2C(g) " " DeltaH = 1410 kJ mol^(-1)` `H_(2)(g) rarr 2H(g)" " DeltaH = 330 kJ mol^(-1)`A. 1165B. 837C. 875D. 815
Answer» Correct Answer - D (i) `2C(s) + H_(2)(g) rarr H-C-= C-H(g), DeltaH=225kJ mol^(-1)` (ii) `2C(s) rarr 2C(g) , DeltaH = 1410kJ mol^(-1)` (iii) `H_92)9g)rar 2H(g), DeltaH = 330 kJ mol^(-1)` From eq. (i) `225= [2 xxDelta_(C(s) rarr C(g)) + BE_(H-H)]-[2BE_(C-H)+BE(C-=C)]` or `225=[ 1410 + 330]-[2 xx 350 k+ BE_(C-=C)]` or ` 225=1740- 700- BE_(C-=C)` or`BE_(C-=C) 1040-225 = 815kJ mol^(-1)`

Discussion

3155.	Out of boiling point (P), entropy (Q), pH (R ) and e.m.f. of cell (S), intensive properties are:A. `P,Q`B. `P,Q,R`C. `P,R,S`D. All of these
Answer» Correct Answer - C

Discussion

3156.	An ideal gas is allowed to expand against a constant pressure of 2 bar from 10L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case ? (Give that, 1 L bar = 100 J)
Answer» In the first case, as the expansion is against constant external pressure `W = p_("ext") (V_(2) - V_(1)) = - 2 "bar" xx (50 - 10) L` `= - 80 L "bar" " " (1 L "bar" = 100 J)` `= - 80 xx 100 J` `= - 8 kJ` If the given expansion was carried out reversibly, the internal pressure of the gas should be greater than the external pressure at every stage. Hence, the work done will be more

Discussion

3157.	An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one stop. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly,will the work done be higher or lower than the earlier case ?
Answer» In the first case, as the expansion is against constant external pressure, ,brgt `w= - P_(ext) (V_(2)-V_(1)0 = - 2 "bar" xx ( 50-10) L= -80 L "bar" = - 80 xx 100 J = - 8 kJ ( 1L `bar` = 100 J) ` If the above expansion were carried out reverisbly, the internal pressure of the gas should be infinitesimally greater than the external pressure at every stage. Hence, the work done will be more.

Discussion

3158.	Statement -1: Entropy change in reversible adiabatic expansion of an ideal gas is zero. Statement-2: The increase in entropy due to volume increase just compensates the decrease in entropy due to fall in temperature.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-6B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-6C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE
Answer» Correct Answer - A

Discussion

3159.	5 moles of an ideal monoatomic gas absorbes x joule when heated from `25^(@)` C to `30^(@)` C at a constant volume. The amount of heat absorbed when 2 moles of the same gas is heated from `25^(@)` C to `30^(@)` C at constant pressure, is :A. `(3)/(5)xJ`B. `(5)/(3)xJ`C. `(2)/(3)xJ`D. `(25)/(6)xJ`
Answer» Correct Answer - C

Discussion

3160.	Identify the correct statement for change of Gibbs energy for a system `(Delta_(sys)G)` at constant temperature and pressure:A. Must be spontaneous at any temperatureB. Cannot be spontaneous at any temperatureC. Will be spontaneous only at low tempertureD. Will be spontaneous only at high temperature
Answer» `DeltaC = DeltaH - T Deltas` `DeltaG =- ve` at high temperature

Discussion

3161.	One mole of an ideal mono-atomic gas is taken round cyclic process `ABCD` as shown in figure below. Calculate
Answer» Path `AB` is isochoric `(w_(1)=0)`, path `BC` is isothermal `(w_(2)=-ve)`, path `CA` is isobaric `(w_(3)=+ve)` Total work done by gas `(w)` `=w_(1)+w_(2)+w_(3)` `=0+2.303nRT"log"(V_(B))/(V_(C))+P(V_(C)-V_(A))` `=0+2.303P_(B)V_(B)"log"(V_(B))/(V_(C))+P(V_(C)-V_(A))` `=2.303xx3P_(0)V_(0)"log"(V_(0))/(2V_(0))+P_(0)(2V_(0)-V_(0))` `=-2P_(0)V_(0)+P_(0)V_(0)=-P_(0)V_(0)` Also, `w_(2)=-2P_(0)V_(0)` and `w_(3)=P_(0)V_(0)` also, For the path `AB`, ie. isochoric `q_(1)=nxxC_(P)xx(T_(B)-T_(A))=1xx3/2R[(P_(B)V_(B)-P_(A)V_(A))/R]` `=3/2[3P_(0)V_(0)-P_(0)V_(0)]=+3P_(0)V_(0)` For the path `CA`, i.e., isochoric: `q_(3)=nxxC_(P)xx(T_(A)-T_(B))` `=1xx5/2R[(P_(A)V_(A)-P_(B)V_(B))/R]` `=5/2[P_(0)V_(0)-2P_(0)V_(0)]` `q_(3)=-5/2 P_(0)V_(0)` Also net heat absorbed `=3P_(0)V_(0)-5/2P_(0)V_(0)-=(P_(0)V_(0))/2` `:. q_("net")=(P_(0)V_(0))/2` Also, `(P_(0)V_(0))/(T_(1))=(3P_(0)V_(0))/(T_(2)):. T_(2)=3T_(1)=(3P_(0)V_(0))/R`

Discussion

3162.	Identify the option which correctly represents set of true (T) false (F) statements: Statement :1 In an adiabatic free expansion, entropy of system remains constant. Statement-2: For every isothermal process, internal energy of the system remains contant. Statement-3: Molar enthalpy is an intensive parameter. Statement-4: For every reversible cyclic process, final state of surroundings is same as that of initial state of surroundings.A. TTFTB. FFTFC. FFTTD. TFTF
Answer» Correct Answer - C

Discussion

3163.	One mole of an ideal mono-atomic gas is taken round cyclic process `ABCD` as shown in figure below. Calculate A. The work done by the gas.B. The heat rejected by the gas in the path `CA`and the heat absorbed by the gas in the path `AB`.C. The net heat absorbed by the gas in the path `BC`.D. The maixmum temperature attained by the gas during the cycle.
Answer» Path `AB` is isochoric `(w = 0)`, path `BC` is isothermal `(w_(2) =- ve)`, path `CA` is isobaric `(w_(3) = +ve)`. Total work done by gas, `w = w_(1) +w_(2) +w_(3)` ` = 0+2.303nRT log .(V_(B))/(V_(C)) +P_(0) (V_(C) - V_(A))` `= 0 +2.303 P_(B)V_(B) log.(V_(B))/(V_(C)) +P_(0) (V_(C)-V_(A))` `= 2.303 xx 3P_(0) xx V_(0) log.(V_(0))/(2V_(0)) +P_(0) (2V_(0)-V_(0))` `=- 2P_(0)V_(0) +P_(0)V_(0) =- P_(0)V_(0)` Also, `w_(2) =- 2P_(0)V_(0)` and `w_(3) = P_(0)V_(0)` Also, For the path `AB`, i.e., isochoric `q_(1) = n xx C_(V) xx (T_(B)-T_(A)) = 1 xx(3)/(2)R [(P_(B)V_(B)-P_(A)V_(A))/(R)]` `= (3)/(2) [3P_(0)V_(0) - P_(0)V_(0)] =+3P_(0)V_(0)` For the ptha `CA`, i.e., isochoric: `q_(3) = nxx C_(P) xx (T_(A) - T_(B)) = 1xx(5)/(2)R [(P_(A)V_(A)-P_(B)V_(B))/(R)]` `= (5)/(2) [P_(0)V_(0) - 2P_(0)V_(0)]` `q_(3) =- (5)/(2)P_(0)V_(0)` Also, net heat aboserbed `= 3P_(0)V_(0) - (5)/(2)P_(0)V_(0) = (P_(0)V_(0))/(2)` `:. q_(net) = (P_(0)V_(0))/(2)` Also, `(P_(0)V_(0))/(T_(1)) = (3P_(0)V_(0))/(T_(2)) rArr T_(2) = 3T_(1) = (3P_(0)V_(0))/(R )`

Discussion

3164.	The only incorrect statement for the value of `gamma` for `NH_(3)` gas is : (Assume ideal gas behaviour)A. `gamma=(7)/(5)` at moderate temperatureB. `gamma=(5)/(9)` at very low temperatureC. `gamma=(10)/(9)` at very high temperatureD. `gamma=(7)/(6)` considering only `50%` contribution of vibrational energy.
Answer» Correct Answer - A

Discussion

3165.	2 mol of an ideal gas at `27^(@)C` temperature is expanded reversibly from `2L` to `20L`. Find entropy change `(R = 2 cal mol^(-1) K^(-1))`A. 92.1B. 0C. 4D. 9.2
Answer» Correct Answer - D `DeltaS` (enthropy change) `=2.303 nR "log"_(10)(V_(2))/(V_(1))` `=2.303 xx 2xx2xx"log"_(10)(20)/(2)` `=2.303 xx 2xx2xx1=9212` cal

Discussion

3166.	The heat change at constant volume for the decomposition of silver (I) oxide is found to be `30.66 kJ`. The heat change at constant pressure will beA. `30.66 kJ`B. `gt30.66 kJ`C. `lt30.66 kJ`D. Unpredicatable
Answer» `Ag_(2)O(s) rarr 2Ag(s) +(1)/(2)O_(2)(g)`, `(DeltaU)_(V) = 30.66 kJ` We know that `DeltaH = DeltaU + DeltanRT`, `=30.66 +(3)/(2)xxR xx T ({:(Deltan=n_(P)-n_(R)),(=(5)/(2)-1=(3)/(2)):})` `:. DeltaH gt DeltaU rArr (b)`

Discussion

3167.	The entropy change when 2 moles of an ideal monoatomic gas us subjective to change in state from `(1atm,10L)` to ` (2atm,5L)` will be : `(In 2=0.7)`A. `-2.8cal//K`B. `2.8J//K`C. `-1.4cal//K`D. `5.6cal//K`
Answer» Correct Answer - A

Discussion

3168.	An ideal gas absorbs `600cal` of heat during expansion from `10L` to `20L` against the constant pressure of `2atm`. Calculate the change in internal enegry.
Answer» `w = P(V_(2)-V_(1))` `=2(20-10) L atm = 2xx 10 xx 24.2cal = 484 cal` `:. DeltaU = q - w = 600 - 484 = 116 cal`

Discussion

3169.	Five moles of gas is put through a series of changes as shown graphically in a cyclic process the `A rarr B , B rarr C` and `C rarr A` respectively areA. Isochoric, Isobaric, IsothermalB. Isobaric, Isochoric, IsothermalC. Isothermal, Isobaric, IsochoricD.
Answer» Correct Answer - A `A rarr B`, volume is not changing (Isochoric) `B rarr C` Isobaric `C rarr A` Temperature is constant (Isothermal)

Discussion

3170.	The enthalpy of combustion of `H_(2)` , cyclohexene `(C_(6)H_(10))` and cyclohexane `(C_(6)H_(12))` are `-241` , `-3800` and `-3920KJ` per mol respectively. Heat of hydrogenation of cyclohexene isA. `-121 kJ mol^(-1)`B. `+121 kJ mol^(-1)`C. `-242 kJ mol^(-1)`D. `+242 kJ mol^(-1)`
Answer» Correct Answer - A `H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)O(l) (Delta H = -241 kJ)` ...(i) `C_(6)H_(10) +(17)/(2)O_(2)(g) rarr 6CO_(2)(g)+5H_(2)O(l)` `(Delta H = -3800 kJ)` ...(ii) `C_(6)H_(12)+9O_(2)(g) rarr 6CO_(2)(g)+6H_(2)O(l)` `Delta H = -3920 kJ)` ...(iii) Eq. (i) + eq. (ii) -eq. (iii) gives `Delta H = -241-3800-(-3920) = -121 kJ` for `C_(6)H_(10) + H_(2) rarr C_(6)H_(12)`

Discussion

3171.	Among the following , the state funcation (s) is (are)A. Internal EnergyB. Irresversible Expansion workC. Reversible Expansion workD. Molar Enthylpy
Answer» Correct Answer - A::D State functions do not depend on path followed.

Discussion

3172.	The standard enthalpy of formation of octane `(C_(8)H_(18))` is `-250kJ //mol`. Calculate the enthalpy of combustion of `C_(8)H_(18)`. The enthalpy of formation of `CO_(2)(g)` and `H_(2)O(l)` are `-394 kJ//mol` and `-286kJ//mol` respectively.A. `- 5200` kJ/molB. `- 5726` kJ/molC. `- 5476` kJ/molD. `- 5310` kJ/mol
Answer» Correct Answer - C `C_(8)H_(18)(g)+(25)/(2)O_(2)(g)rarr8CO_(2)(g)+9H_(2)O(l)` `Delta_(r )H^(@)=8xx(-394)+9xx(-286)-(-250)` `=-5476 " kJ"//"mol"`

Discussion

3173.	Benzene and naphthalene form an ideal solution at room temperature. For this process,the true statement(s) is (are)A. `Delta G_("system")` is positiveB. `Delta S_("system")` is positiveC. `Delta S_("surrounding") = 0`D. `Delta H = 0`
Answer» Correct Answer - B::C::D `Delta G = -ve, Delta H = 0, Delta S_("sys") = + ve, Delta S_("surr") = 0`

Discussion

3174.	What is the value of change in internal energy at 1 atm in the process? `H_(2)O(l,323K)rarrH_(2)O(g,423K)` Given : `C_(v,m)(H_(2)O,l)=75.0JK^(-1) mol^(-1):" "C_(p,m)(H_(2)O,g)=33.314JK^(-1)mol^(-1)` `DeltaH_(vap)"ar 373 K"=40.7KJ//mol`A. `52.91 kJ//mol`B. `43086 kJ//mol`C. `42.6kJ//mol`D. `49.6kJ//mol`
Answer» Correct Answer - C `H_(2)O(l,323 K) overset(Delta U_(1))rarr H_(2)O(l, 373 K)` , `Delta U_(2)` `H_(2)O(g, 323K) overset(Delta U_(3))larr H_(2)O(g, 373 K)` `C_(V, m) (H_(2)O, g) = 33.314-8.314` `Delta U_(2) = Delta n_(g)RT = 376`, `Delta U_("total") = Delta U_(1) + Delta U_(2) + U_(3)` `= C_(v, m)(l)Delta T+Delta U_(vap) + C_(vap) + C_(vm)(g) Delta T` `= (75 xx 50)/(1000) + 37.6 + (25 xx 50)/(1000) = 42.6 kJ//mol`

Discussion

3175.	The standard enthalpy of formation of octane `(C_(8)H_(18))` is `-250kJ //mol`. Calculate the enthalpy of combustion of `C_(8)H_(18)`. The enthalpy of formation of `CO_(2)(g)` and `H_(2)O(l)` are `-394 kJ//mol` and `-286kJ//mol` respectively.A. `-5200kJ//mol`B. `-5726 kJ//mol`C. `-5476 kJ//mol`D. `-5310kJ//mol`
Answer» Correct Answer - C `Delta H^(@) = 8 xx (-394)+9 xx (-286)-(-250)` `= -5476 kJ//mol`

Discussion

3176.	For an ideal gas, consider only `P-V` work in going from an initial state `X` to the final state `Z`. The final state `Z` can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct? [Take `DeltaS` as change in entropy and `w` as work done] A. `Delta S_(x rarr z) = Delta S_(x rarr y) + Delta S_(y rarr z)`B. `W_(x rarr z) = W_(x rarr y) + W_(y rarr z)`C. `W_(x rarr y rarr z) = W_(x rarr y)`D. `Delta S_(x rarr y rarr z) = Delta S _(x rarr y)`
Answer» Correct Answer - A::C::D `Delta S_(x rarr z) = Delta S_(x rarr y) + Delta S _(Y + z)` [Enthropy (S) is a state fuction, hecne additive] `W_(X rarr Y rarr Z) = W_(X rarr Y)` (work done in `Y rarr Z` is zero as it is an isochoric process)

Discussion

3177.	In thermodynamics, which one of the following properties is not an intensive property?A. PresssureB. TemperatureC. VolumeD. Density
Answer» Correct Answer - C

Discussion

3178.	Among the following, the intensive property is (properties are):A. molar conductivityB. eletromotive forceC. resistanceD. heat capacity
Answer» Correct Answer - A::B Molar conductivity and electromotive force `(emf)` are intensive properties as these are size independent.

Discussion

3179.	For an ideal gas, consider only `P-V` work in going from an initial state `X` to the final state `Z`. The final state `Z` can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct? [Take `DeltaS` as change in entropy and `w` as work done] A. `DeltaS_(X)to _(z) = DeltaS_(X) to_(Y) + DeltaS_(Y)to _(Z)`B. `W_(X)to _(z) = W_(X) to_(Y) + W_(Y)to _(Z)`C. `W_(X)to _(z)to_(Z) = W_(X) to_(Y)`D. `Delta_(X)to _(Y)to_(Z) = DeltaS_(X) to_(Y)`
Answer» Correct Answer - a,c,d

Discussion

3180.	Among the following , the state funcation (s) is (are)A. internal energyB. irreversible expansion workC. reversible expansion workD. molar enthalpy
Answer» Correct Answer - a,c,d

Discussion

3181.	Identify the extensive quantities from the following (i) refractive (ii) volume (iii) temperature (iv) enthalpyA. enthalpyB. temperatureC. volumeD. refracative index
Answer» Correct Answer - b,d

Discussion

3182.	In the equationn pV = nRT, how many state varibles are present?
Answer» Correct Answer - Dependent varible.

Discussion

3183.	A gas expands against a variable pressure given by `P=(20)/(V)` bar. During expainsion from volume of `1` litre to `10` litre, the gas undergoes a change in internal energy of `400 J`. Heat absorbed by the gas during expansion ( in `KJ`):
Answer» Correct Answer - 5 `Delta W=underset(1)overset(-10)intPdv= -20underset(1)overset(10)int(dv)/(V)=20[ l n(V)]_(1)^(10)= -20 ln(10)= -4605 J` `Delta q=Delta U-Delta w` `= 400-(-4605)=5005 J ~~ 5 KJ`

Discussion

3184.	In an isochoric process if `T_(1)=27^(@)C` and `T_(2)=127^(@)C` then `P_(1)//P_(2)` will be equal toA. `9/59`B. `2/3`C. `3/4`D. None of these
Answer» Correct Answer - D

Discussion

3185.	Calculate the average molar heat capacity at constant volume of a mixture containing 2 moles of monoatomic and 3 moles of diatomic ideal gas.A. RB. 2.1 RC. 3.2 RD. 4 R
Answer» Correct Answer - B

Discussion

3186.	One mole of an ideal gas undergoes a cyclic change ABCD. From the p-V diagram shown below, calculate the net work done in the process.1 atm = 106 dyne cm-2.
Answer» The work done in cyclic process is numerically equal to the area of the cycle. As the cycle ABCD is traced in clock- wise direction, the work done is positive. W = + area ABCD = DC x AD Now, DC = 4 - 3 = 3 litres = 3 x 10³ cm³ ( ∴1 lit = 10³ cm³) and AD = 5 - 2 = 3 atm = 3 x 10⁶dyne cm^-1 work done = 3 x 10³ x 3 x 10⁶ = 9 x 10⁹ ergs.

3186.

One mole of an ideal gas undergoes a cyclic change ABCD. From the p-V diagram shown below, calculate the net work done in the process.1 atm = 106 dyne cm-2.

Answer»

The work done in cyclic process is numerically equal to the area of the cycle.

As the cycle ABCD is traced in clock- wise direction, the work done is positive.

W = + area ABCD

= DC x AD

Now, DC = 4 - 3 = 3 litres

= 3 x 10³ cm³

( ∴1 lit = 10³ cm³)

and AD = 5 - 2 = 3 atm

= 3 x 10⁶dyne cm^-1

work done = 3 x 10³ x 3 x 10⁶

= 9 x 10⁹ ergs.

Discussion

3187.	`84.12g "of gold at" 120.1^(@)C`. Is placed in `106.4g` of `H_(2)O`g at `21.4^(@)C`. What is the final temperature of this system? A. `70.8`B. `65.0`C. `27.8`D. `23.7`
Answer» Correct Answer - D

Discussion

3188.	Refrigerator A works between- `10^(@)C and 27^(@)C` while refrigerator B works between `-20^(@)C and 17^(@)C`, both removing `2000J` from the freezer. Which of the two is better refrigerator?
Answer» Correct Answer - A is better For refrigerator A, `T_(2)=10^(@)C= (-10+273)K=263K` `T_(2)= 27^(@)C=27+273= 300K` `COP=(T_(2))/(T_(1)-T_(2))= (263)/(300-263)= (263)/(37)` For refrigerator B `T_(2)= -20^(@)C=(-20+273)K=253K` `T_(1)= 17^(@)C= (17+273)K=290K` `COP=(T_(2))/(T_(1)-T_(2))= (253)/(290-253)= (253)/(37)` As `(COP)_(A)gt(COP)_(B)` `:.` Refrigeraotr A is better than B.

Discussion

3189.	A factory, producing methanol, is based on the reaction : `CO+2H_(2)rarr CH_(3)OH` ltBRgt Hydrogen and carbon monoxide are obtained by the reaction `CH_(4)+H_(2)Orarr CO+3H_(2)` Three units of factory namely, the "reformer" for the `H_(2)` and CO production, the "methanol reactor" for production of methonol adn a "separator" to separate `CH_(3)OH` form `CO and H_(2)` are schematically shown in figure. The flow of methonal from valve-3 is `10^(3)mol//sec.` The factory is so designed that `(2)/(3)` of the CO is converted to `CH_(3)OH.` Assume that the feromer reaction goes to completion. `CO+2H_(2)rarrCH_(3)OH Delta_(r) H=-100 R` Amount of energy released in methanol reactor in1 minute :A. 1200 kcalB. 12000 kcalC. 6000 kcalD. none of these
Answer» Correct Answer - b

Discussion

3190.	For the reaction `2CO(g) +O_(2)(g) rarr 2CO_(2)(g) ,` `DeltaH=-560 kJ mol^(-1)` In 1 litre vessel at 500 K, the initial pressure is 70 atm and after the reaction, it becomes 40 atm at constant volume of one litre. All the above gases show significant deviation from ideal behavious (1 L atm =0.1 kJ). the change in internal energy will beA. `+10 kJ mol^(-1)`B. `-557 kJ mol^(-1)`C. `-106 kJ mol^(-1)`D. `+106 kJ mol^(-1)`
Answer» Correct Answer - B `DeltaH =DeltaU+Delta nRT`

Discussion

3191.	The enthalpy of fusion of water is`1.435kcal//mol`.The molar entropy change for the melting of ice at`0^(@)C`isA. 10.52 cal/mol KB. 21.04 cal/mol KC. 5.260 cal/mol KD. 0.526 cal/mol K
Answer» Correct Answer - C Molar entropy change for the melting ice, `DeltaS_("melt")=(DeltaH_("fussion"))/(T)=(1.435 kcal//"mol")/((0+273)K)` `=526 xx 10^(-3)` kcal/mol K =526 cal/mol K

Discussion

3192.	The latent heat of fusion of ice is 5.99 kJ/mol at its melting point. Then (i) `DeltaS` for fusion of 900 g ice and (ii) `DeltaS` for freezing of liquid water are respectivelyA. `8698 JK^(-1) , -16.8 J mol^(-1) K^(-1)`B. `9269 JK^(-1) , -12.6 J mol^(-1) K^(-1)`C. `1097 JK^(-1) , -21.9 Jmol^(-1) K^(-1)`D. `1236 JK^(-1) , -28.6 J mol^(-1) K^(-1)`
Answer» Correct Answer - C `DeltaS=(DeltaH)/(T)`

Discussion

3193.	A factory, producing methanol, is based on the reaction : `CO+2H_(2)rarr CH_(3)OH` ltBRgt Hydrogen and carbon monoxide are obtained by the reaction `CH_(4)+H_(2)Orarr CO+3H_(2)` Three units of factory namely, the "reformer" for the `H_(2)` and CO production, the "methanol reactor" for production of methonol adn a "separator" to separate `CH_(3)OH` form `CO and H_(2)` are schematically shown in figure. The flow of methonal from valve-3 is `10^(3)mol//sec.` The factory is so designed that `(2)/(3)` of the CO is converted to `CH_(3)OH.` Assume that the feromer reaction goes to completion. `CO+2H_(2)rarrCH_(3)OH Delta_(r) H=-100 R` What is the flow of `CO and H_(2)` at valve-2 ?A. `CO:500 mol//sec,` `H_(2):1000 mol//sec`B. `CO:1500 mol//sec,` `H_(2):2500 mol//sec`C. `CO:500 mol//sec,` `H_(2):2000 mol//sec`D. `CO:500 mol//sec,` `H_(2):1500 mol//sec`
Answer» Correct Answer - b

Discussion

3194.	Calculate the change in entropy for the following reaction `2CO(g) +O_(2)(g) rarr 2CO_(2)(g)` Given: `S_(CO)^(Theta)(g)=197.6 J K^(-1)mol^(-1)` `S_(O_(2))^(Theta)(g)=205.03 J K^(-1)mol^(-1)` `S_(CO_(2))^(Theta)(g)=213.6 J K^(-1)mol^(-1)`
Answer» `DeltaS^(Theta) =sum (S^(Theta)) ("products") -sum (S^(Theta)) ("reactants")` `=[2 xx S_(CO_(2))^(Theta)(g) - 2 xx S_(CO)^(Theta)+S_(O_(2))^(Theta)]` `= (213.6 xx2) - (2xx197.6 +205.03)` `= 427.2 - 600.23 =- 173.03 J K^(-1) mol^(-1)`

Discussion

3195.	Calculate the entropy change `(DeltaS)` when `1 mol` of ice at `0^(@)C` is converted into water at `0^(@)C`. Heat of fusion of ice at `0^(@)C` is `1436 cal` per mol.
Answer» Entropy change `(DeltaS)` for the change of state of a substance is given by dividing the heat change by the absolute temperature at which the change takes place reversibly, i.e., at the melting point in this case. Hence, `DeltaS = (q_(rev))/(T) = (1436)/(237) = 5.26 cal deg^(-1) mol^(-1)`

Discussion

3196.	One mol of an ideal diatomic gas underwent an adiabatic expansion form `298K, 15.00atm`, and `5.25L` to `2.5atm` against a constant external pressure of `1.00atm`. What is the final temperature of the system?
Answer» This is an isobaric adiabatic expansion against constant externla pressure, but overall pressure decreases (volume increases, gas expands). Final temperature `T_(2)` is given by `P-V-T` relation as: `T_(2)=T_(1)((C_(v)+P_(ex)(R )/(P_(1)))/(C_(v)+P_(ex)(R )/(P_(2))))` For diatomic gas, `C_(v) = (5)/(2) R, T_(1) = 298 K, T_(2) = ?` `P_(2) = 2.50atm, P_(1) = 15.00atm. P_(ex) = 1.900 atm` `:. T_(2) = 298 (((5)/(2)R+(R)/(15))/((5)/(2)R+(R)/(2.5))) = 263.7 K`

Discussion

3197.	Statement-1: The heat absorbed during the adiabatic expansion of an a ideal gas against vacuum is zero. Statement-2:The volume occupied by an ideal gas is zero.A. Statement-1 is True, Statement -2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement -1 is True ,Starement -2 is True ,Statement-2 is not a correct explanation for Statement-1C. Statement-1 is True ,Statement-2 is False.D. Statement-1 is False ,Statement-2 is True.
Answer» Correct Answer - c

Discussion

3198.	In thermodynamics, a process is called reversible whenA. The surroundings and system change into each other.B. There is no boundary between the system and the surroundings.C. The surroundings are always in equilibrium with the system.D. The system changes into the surroundings spontaneously.
Answer» In thermodynamics, a process is called reversible when the surroundings are always in equilibrium with the system.

Discussion

3199.	A gas originally at `1.10atm` and `298K` underwent a reversible adiabatic expansion to `1.00atm` and `287K`. What is the molar heat capacity of the gas?
Answer» Correct Answer - 5 For reversible adiabatic process `((T_(2))/(T_(1)))^(C_(p)//R)=(P_(2))/(P_(1))` `implies (287/298)^(C_(p)//2)=1.0/1.1` `(C_(p))/2xx0.01633=-0.04139` `implies C_(p)=5 "cal "mol^(-1)`

Discussion

3200.	Calculate the `DeltaH` for the isothermal reversible expansion of `1` mole of an ideal gas from initial pressure of `0.1 bar` at a constant temperature at `273K`.
Answer» `P_(1)V_(1)=P_(2)V_(2)` or `PV`=const. (at constant temperature) Thus, `DeltaH=DeltaU+Delta(PV)` `implies DeltaH=0`

Discussion

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