4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 62 InterviewSolution

3051.	The pressure `P_(1)and density d_(1)` of a diatomic gas `(gamma=(7)/(5))` change to `P_(2) and d_(2)` during an adiabatic operation .If `(d_(2))/(d_(1))=32, then (P_(2))/(P_(1))`isA. 76B. 128C. 168D. 298
Answer» Correct Answer - B In an adiabatic operation `P_(1)V_(1)^(gamma)=P_(2)V_(2)^(gamma)` `(P_(2))/(P_(1))=(V_(1))/(V_(2))^(gamma)=((m)/(d_(1)))xx(d_(2))/(m)^(gamma)=((d_(2)/(d_(1)))^(gamma)` `P_(2)/(P_(1)=(32)^(7//5)=(2^(5))^(7//5)=26(7)=128` `(P_(2))/(P_(1))=128`

Discussion

3052.	One mol an ideal gas is expanded from`(10 atm, 10lit)`. `(2 atm , 50 lit)` isothermally . First against 5 atm then against 2 atm . Calculate work done in each step and compare it with single step work done.
Answer» `P_(1)" "V_(2) " "overset(Isothermal)to" "P_(1)" "V_(1)` 10 atm 10 lit `" "`2 atm 50 lit. (i) Work done against 5 atm pressure `(underset("10 atm")(P_(1))" "underset("10 lit")(V_(1)) to underset("5 atm ")(P_(2))underset(" 20 lit")" "V_(2))` Vol. of system at 5 atm `=(10 xx 10)/(5)= 20 lit`. `W_("irrev") = - P_("ext") (V_(2)- V_(1)) = - 5(20-10) =- 50 "atm lit"` (ii) Work done agnist 2 atm `underset("5 atm") (P_(1))" "underset("20 lit")(V_(1))to underset("2 atm") (P_(2))" "underset("50 lit")(V_(2))` `W_("irr")= - P_("ext") (V_(2)- V_(1)) = - 2xx (50-20)= - 60 " atm lit"` `W_("total")= - 50- 60 = - 110" atm lit"` Total work done aganist 2 atm `underset("10 atm") (P_(1))" "underset("10 lit")(V_(1))to underset("2 atm") (P_(2))" "underset("50 lit")(V_(2))` `W = - 2(50-10) = - 80" atm lit"`. Magntiude of work done in more than one step is more than single step work done.

Discussion

3053.	An average personneeds about 10,000 kJ per day. How much carbohydrates ( in mass) will he haveto consume, assuming that all his energy needs are met only by carbohydrates in the form glucose ? Given that the heat of combustion of glucose is 2900 kJ `mol^(-1)`
Answer» Correct Answer - `620.7g` Molar mass of glucose `(C_(6)H_(12)O_(6)) = 180 g mol^(-1)` . Glucose requried per day `=( 180 // 2900 ) xx10,000 g `

Discussion

3054.	Reaction between red phosphorus and liquid bromine is an exothermic reaction represented as follows`:` `2P (s) +3Br_(2)(l) rarr 2P Br_(3)(g),DeltaH^(@) = - 243kJ mol^(-1)` What will be the enthalpy change when 2.63g of phosphorus reacts according to the above reaction. Take atomic mass of phosphorus as 31.0 .
Answer» Correct Answer - `10.3kJ` Required `Delta H = ( 243 ) /( 62)xx 2.63 J = 10.3 kJ`

Discussion

3055.	What is the total number of intensive properties in the giben list ? Internel energy Pressure Molar entropy Volume Density Boiling point Molality
Answer» Correct Answer - 5

Discussion

3056.	Two carnot engine A and B operate respectively between `500 K` and `400 K` , and `400 K` and `300 K`. What is the percentage difference in their efficiencies ?
Answer» `eta_(A)= 1-((T_(2))/(T_(1)))_(A)= 1-(400)/(500)` `1/5= 1/5xx100%= 20%` `eta_(B)= 1-((T_(2))/(T_(1)))_(B)- 1-(300)/(400)` `= 1/4= 1/4xx100= 25%` `eta_(B)-eta_(A)= 25-20 = 5%`

Discussion

3057.	A carnote engine absorbs `8 KJ` of energy at `400K`. If sink is maintained at `300 K`, Calculate useful work done per cyclic (in joule) by the engine.
Answer» Correct Answer - B Here, `T_(1)= 400K, Q_(1)= 8 KJ, W=?, T_(2)= 300 K` As `Q_(2)/(Q_(1))=(T_(2))/(T_(1))` `:. Q_(2)=Q_(1)xx(T_(2))/(T_(1))= 8xx(300)/(400)= 6 J` `W= Q_(1)-Q_(2)= 8-6= 2KJ`

Discussion

3058.	Which of the following reaction is endothermic?A. `CaCO_(3) rarr CaO +CO_(2)`B. `Fe +S rarr FeS`C. `NaOH +HCI rarr NaCI +H_(2)O`D. `CH_(4)+2O_(2) rarr CO_(2) +2H_(2)O`
Answer» Dissociation of `CaCO_(3)` required enegry

Discussion

3059.	Evaporation of water isA. An exothermic changeB. An endothermic changeC. A process where no heat changes occurD. A process accompained by chamical reaction.
Answer» Evaporation of water required heat energy to proceed the reaction.

Discussion

3060.	In an adiabatic expansion of air (assume it a mixture of `N_2` and `O_2`), the volume increases by `5%`. The percentage change in pressure is:
Answer» Correct Answer - 7 `:. PV^(gamma)=`const. `P_(1)xx[(105V)/100]^(gamma)`=const. `:. (P_(1))/Pxx[105/100]^(gamma)` or `(P_(1))/P=[100/105]^(gamma)` or `(P_(1))/P=[100/105]^(7//5)` `:. P_(1)=0.93 P :. 7%` decrease in `P`

Discussion

3061.	In an adiabatic expansion of air (assume it a mixture of `N_2` and `O_2`), the volume increases by `5%`. The percentage change in pressure is:A. `3%`B. `4%`C. `6%`D. `7%`
Answer» Correct Answer - D `PV^(gamma)=` constant `P_(1)xx[(105 V)/100]^(gamma)=` constant `:.(P_(1))/Pxx[105/100]^(gamma)=1` or `(P_(1))/P=[100/105]^(gamma)` or `(P_(1))/P=[100/105]^(7//5)` `:. P_(1)=0.93 P` `:. 7%` decrease in `P`.

Discussion

3062.	In the pressure-volume diagram given below, the isochoric, isothermal, isobaric, and isoentropic parts, respectively, are: A. `BA,AD,DC,DB`B. `DC,CB,BA,AD`C. `AB,BC,CD,DA`D. `CD,DA,AB,BC`
Answer» `CD-` Isochroic (at constant volume) `DA-`Isothermal compression `(DeltaT = 0)` `AB-`Isobaric (at constant pressure) `BC-`Isoentropic (at constant entropy)

Discussion

3063.	An ideal gas with `C_(v)=3 R` expands adiabatically into a vaccum thus doubling its volume. The final temeperature is given by :A. `T_(2)=T_(1)[2^(-1//3)]`B. `T_(2)=T_(1)`C. `T_(2)=2T_(1)`D. `T_(2)=(T_(1))/(2)`
Answer» Correct Answer - B

Discussion

3064.	Molar heat capacity of `CD_(2)O` (deuterated form of formaldehyde) at constant pressure in `9 cal mol^(-1) K^(-1) at 1000K`. Calculate the entropy change associated with cooling of `3.2 g` of `CD_(2)O` vapour from `1000 to 900K`.
Answer» `DeltaH` for cooling `CD_(2)O = mS DeltaT = (3.2)/(32) xx 9 xx 100 = 90 cal` Therefore, cooling is exothermic, `DeltaH =- 90 cal` Now, `DeltaS = (DeltaH)/(T) = - (90)/(900) =- 0.1 cal deg^(-1)`

Discussion

3065.	Calculate the entropy change for vaporization of `1mol` of liquid water to stem at `100^(@)C`, if `Delta_(V)H = 40.8 kJ mol^(-1)`.
Answer» For entropy change of vaporisation, `Delta_(V)S^(Theta) = (Delta_(V)H^(Theta))/(T)` Given, `Delta_(V)H^(Theta) = 40.8 xx 10^(3)J ,T = 373 K` `Delta_(V)S^(Theta) = (40.8 xx 10^(3))/(373) = 109.38 J K^(-1) mol^(-1)`

Discussion

3066.	What will be the final volume of a mole of an idela gas at `20^(@)C` when it expands adiabatically from a volume of `5L` at `30^(@)C? C_(V)` of the gas `=5 cal//degree`.
Answer» Given, `n = 1, T_(2) = (273 +20) = 293 K, V_(1) = 5L, V_(2) = ?, T_(1) = 303K` From the first law, `dq +dw = dU` `dq = 0` (adiabatic process) or `dU =dw` or `C_(V) = int_(1)^(2) (dT)/(T) =- R int_(1)^(2) (dV)/(V)` or `C_(V). In (T_(2))/(T_(1)) =- R In (V_(2))/(V_(1)) = R In (V_(1))/(V_(2))` or `C_(V) log_(10).(T_(2))/(T_(1)) = R log_(10) .V_(1)//V_(2)` or `5 xx log_(10).(303)/(293) = 2 xx log_(10) V_(2)//5` or `5 xx log_(10) 1.034 = 2 xx log_(10) .V_(2)//5` or `0.0725 = 2 xx log_(10) .V_(2)//5` or `0.0362 = log_(10) .V_(2) - log_(10) 5` or `0.0362 - 0.6990 = log_(10) V_(2)` `V_(2) = 5.43L` The specific heat capacity (or specific heat) of a substance is defined as the amount of heat required to raise the temperature of `1g` of the substance through `1^(@)C`.

Discussion

3067.	Calculate the work done by `1.0 mol` of an idela gas when it expands from `10atm` to `2atm` at `27^(@)C`.
Answer» Process is reversible, since the pressure is decreased from `10 atm` to `2atm` in one step. `w_(("irr,isotherm)) =- P_(ex)(V_(2)-V_(1))` `=- P_(ex) ((nRT)/(P_(2))-(nRT)/(P_(1)))` `=- P_(ex) xx nRT ((1)/(P_(2))-(1)/(P_(1)))` `= - 2xx1xx 0.082 xx 300 ((1)/(2)-(1)/(10))` `=- 2 xx 0.082 xx 300 xx (4)/(10)` `= 19.68 L-atm` `= 119.68 xx 101.3J =- 1993.58J` (since `1L-atm = 101.3J)`

Discussion

3068.	Two moles of an idela gas at `2atm` and `27^(@)C` is compressed isothermally to one-half of its volume by an external pressure of `4atm`. Calculate `q,w`, and `DeltaU`.
Answer» Work done on the system, `w =- p DeltaV =- (V_(f)-V_(i))` `p = 4 atm, V_(i) = (nRT)/(p)` `n = 2 mol R = 0.082 atm L mol^(-1) K^(-1)`, `T = 273 +27 = 300K,p = 2 atm` `V_(i) = (2xx 0.082 xx 300)/(2) = 24.6L` and `V_(f) = (V_(i))/(2) = (24.6)/(2) = 12.3L` `w =- 4 atm xx (12.3 - 24.6)L` `= 49.2 L-atm = 49.2 xx 101.3 J = 4984J`. Since, it is isothermal compression, `DeltaU = 0` Now, `DeltaU = q +w` `0 = q + 4984J` or `q =- 4984 J`

Discussion

3069.	Two moles of an ideal gas at 2 atmand `27^(@)C` are compressed isothermally to half the volume by an external pressure of4 atm. Calculate w, q and `DeltaU`.
Answer» Correct Answer - `w= 5150J,Delta U =0, q = - 5150J` `V_("initial")= (nRT)/(P) = (2 mol xx 0.0821 L atm K^(-1) mol^(_1) xx 300 K)/(2 atm) =25.42 L, V_("final") = (25.42)/(2) = 12.71L` `:. w = -P_(ext) DeltaV=- 4 atm xx ( 12.71 -25.42) L= 4 xx 12.71 L atm = 50.84 Latm=50.84 xx 101.3 J= 5150 J`

Discussion

3070.	`2` moles of an ideal gas is compressed from (`1` bar, `2` L) to `2` bar isothermally. Calculate magnitude of minimum possible work in change (in joules ). (Given : `1` bar L = `100` J, ln`2=0.7`)
Answer» Correct Answer - 140

Discussion

3071.	A certain mass of gas expanded from ( 1L, 10 atm) to (4 L, 5 atm) against a constant external pressure of 1 atm. If initial temperature of gas is 300 K and the heat capacity of process is `50 J//^(@)C`. Then the enthalpy change during the process is : (`1"L" "atm" underline ~ 100 J`)A. `Delta H = 15 KJ`B. `Delta H= 15.7 KJ`C. `Delta H=14.4 KJ`D. `DeltaH=14.7 KJ`
Answer» Correct Answer - B `Delta H=Delta E+Delta(PV)& Delta E=q+W=(50xx300-3xx100)J ["as" T_(f)=2xx300 K=600K]=14.7 KJ` `Delta H=14700+10xx100=15700 J=15.7 KJ`.

Discussion

3072.	A certain mass of a gas is expanded from (1L, 10 atm) to (4L, 5 atm) against a constant external pressure of 1 atm . If initial temperature of gas is 300 K and the heat capacity of process is `50 J//^(@)C. Calculate `DeltaH`(in kJ) of the process .(Given 1 L atm = 100 J). (Fill your answer by multiplying it with 100 )
Answer» `DeltaU=Q+W" "(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `=50 xx(600-300)-1(4-1)xx100 " "or" "(10xx1)/(300)=(5xx4)/(T_(2))` `15000 - 300 J " or " " T_(2)=5xx4 xx 30 = 600 K` `=14700 J` `= 14.7 J ` `DeltaH=DeltaU +(P_(2)V_(2)-P_(1) V_(1))` `=14.7 +((20-10)xx100)/(1000) ` `14.7 +1= 15.7 kJ= 1570`

Discussion

3073.	The constant volume molar heat capacity of an ideal gas is expressed by `C_(v, m) = 16.5 +10^(-2)T` (All values are in SI units). If 2.5 mol of this gsa at constant value is heated from `27^(@)C` ro `127^(@)C`, the internal energy increases by 'x' kJ. Hence, x is
Answer» Correct Answer - 5 `Deltau= underset(T_(1))overset(T_(2))(int)nc_(v)dT=n underset(T_(1))overset(T_(2))(int) adT+bTdT` `Deltau=na(T_(2)-T_(1))+(nb)/(2) (T_(2)^(2)-T_(1)^(2))` Given `C_(vm)=a +bT=16.5+10^(-2) T`

Discussion

3074.	A certain gas in expanded from `(1L, 10 atm)` to `(4L, 5 atm)` against a constant external pressure of `1atm`. If the initial temperature of gas is `300K` and heat capacity for the process is `50J^(@)C^(-1)`, the enthalpy change during the process is: (use: `1L-atm = 100J)`
Answer» Correct Answer - 157

Discussion

3075.	Calculate the proton affinity of `NH_(3)(g)` from the following data (in `kJ mol^(-1))`: `DeltaH^(Theta)` dissociation: `H_(2)(g) = 218` `DeltaH^(Theta)` formation: `NH_(3)(g) =- 46` Lattice energy of `NH_(4)CI(s) = 683` Ionisation energy of `H = 1310` Electron affinity of `CI =- 348` Bond dissociation energy `CI_(2)(g) = 124` `Delta_(f)H^(Theta) (NH_(4)CI) =- 314`
Answer» We have to calculate `DeltaH` for the following equation `NH_(3)(g)+H^(o+)(g)rarr 2H(g)` Given `H_(2)(g) +2H(g) rarr 2H(g) DeltaH_(1) = 218 kJ mol^(-1)` `(1)/(2)N_(2)(g) +(3)/(2)H_(2)(g) rarr NH_(3)(g) DeltaH_(2) =- 46 kJ mol^(-1)` `NH_(4)CI(s)rarr NH_(4)^(o+)(g)+CI^(Theta)(g) DeltaH_(3) = 683kJ mol^(-1)` `H(g)rarr H(g)^(**) DeltaH_(4) = 1310 kJ mol^(-1)` `CI(g) rarr CI^(Theta) DeltaH_(5) =- 348 kJ mol^(-1)` `CI_(2)(g) rarr 2CI(g) DeltaH_(6) = 124 kJ mol^(-1)` `(1)/(2)N_(2)(g) +2H_(2)(g)+ (1)/(2)CI_(2)(g) rarr NH_(4)CI(g) DeltaH_(7) = - 314 kJ mol^(-1)` `DeltaH = (1)/(2) DeltaH_(1) - DeltaH_(2) +DeltaH_(3) - DeltaH_(4) - DeltaH_(5) - (1)/(2)DeltaH_(6) +DeltaH_(7)` `=- 718 kJ mol^(-1)`

Discussion

3076.	In certan areas where coal is cheap, artificial gas is produced for household use by the 'water gas' raction `C(s) +H_(2)O(g) underset(600^(@)C)rarr H_(2)(g) +CO(g)` Assuming that coke is `100%` carbon, calculate the maximum heat obtainable at `298K` from the combustion of `1kg` of coke, and compare this value to the maximum heat obtainable at `298K` from burning the water was produced from `1.00 kg` of coke. GIven: `Delta_(f)H^(Theta), H_(2)O(l) =- 68.32 kcal//mol` `Delta_(f)H^(Theta),CO_(2) (g) =- 94.05 kcal//mol` `Delta_(f)H^(Theta),CO(g) =- 26.42 kcal//mol`
Answer» `C(s) +O_(2)(g) rarr CO_(2)(g), DeltaH =- 94.05 kcal mol^(-1)` Maximum heat obtainable form the combustion of `1.0 kf of coke =- (94.05)/(12) xx 1000 =- 7840 cal` Moles of carbon in `1kg` of coke `= (1000)/(12) = 83.3` Maximum heat obtainable from burning the water gas produced from `1kg` of coke `= (83.3) [(-68.32) +(67.6)]` `=- 11320.47 kcal` More energy is obtainable from water gas.

Discussion

3077.	A certain gas in expanded from `(1L, 10 atm)` to `(4L, 5 atm)` against a constnt external pressure of `1atm`. If the initial temperature of gas is `300K` and heat capacity for the process is `50J^(@)C^(-1)`, the enthalpy change during the process is: (use: `1L-atm = 100J)`A. `15 kJ`B. `15.7 kJ`C. `14.3 kJ`D. `14.7kJ`
Answer» Use: `DeltaH = DeltaU + Delta(PV) = DeltaU + (P_(2)V_(1) - P_(1)V_(1))` `DeltaU` can be calculated as follows: i. Use: `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2)) rArr (10 xx 1)/(300) = (5 xx 4)/(T_(2)) rArr T_(2) = 600K` ii. `q = C DeltaT = 50 xx (600 - 300) = 15 kJ` iii. `-w = P_(ex) DeltaV = 1(14-1) = 3 L-atm -= 0.3 kJ` iv. `q = DeltaU +(-w) rArr 15 = DeltaU = 0.3 rArr DeltaU = 14.7 kJ` `rArr DeltaH = 14.7 +(5 xx 4 - 10 xx1) xx 100 xx 10^(-3) = 15.7kJ`

Discussion

3078.	Graph for specific heat at constant volume for a monoatomic gasA. B. C. D.
Answer» For mono-atomic gas `C_(V) = (3)/(2)R`

Discussion

3079.	The following is not an endothermic reactions:A. Combustion of methaneB. Decomposition of waterC. Dehydrogenation of enthane of diamondD. Conversion of graphite to diamond
Answer» All of these three are endothermic processes.

Discussion

3080.	Calculate the entropy change accompanying the following change of state `5 mol` of `O_(2) (27^(@)C, 1 atm) rarr 5 mol of O_(2) (117^(@)C, 5 atm)` `C_(P)` for `O_(2) = 5.95 cal deg^(-1) mol^(-1)`
Answer» `DeltaS_(1) = (q_(rev))/(T) = (W_(rev))/(T) = (1)/(T) xx int_(V_(1))^(V_(2)) P dV` `= (1)/(T) nRT int_(V_(1))^(V_(2)) (dV)/(V) =nR xx 2.303 "log" (V_(2))/(V_(1))` `= 5 xx 1.987 xx 2.303 "log"(P_(1))/(P_(2))` `= 5 xx 1.987 xx 2.303 "log"(1)/(5) =- 16.0 cal deg^(-1)` `DeltaS_(2) =` for the change of temperature form `27^(@)C to 117^(@)C` `DeltaS_(2) nC_(P)int_(T_(1))^(T_(2))(dT)/(T)` `= 5 xx 6.95 xx 2.303 "log" (390)/(300)` `= 5 xx 6.95 xx 2.303 xx 0.1139` `= 9.12 cal deg^(-1)` `DeltaS =- 16.0 + 9.12 =- 6.88 cal deg^(-1)`

Discussion

3081.	Oxygen gas weighing `128 g` is expanded form `1atm` to `0.25 atm` at `30^(@)C`. Calculate entropy change, assuming the gas to be ideal.
Answer» `n = (W)/("Molecular weight") = (128)/(32) = 4` `DeltaS = 2.303 nR log_(10) ((P_(1))/(P_(2)))` `= 2.303 xx 4xx 8.314 log_(10) ((1)/(0.25))` `= 46.10 J K^(-1)`

Discussion

3082.	The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of `10 dm^(3)` to a volume of `100dm^(3)` at `27^(@)C` isA. `38.3 J mol^(-1) K^(-1)`B. `35.8 J mol^(-1) K^(-1)`C. `32.3 J mol^(-1) K^(-1)`D. `42.3 J mol^(-1) K^(-1)`
Answer» Correct Answer - a Entropy change for an isothermal process is `DeltaS=2.303R log((V_(2))/(V_(1)))` `DeltaS=2.303xx2xx8.314xxlog(100/10)` `=38.294 J mol^(-1) K^(-1)`

Discussion

3083.	Calculate the entropy change accompanying the following change of state `H_(2)O (s, -10^(@)C, 1 atm) rarr H_(2)O(l, 10^(@)C, 1atm)` `C_(P)` for ice `= 9 cla deg^(-1) mol^(-1)` `C_(P)` for `H_(2)O = 18 cal deg^(-1) mol^(-1)` Latent heat of fustion of ice `= 1440 cal mol^(-1) at 0^(@)C`.
Answer» Step 1. (using the third law of thermodynamics): (For changing `H_(2)O(s) (-10^(@)C, 1 atm) rarrH_(2)O(s, 0^(@)C1 atm)` `DeltaS_(1) =int_(-10)^(0) n(C_(P))/(T)dT = 1 xx 9 xx 2.3 xx log(273)/(263)` `= 0.336 cal deg^(-1) mol^(-1)` Step 2 (using the second law of thermodynamics): `H_(2)O(s) (0^(@)C, 1atm) rarr H_(2)O(l) (10^(@)C, 1atm)` `DeltaS_(2) = (q_(rev))/(T) = (1440)/(273) = 5.258 cal deg^(-1) mol^(-1)` Step 3 (using the third law of themrodynamics): `H_(2)O(l) (0^(@)C, 1atm) rarr H_(2)O(l)(10^(@)C, 1 atm)` `DeltaS_(3) =int_(0)^(10) n(C_(P))/(T) dT = 1 xx 18 xx 2.3 xx log (283)/(273)` `= 0.647 cal deg^(-1) mol^(-1)` `DeltaS = DeltaS_(1) +DeltaS_(2) +DeltaS_(3) = 0.336 +5.258 +0.647` `=6.258 cal deg^(-1) mol^(-1)`

Discussion

3084.	Identify intensive property from the followingA. EnthalpyB. TemperatureC. VolumeD. Refractive index
Answer» The quantities that do not depend upon the quantity or mass of a substance are called intensive properties.

Discussion

3085.	The species which by definition has zero standard molar enthalpy of formation at `298K` isA. `Br_(2(g))`B. `Cl_(2(g))`C. `H_(2)O_((g))`D. `CH_(4(g))`
Answer» Correct Answer - b `Cl_(2(g))` is considered as standard state of `Cl_(2)` and by assumption standard molar enthalpy of formation in standard state is taken as zero.

Discussion

3086.	Which of the following are not state functions? (I) `q+w` (II)`q` (III) `w` (IV) `H-TS`A. (i) and (iv)B. (ii),(iii),(iv)C. (i),(ii),(iii)D. (ii) and (iii)
Answer» Correct Answer - D According to thermodynamics, `Delta U = q + w` `G = H - TS` Internal energy `(U)` and Gibbs energy `(G)` are state functions. Heat `(q)` and work `(w)` are not state functions, they are path functions.

Discussion

3087.	For the gas phase reaction `PCl_(5)rarrPCl_(3)(g)+Cl_(2)(g)` which of the following conditions are correct?A. `Delta H = lt` and `Delta S lt 0`B. `Delta H gt 0` and `Delta S gt 0`C. `Delta H lt 0` and `Delta S lt 0`D. `Delta H gt 0` and `Delta S lt 0`
Answer» Correct Answer - B The reaction involves the dissociation of `PCl_(5)`, an energy consuming. Thus, `Delta H` is positive `(gt 0)`. During the course of reaction, there is an increase in the number of gaseous species and, hence `Delta S` is also positive `(gt 0)`.

Discussion

3088.	The species which by definition has zero standard molar enthalpy of formation at `298K` isA. `Br_(2)(g)`B. `CI_(2)(g)`C. `H_(2)O(g)`D. `CH_(4)(g)`
Answer» This is possible only for elements, chlorine is a gas at this temperature, but bromines is a liquid, so it is possible only for chlorine .

Discussion

3089.	Assertion (A) : There is a natural asymmetry between converting work to heat and converting heat of work. Reason (R ) : No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.A. If both (A) and (R ) are correct, and (R ) is the correct explanation for (A).B. If the both (A) and (R ) are correct, but(R ) is not a correct explanation for (A).C. If (A) is correct, but (R ) is incorrect.D. If both (A) and (R ) are incorrect.
Answer» (A) is true, (R ) is true, (A) is the correct explanation for (R ). There is a natural asymmetry between converting work to heat and converting heat to work. No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. this is in accordance with the second law of thermodynamics.

Discussion

3090.	Assertion (A): For every chemical reaction at equilibrium, standard Gibbs enegry of the reaction is zero. Reason (R ) : At constant temperature and pressure chemical reactions are spontaneous in the direction of the decreasing Gibbs energy.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` are correct and `E`C. Both `S` and `E` are correct but `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`.
Answer» Correct Answer - b AT equilibrium `DeltaG=0`, but standard Gibbs energy `(DeltaG^(2))` of a reaction may or may not be zero. For reaction to be spontaneousn `DeltaG`(Gibbs energy) should be more negative, i.e., `DeltaGgt0`.

Discussion

3091.	Assertion: Work and internal energy are not state functions. Reason: The sum of `q+w` is a state function.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` are correct and `E`C. Both `S` and `E` are correct but `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`.
Answer» Correct Answer - b `w` and `q` are not state functions whereas `DeltaU=q+w` is state function.

Discussion

3092.	Statement: Internal energy of a system is an extensive property. Explanation: The internal energy of a system depends upon the amount and physical state of the susbtance.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` are correct and `E`C. Both `S` and `E` are correct but `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`.
Answer» Correct Answer - c Reson is correct explanation for statement.

Discussion

3093.	Assertion (A): For every chemical reaction at equilibrium, standard Gibbs enegry of the reaction is zero. Reason (R ) : At constant temperature and pressure chemical reactions are spontaneous in the direction of the decreasing Gibbs energy.A. Both (A) and (R ) are correct, and (R ) is the correct explanation for (A).B. Both (A) and (R ) are correct, but(R ) is not a correct explanation for (A).C. (A) is correct, but (R ) is incorrect.D. (A) is incorrect, but (R ) is correct.
Answer» (A) is false, (R ) is true. For every chemical reaction at equilibrium standard Gibbs energy of the reaction is not zero. It is zero only for the reactions at equilibrium. At constant temperature and pressure, the chemical reactions are spontaneous in the direction of the decreasing Gibbs energy. For a reaction to be spontaneous, the free energy should decreases.

Discussion

3094.	Assertion (A): For every chemical reaction at equilibrium, standard Gibbs enegry of the reaction is zero. Reason (R ) : At constant temperature and pressure chemical reactions are spontaneous in the direction of the decreasing Gibbs energy.A. statement I is true, Statement II is true, Statement II is the correct explanation of Statement IB. Statement I is true , statement II is true , Statement II is not the correct explanation of Statement IC. Statement I is true, Statement II is falseD. Statement I is false, Statement II is true
Answer» Correct Answer - d

Discussion

3095.	Statement -1. For every chemical reaction at equilibrium ,standard Gibbs energy of reactions is zero. Statement-2. At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energyA. Statement -1 is True, Statement-2 is True, Statement-2 is a correct explanation of Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement -1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True
Answer» Correct Answer - d

Discussion

3096.	Two what type of system the following belong ? (i) Tree (ii) Pond(iii) Animals (iv) Tea placed in a kettle (v) Tea placed in thermos flask(vi) Tea placed in a cup.
Answer» (i) Open system (ii) Open system (iii) Open system (iv) closed system (v) isolated system (vi) open system.

Discussion

3097.	What are spontaneous reactions? What are the conditions for the spontaneity of a process?
Answer» 1. A reaction that does occur under the given set of conditions is called a spontaneous reaction. 2. The expansion of a gas into a evacuated bulb is a spontaneous process, the reverse process that is gathering of all molecules into one bulb is not spontaneous. This example shows that processes that occur spontaneously in one direction cannot take place in opposite direction spontaneously. 3. Increase in randomness favours a spontaneous change. 4. If enthalpy change of a process is negative, then the process is exothermic and occurs spontaneously. Therefore ∆H should be negative. 5. if entropy change of a process is positive, then the process occurs spontaneously, therefore ∆S should be positive. 6. If free energy of a process is negative, then the process occurs spontaneously, therefore ∆G should be negative. 7. For a spontaneous. irreversible process. ∆H 0, ∆G < 0. i.e., ∆H = -ve, ∆S = +ve and ∆G = -ve.

3097.

What are spontaneous reactions? What are the conditions for the spontaneity of a process?

Answer»

1. A reaction that does occur under the given set of conditions is called a spontaneous reaction.

2. The expansion of a gas into a evacuated bulb is a spontaneous process, the reverse process that is gathering of all molecules into one bulb is not spontaneous. This example shows that processes that occur spontaneously in one direction cannot take place in opposite direction spontaneously.

3. Increase in randomness favours a spontaneous change.

4. If enthalpy change of a process is negative, then the process is exothermic and occurs spontaneously. Therefore ∆H should be negative.

5. if entropy change of a process is positive, then the process occurs spontaneously, therefore ∆S should be positive.

6. If free energy of a process is negative, then the process occurs spontaneously, therefore ∆G should be negative.

7. For a spontaneous. irreversible process. ∆H 0, ∆G < 0.

i.e., ∆H = -ve, ∆S = +ve and ∆G = -ve.

Discussion

3098.	What is the condition spontaneity in terms of free energy change?
Answer» 1. If ∆G is negative, process is spontaneous. 2. if ∆G is positive, process is non-spontaneous. 3. if ∆G = 0, the process is in equilibrium.

3098.

What is the condition spontaneity in terms of free energy change?

Answer»

1. If ∆G is negative, process is spontaneous.

2. if ∆G is positive, process is non-spontaneous.

3. if ∆G = 0, the process is in equilibrium.

Discussion

3099.	What is the mathematical from of the first law of the thermodynamics ?
Answer» The first law of thermodynamics is the law of conservation of energy. It states that if a system absorbs heat dQ and as a result the internal energy of the system changes by dU and the system does a work dW, then dQ = dU + dW But, dW = PdV Hence, dQ = dV + pdV.

3099.

What is the mathematical from of the first law of the thermodynamics ?

Answer»

The first law of thermodynamics is the law of conservation of energy. It states that if a system absorbs heat dQ and as a result the internal energy of the system changes by dU and the system does a work dW, then

dQ = dU + dW

But, dW = PdV

Hence, dQ = dV + pdV.

Discussion

3100.	In a refrigerator, heat from inside at `277 K` is transferred to a room at `300K`. What is the cofficient of performance of the refrigerator. How many joule of heat will be deliverd to the room for each joule of electric energy consumed ideally?
Answer» Correct Answer - `12.04 ; 13.04J` Here, `T_(2)= 277K` `T_(1)= 300K, COP=? Q_(1)=?` `W=1J` `COP= (Q_(2))/W=(T_(2))/(T_(1)-T_(2))` `COP=(273)/(300-277)= (273)/(23)=12,04` Again, as `COP= (Q_(2))/W=(Q_(2))/(1J)= 12.4` `Q_(2)= 12.04J` `Q_(1)=Q_(2)+W= 12.04+1= 13.04J`

Discussion

Explore topic-wise InterviewSolutions in .

The pressure `P_(1)and density d_(1)` of a diatomic gas `(gamma=(7)/(5))` change to `P_(2) and d_(2)` during an adiabatic operation .If `(d_(2))/(d_(1))=32, then (P_(2))/(P_(1))`isA. 76B. 128C. 168D. 298

One mol an ideal gas is expanded from`(10 atm, 10lit)`. `(2 atm , 50 lit)` isothermally . First against 5 atm then against 2 atm . Calculate work done in each step and compare it with single step work done.

An average personneeds about 10,000 kJ per day. How much carbohydrates ( in mass) will he haveto consume, assuming that all his energy needs are met only by carbohydrates in the form glucose ? Given that the heat of combustion of glucose is 2900 kJ `mol^(-1)`

What is the total number of intensive properties in the giben list ? Internel energy Pressure Molar entropy Volume Density Boiling point Molality

Two carnot engine A and B operate respectively between `500 K` and `400 K` , and `400 K` and `300 K`. What is the percentage difference in their efficiencies ?

A carnote engine absorbs `8 KJ` of energy at `400K`. If sink is maintained at `300 K`, Calculate useful work done per cyclic (in joule) by the engine.

Which of the following reaction is endothermic?A. `CaCO_(3) rarr CaO +CO_(2)`B. `Fe +S rarr FeS`C. `NaOH +HCI rarr NaCI +H_(2)O`D. `CH_(4)+2O_(2) rarr CO_(2) +2H_(2)O`

Evaporation of water isA. An exothermic changeB. An endothermic changeC. A process where no heat changes occurD. A process accompained by chamical reaction.

In an adiabatic expansion of air (assume it a mixture of `N_2` and `O_2`), the volume increases by `5%`. The percentage change in pressure is:

In an adiabatic expansion of air (assume it a mixture of `N_2` and `O_2`), the volume increases by `5%`. The percentage change in pressure is:A. `3%`B. `4%`C. `6%`D. `7%`

In the pressure-volume diagram given below, the isochoric, isothermal, isobaric, and isoentropic parts, respectively, are: A. `BA,AD,DC,DB`B. `DC,CB,BA,AD`C. `AB,BC,CD,DA`D. `CD,DA,AB,BC`

An ideal gas with `C_(v)=3 R` expands adiabatically into a vaccum thus doubling its volume. The final temeperature is given by :A. `T_(2)=T_(1)[2^(-1//3)]`B. `T_(2)=T_(1)`C. `T_(2)=2T_(1)`D. `T_(2)=(T_(1))/(2)`

Molar heat capacity of `CD_(2)O` (deuterated form of formaldehyde) at constant pressure in `9 cal mol^(-1) K^(-1) at 1000K`. Calculate the entropy change associated with cooling of `3.2 g` of `CD_(2)O` vapour from `1000 to 900K`.

Calculate the entropy change for vaporization of `1mol` of liquid water to stem at `100^(@)C`, if `Delta_(V)H = 40.8 kJ mol^(-1)`.

What will be the final volume of a mole of an idela gas at `20^(@)C` when it expands adiabatically from a volume of `5L` at `30^(@)C? C_(V)` of the gas `=5 cal//degree`.

Calculate the work done by `1.0 mol` of an idela gas when it expands from `10atm` to `2atm` at `27^(@)C`.

Two moles of an idela gas at `2atm` and `27^(@)C` is compressed isothermally to one-half of its volume by an external pressure of `4atm`. Calculate `q,w`, and `DeltaU`.

Two moles of an ideal gas at 2 atmand `27^(@)C` are compressed isothermally to half the volume by an external pressure of4 atm. Calculate w, q and `DeltaU`.

`2` moles of an ideal gas is compressed from (`1` bar, `2` L) to `2` bar isothermally. Calculate magnitude of minimum possible work in change (in joules ). (Given : `1` bar L = `100` J, ln`2=0.7`)

The constant volume molar heat capacity of an ideal gas is expressed by `C_(v, m) = 16.5 +10^(-2)T` (All values are in SI units). If 2.5 mol of this gsa at constant value is heated from `27^(@)C` ro `127^(@)C`, the internal energy increases by 'x' kJ. Hence, x is

A certain gas in expanded from `(1L, 10 atm)` to `(4L, 5 atm)` against a constant external pressure of `1atm`. If the initial temperature of gas is `300K` and heat capacity for the process is `50J^(@)C^(-1)`, the enthalpy change during the process is: (use: `1L-atm = 100J)`

Graph for specific heat at constant volume for a monoatomic gasA. B. C. D.

The following is not an endothermic reactions:A. Combustion of methaneB. Decomposition of waterC. Dehydrogenation of enthane of diamondD. Conversion of graphite to diamond

Calculate the entropy change accompanying the following change of state `5 mol` of `O_(2) (27^(@)C, 1 atm) rarr 5 mol of O_(2) (117^(@)C, 5 atm)` `C_(P)` for `O_(2) = 5.95 cal deg^(-1) mol^(-1)`

Oxygen gas weighing `128 g` is expanded form `1atm` to `0.25 atm` at `30^(@)C`. Calculate entropy change, assuming the gas to be ideal.

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of `10 dm^(3)` to a volume of `100dm^(3)` at `27^(@)C` isA. `38.3 J mol^(-1) K^(-1)`B. `35.8 J mol^(-1) K^(-1)`C. `32.3 J mol^(-1) K^(-1)`D. `42.3 J mol^(-1) K^(-1)`

Calculate the entropy change accompanying the following change of state `H_(2)O (s, -10^(@)C, 1 atm) rarr H_(2)O(l, 10^(@)C, 1atm)` `C_(P)` for ice `= 9 cla deg^(-1) mol^(-1)` `C_(P)` for `H_(2)O = 18 cal deg^(-1) mol^(-1)` Latent heat of fustion of ice `= 1440 cal mol^(-1) at 0^(@)C`.

Identify intensive property from the followingA. EnthalpyB. TemperatureC. VolumeD. Refractive index

The species which by definition has zero standard molar enthalpy of formation at `298K` isA. `Br_(2(g))`B. `Cl_(2(g))`C. `H_(2)O_((g))`D. `CH_(4(g))`

Which of the following are not state functions? (I) `q+w` (II)`q` (III) `w` (IV) `H-TS`A. (i) and (iv)B. (ii),(iii),(iv)C. (i),(ii),(iii)D. (ii) and (iii)

For the gas phase reaction `PCl_(5)rarrPCl_(3)(g)+Cl_(2)(g)` which of the following conditions are correct?A. `Delta H = lt` and `Delta S lt 0`B. `Delta H gt 0` and `Delta S gt 0`C. `Delta H lt 0` and `Delta S lt 0`D. `Delta H gt 0` and `Delta S lt 0`

The species which by definition has zero standard molar enthalpy of formation at `298K` isA. `Br_(2)(g)`B. `CI_(2)(g)`C. `H_(2)O(g)`D. `CH_(4)(g)`

Two what type of system the following belong ? (i) Tree (ii) Pond(iii) Animals (iv) Tea placed in a kettle (v) Tea placed in thermos flask(vi) Tea placed in a cup.

What are spontaneous reactions? What are the conditions for the spontaneity of a process?

What is the condition spontaneity in terms of free energy change?

What is the mathematical from of the first law of the thermodynamics ?

In a refrigerator, heat from inside at `277 K` is transferred to a room at `300K`. What is the cofficient of performance of the refrigerator. How many joule of heat will be deliverd to the room for each joule of electric energy consumed ideally?