4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 22 InterviewSolution

1051.	Which one of the following gases possesses the largest internal energyA. 2 moles of helium occupying `1m^(2) at 300 k`B. 56 kg nitrogen at `107Nm^(2)` and 300 kC. 8 grams of oxygen at 8 atm and 300 kD. `6xx10^(26)` moleculer of oxygen occupying `40m^(2)` at 900 k
Answer» Correct Answer - B

Discussion

1052.	If the work done by a system on its surroundings is 100 J and the increase in the internal energy of the system is 100 cal, what must be the heat supplied to the system? (Given : J = 4.186J/cal)
Answer» Data : W = 100 J, ∆ U = 100 cal, J = 4.186 J / cal The heat supplied to the system, Q = ∆ U + W = (100 cal) (4.186 J/cal) + 100 J = 418.6J + 100J = 518.6 J

1052.

If the work done by a system on its surroundings is 100 J and the increase in the internal energy of the system is 100 cal, what must be the heat supplied to the system? (Given : J = 4.186J/cal)

Answer»

Data : W = 100 J, ∆ U = 100 cal, J = 4.186 J / cal

The heat supplied to the system,

Q = ∆ U + W = (100 cal) (4.186 J/cal) + 100 J = 418.6J + 100J = 518.6 J

Discussion

1053.	Which one of the following gases possesses the largest internal energyA. 2 moles of helium occupying 1 `m^(3)` at 300 KB. 56 kg of nitrogen at `10^(7) Nm^(-2)` and 300 KC. 8 grams of oxygen at 8 atm and 300 KD. `6 xx 10^(26)` molecules of argon occupying `40 m^(3)` at 900 K
Answer» Correct Answer - c

Discussion

1054.	When air of the atmosphere rises up, it cools. Why?
Answer» When air rises up, it expands due to decrease in the atmospheric pressure. Therefore, temperature falls.

Discussion

1055.	What is the value of specific heat of water in S.I. units? Does it vary with temperature?
Answer» Specific heat of water is 4180 J kg^-1 K^-1.Yes, it does vary very little with temperature.

Discussion

1056.	Define Radiation.
Answer» The process of the transfer of heat from one place to another place without heating the intervening medium is called radiation.

Discussion

1057.	State second law of thermodynamics.
Answer» Claussius statement-" It is impossible for a self-acting machine, unaided by any external agency, to conduct heat from one body to another at higher temperature".

Discussion

1058.	What is the value of work done in cyclic process?
Answer» The work done in a cyclic process is equal to the area of the closed curve on a p-V diagram.

Discussion

1059.	Find the fraction of molecules whose velocity projections on the `x` axis fall within the inerval from `v_x` to `v_x + dv_x`, while the moduli of perpendicular velocity components fall within the inerval from `v_( _\|_)` to `v_(_\|_) + dv_(_\|_)`. The mass of each molecule is `m`, and the temperature is `T`.
Answer» `d^3 v = 2 pi v_(_\|_) dv_(_\|_) dx_x` Thus `d n(v) = N((m)/(2 pi kT))^(3//2) e-(m)/(2 kT) (v_x^2 + v_(_\|_)^2) dv_x 2 pi v_(_\|_) dv _(_\|_)`.

Discussion

1060.	Unit mass of liquid of volume `V_(1)` completely turns into a gas of volume `V_(2)` at constant atmospheric pressure P and temperature T. The latent heat of vaporization is "L". Then the change in internal energy of the gas isA. ZeroB. `P(V_(1)-V_(2))`C. `L-P(V_(2)-V_(1))`D. L
Answer» Correct Answer - C

Discussion

1061.	A gas expands `0.25m^(2)` at constant perssure `10^(3)N//m^(2),` the work done isA. 2.5 ergsB. 250 jC. 250 WD. 340 N
Answer» Correct Answer - C

Discussion

1062.	Calculate the proton affinity of `NH_(3)` (g) from the following data (in kJ/mol): `{:(DeltaH^(@)"dissociation":H_(2)(g),=436),(DeltaH^(@)"formation":NH_(3)(g),=-46),("Lattic energyof"NH_(4)Cl(s),=-683),("Ionisation enorgy of"H,=130),("Electron affinity of"Cl,=380),(DeltaH^(@)"dissociation" :Cl_(2)(g),=240),(DeltaH^(@)"formation of"NH_(4)Cl(s),=-314):}`
Answer» Correct Answer - 327

Discussion

1063.	Determine enthalpy change for, `C_(3)H_(8(g))+H_(2(g))rarr C_(2)H_(6(g))+CH_(4(g))` at `25^(@)C` using heat of combustion values under standard condition. `{:(Compounds,H_(2(g)),CH_(4(g)),C_(2)H_(6(g)),C_((Graph ite))),(DeltaH^(@)i n kJ//mol,-285.8,-890.0,-1560.0,-393.5):}` The standard heat of formation of `C_(3)H_(8(g))` is `-103.8 kJ mol^(-1)`.
Answer» Correct Answer - 55

Discussion

1064.	Consider a Carnot’s cycle operating between T1 = 500 K and T2 = 300 K producing 1 KJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.
Answer» Given : Temperature of the source, T₁= 500 K Temperature of the sink, T₂ = 300 K Work done per cycle, W = 1 kJ = 1000 J Heat transferred to the engine per cycle Q₁ = ? Efficiency of Carnot engine (η) = 1 − \(\frac{T_2}{T_1}\) = 1 − \(\frac{300}{500}\) = \(\frac{2}{5}\) and η = \(\frac{W}{Q_1}\) or Q₁ = \(\frac{W}{η}\) = \(\frac{1000}{2/5}\)= 2500 J.

1064.

Consider a Carnot’s cycle operating between T1 = 500 K and T2 = 300 K producing 1 KJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.

Answer»

Given :

Temperature of the source,

T₁= 500 K

Temperature of the sink,

T₂ = 300 K

Work done per cycle,

W = 1 kJ = 1000 J

Heat transferred to the engine per cycle

Q₁ = ?

Efficiency of Carnot engine

(η) = 1 − \(\frac{T_2}{T_1}\) = 1 − \(\frac{300}{500}\) = \(\frac{2}{5}\)

and η = \(\frac{W}{Q_1}\)

or Q₁ = \(\frac{W}{η}\) = \(\frac{1000}{2/5}\)= 2500 J.

Discussion

1065.	The correct thermodynamics conditions for the spontaneous reaction at all temperature isA. `Delta H lt 0` and `Delta S gt 0`B. `DeltaH lt 0` and `DeltaS lt 0`C. `Deltah lt 0` and `DeltaS =0`D. `DeltaH gt 0` and `Delta S lt 0`
Answer» Correct Answer - a,c A reaction is spontaneous at all temperatures,if both energy factor an anetropy factor favour the process, i.e., `DeltaHlt0` and`DeltaSgt 0` or when energy factor favours but entropy factor has no roleto play, i.e.,`DeltaHlt0` and`DeltaS = 0`.

Discussion

1066.	What is an adiabatic process ?
Answer» A process during which no heat flows between the system and the surroundings is called an adiabatic process.

Discussion

1067.	For a sample of perfect gas when its pressure is changed isothermally from `p_(i)` to `p_(f)` , the entropy change is given byA. `DeltaS= nRln((p_(f))/( p_(i)))`B. `DeltaS= nRln((p_(i))/( p_(f)))`C. `DeltaS= nRTln((p_(f))/( p_(i)))`D. `DeltaS= nRTln((p_(i))/( p_(f)))`
Answer» Correct Answer - b For n moles of an ideal gas undergoing reversible expansion when temperature changes from`T_(1)` to`T_(2)`and pressure changes from `p_(1)` to `p_(2)`, entropy change is given by `DeltaS = n C_(p) ln. (T_(2))/(T_(1)) + nRln. (P_(1))/(P_(2)) ` or `DeltaS = nC_(p)ln. (T_(f))/(T_(i))+ nRTln. (p_(i))/(p_(f))` For isothermal expansion, `T_(i) = T_(f)` so that `ln. (T_(f))/(T_(i)) = ln 1= 0`. Hence, `DeltaS = nR ln. (p_(i))/( p_(f))`

Discussion

1068.	Why internal energy is a state function but work is not ?
Answer» The change in internal energy during a process depends only upon the initial state and final state while work depends upon the path followed.

Discussion

1069.	Solutions A and B are both clear and colourles. When solution A is mixed with solution B, the temperature of the mixture increases and a yellow precipitate is observed. What can be concluded from these observations?A. The reaction is thermodynamically favoured ( spontaneous) at all temperaturesB. The reaction is thermodynamically favoured ( spontaneous) only athigh temperatureC. The reaction is thermodynamically favoured ( spontaneous) only at low temperatureD. The reaction is not thermodynamically favoured ( spontaneous) at any temperature.
Answer» Correct Answer - C Since on mixing , temperature increses, this means that reaction is exothermic, i.e., energy factor favours the process.Since a precipitate is formed on mixing, this means entropy decreases,i.e.,entropy factor opposes the process. `DeltaG= DeltaH-T DeltaS`. For reaction to be spontaneous, `DeltaG` should be `-ve`. As `DeltaH` is`-ve` and`DeltaS `is `-ve`, `DeltaG` can be -ve only if T is low.Hence, reaction will be spontaneous at low temeprature.

Discussion

1070.	Which of the following reaction is associated with an increase in entropy?A. `N_(2(g))+3H_(2(g)) rarr 2NH_(3(g))`B. `2H_(2(g))+O_(2(g)) rarr 2H_(2)O_((l))`C. `H_(2(g))+I_(2(g)) rarr 2HI_((g))`D. `C("graphite")+H_(2)O(g) rarr CO_((g))+H_(2(g))`
Answer» Correct Answer - D

Discussion

1071.	The value of `DeltaS` is negative for the processA. Burning of rocket FuelB. Dissolution of sugarC. Sublimation of IodineD. Freezing of water
Answer» Correct Answer - D

Discussion

1072.	`1` mole of an ideal gas at `25^(@)` C is subjected to expand reversibly and adiabatically to ten times of its initial volume. Calculate the change in entropy during expansion (in `J K^(-1) mol^(-1)`)A. `19.15`B. `-19.15`C. `4.7`D. zero
Answer» Correct Answer - D `DeltaS = nC_(v) l"n" (T_(2))/(T_(1)) + nR l"n" (V_(2))/(V_(1))` For adiabatic process `(Q-0)` `DeltaE = W " " implies" "nC_(v) ln(T_(2))/(T_(1)) =-nR ln (V_(2))/(V_(1)) " " therefore" "DeltaS=0`

Discussion

1073.	An insulated container is divided into two equal portions. One portion contains as ideal gas at pressure `P` and tenperature `T`. The other portion is a perfect vaccum. If a hole is opened between the two portions, claculate a. the change in internal energy. b. the change in temperature.
Answer» Correct Answer - No change in `T` and `U`;

Discussion

1074.	`1 mol` of an ideal gas at `25^(@)C` is subjected to expand reversibly `10` times of its initial volume. Calculate the change in entropy of expansions.
Answer» Correct Answer - `19.15 JK^(-1)mol^(-1)`;

Discussion

1075.	If `DeltaH_(f)^(@)` for `Ag^(+)` (infinately diluted), `NO_(3)^(-)` (infinity diluted), `Cl^(-)` (infinitely diluted) and AgCl(s) are 105.579, `-207.36`, `-167.159` and `-127.068` respectively. Calculate the enthalpy change for the reaction `AgNO_(3)(aq)+HCl(aq)toAgCl(s)+HNO_(3)(aq)`A. 21.471 KJ`//`molB. 145.688 KJ`//`molC. 65.488 KJ`//`molD. none of these
Answer» Correct Answer - C

Discussion

1076.	A system has internal energy equal to `U_(1), 450J` of heat is taken out of it and `600J` of work is done on it. The final enegry of the system will beA. `(U_(1) +150)`B. `(U +1050)`C. `(U_(1) - 150)`D. None of these
Answer» Initial internal energy `= U_(1)` Heat taken out `=- 450J` Work done on the system `= 600J` `:.` Internal enegry `- q +w` `=- 450 + 600 = 150J` `:.` Total enegry `=[U_(1) +150]`

Discussion

1077.	The molar heat capacity for a gas at constant `T` and `P` isA. `(3)/(2)R`B. `(5)/(2)R`C. Dependent on the atomicity of the gasD. Infinity
Answer» `C_(P) = ((dH)/(dT))_(P)` Since at constant `T, dT = 9` `:. C_(P) = oo`

Discussion

1078.	The bond enthalpies of `C-C`,`C=C` and `C=C` bonds are 348, 610 and 835 kj`//` mole respectively at 298 K and 1 bar . Calculate enthalpy of polymerisation per mole of butyne at 298 K and 1 bar :A. `-123 kJ`B. `-132 kJ`C. `-139 kJ`D. `-37 kJ`
Answer» Correct Answer - A

Discussion

1079.	The following is (are) endothermic reaction(s):A. Combustion of mehtaneB. Decomposition of waterC. Dehydrogenation of ethane to ethyleneD. Conversion of graphite to diamond
Answer» Correct Answer - B::C::D Combustion reaction are exothermic , hydrogenation of alkenes are exothermic and graphite is the modynamically more stabel.

Discussion

1080.	Which of the reaction defines molar `DeltaH_(f)^(0)`?A. `CaO(s) + CO_(2)(g) rarr CaCO_(3)(s)`B. `(1)/(2)Br_(2)(l) + (1)/(2)H_(2)(g) rarr HBr(g)`C. `N_(2)(g) + 2H_(2)(g) + (3)/(2)O_(2)(g) rarr NH_(4)NO_(3)(s)`D. `(1)/(2)I_(2)(s) + (1)/(2)H_(2)(g) rarr HI(g)`
Answer» Correct Answer - B::C::D

Discussion

1081.	Ethanol can undergo decompostion to form two sets of products. If the molar ratio of `C_(4)H_(4) "to" CH_(3)CHO` is 8:1 in a set of product gases, then the energy involved in the decomposition of 1 mole of ethanol is: A. 65.98 KJB. 48.137 KJC. 48.46 KJD. 57.22 KJ
Answer» Correct Answer - B `Delta_(r)H=(8)/(9)xx45.54 + (1)/(9)xx68.91 = 48.137 KJ.`

Discussion

1082.	In thermo dynamics a process is called reversible when :A. surroundings and system change into each otherB. there is no boundary between system and surroundingsC. the surroundings are always in equilibrium with systemD. the system chnges into surrounding spontanously
Answer» Correct Answer - C

Discussion

1083.	One mole of solid Zn is placed in excess of dilute `H_(2)SO_(4)` at `27^(@)C` in a cylinder fitted with a piston . Find the work done for the process of the area of piston is `500 cm^(2)` and it moves out by 50 cm against a pressure of 1 atm during the reaction. `Zn(s) + 2H^(+)(aq) hArr Zn^(2+) (aq) + H_(2)(g)`A. `-1.53 KJ `B. `-2.53 KJ`C. ZeroD. 2.53 KJ
Answer» Correct Answer - B `W=-P_("ext")(DeltaV)=- (1 atm) xx 500 xx50 xx 10^(-3)L=-25L atm =-25xx101.3 xx 10^(-3)KJ=- 2.53 KJ`

Discussion

1084.	On the basis of the following thermochemical data `:` `(Delta_(f)G^(@)H_((aq.))^(+)=0)` `H_(2)O_((l))rarr H_((aq.))^(+)+OH_((aq.))^(-),DeltaH=57.32kJ` `H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l)),DeltaH=-286.20kJ` The value of enthalpy of formation of `OH^(-)` ion at `25^(@)C` is `:`A. `-228.88` kJB. `228.88` kJC. `-343.52` kJD. `-22.88` kJ
Answer» Correct Answer - A

Discussion

1085.	What is `DeltaU` when 2.0 mole of liquid water vaporise at `100^(@)C ` ? The heat of vaporistaion , `DeltaH` vap. of water at `100^(@)C` is `40.66 kJ mol^(-1)`.
Answer» Correct Answer - `DeltaE = 75.11 k J` `DeltaH= DeltaU + Deltan_(g)RT` `40.66 = DeltaU + 1 xx 8.314 xx 10^(-3) xx 373` `Deltan_(g)=1` `DeltaU = 37.555 xx 2 kJ` For two moles = 37.555 xx 2 = 75.11 kJ

Discussion

1086.	The enhtaply change for the reaction of 50ml of ethylene with 50.0 mL if `H_(2)` at 1.5 atm pressure is `DeltaH=-0.31 KJ` . What is the `DeltaE`?A. `-0.3024`B. `0.6048`C. `0.1.2`D. None
Answer» Correct Answer - A `{:(C_(2)H_(4) ,"+", H_(2), rarr ,C_(2)H_(3)),(50ml, ,50ml,,o),(X,,X,,50ml):}` `DeltaH=DeltaU+P(DeltaV)` `-0.31=DeltaU + 1.5 xx 1.01 xx 10^(5) (-50 xx 10^(-6))`. `DeltaU=- 0.3024 KJ .`

Discussion

1087.	Spontaneous adsorption of a gas on a solid surface is exothermic process becauseA. enthalpy of the system increase.B. enthropy increases.C. entropy decreasesD. free energy change increase.
Answer» Correct Answer - C In adsorption there is bond formation between the gases and solid surface which decrease the entropy.

Discussion

1088.	Sponetaneous adsorption of a gas on solid surface is an exothermic process becauseA. `DeltaH` increases for system.B. `DeltaS` increases for gas.C. `DeltaS` decreases for gas.D. `DeltaG` increases for gas.
Answer» `DeltaG = DeltaH - T DeltaS` Entropy decreases during adsorption. Entropy factor is disfavouring, therefore the enthalpy factor must factor spontaneity, so it must be exothermic and `DeltaH` must be negative.

Discussion

1089.	Calculate the entropy change in surroundings when 1 mol-1 of H2O(I) is formed under standard conditions. Given ∆H = – 286 kJ mol-1
Answer» q_rev = (-Δ_fH^⊖) = -286 KJ mol^-1 = 286000 J mol^-1 ΔS_{(surroundings)}= q_rev/ T = (286000 J mol^-1) / 298 K = 959 J K^-1mol^-1

1089.

Calculate the entropy change in surroundings when 1 mol-1 of H2O(I) is formed under standard conditions. Given ∆H = – 286 kJ mol-1

Answer»

q_rev = (-Δ_fH^⊖) = -286 KJ mol^-1 = 286000 J mol^-1

ΔS_{(surroundings)}= q_rev/ T

= (286000 J mol^-1) / 298 K

= 959 J K^-1mol^-1

Discussion

1090.	Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions, ΔfHΘ = -286 kJ mol-1
Answer» q_rev = (-ΔH^Θ) = -286 kJ mol^-1 = 286000 J mol^-1 ΔS_{(Surrounding)} = q_rev/T = {286000 J mol^-1}/{298 K} = 959 J K^-1 mol^-1

1090.

Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions, ΔfHΘ = -286 kJ mol-1

Answer»

q_rev = (-ΔH^Θ) = -286 kJ mol^-1

= 286000 J mol^-1

ΔS_{(Surrounding)} = q_rev/T = {286000 J mol^-1}/{298 K}

= 959 J K^-1 mol^-1

Discussion

1091.	In a reversible process, the entropy of Universe is ……(a) greater than zero (b) less than zero(c) equal to zero (d) remains constant
Answer» (c) equal to zero

Discussion

1092.	Neither `q` nor `w` is a state function but `q +w` is a state function. Explain why?
Answer» `q+w=Delta E` and `Delta E` is a state function.

Discussion

1093.	Give an example of an isolated system.
Answer» Coffee held in a thermos flask is an isolated system because it can neither exchange energy nor matter with the surroundings.

Discussion

1094.	Neither q nor W is a state function but q + W is a state function. Why?
Answer» q + W is equal to ΔE, which is a state function.

Discussion

1095.	What will happen to internal energy if work is done by the system?
Answer» The internal energy of the system will decrease if work is done by the system.

Discussion

1096.	What is Gibb‘s Helmhaltz equation?
Answer» ∆G = ∆H – T. ∆S

Discussion

1097.	Predict the sign of S for the following reactions. CaCO3(s) + CO2(g) heat→ CaO(s) + CO2(g)
Answer» ∆S is positive (entropy increases)

Discussion

1098.	For a reaction to occur spontaneouslyA. `DeltaS` must be negativeB. `(-DeltaH+T DeltaS)` must be positiveC. `DeltaH + TDeltaS` must be negativeD. `DeltaH` must be negative
Answer» Correct Answer - B `(DeltaG)_("system")=DeltaH-T.DeltaS` and `" "DeltaH-T.DeltaSlt0`

Discussion

1099.	State the first law of thermodynamic.
Answer» The first law of thermodynamics states that “the total energy of an isolated system remains constant though it may change from one form to another” (or) Energy can neither be created nor destroyed, but may be converted from one form to another.

1099.

State the first law of thermodynamic.

Answer»

The first law of thermodynamics states that “the total energy of an isolated system remains constant though it may change from one form to another”

(or)

Energy can neither be created nor destroyed, but may be converted from one form to another.

Discussion

1100.	The correct thermodynamic conditions for the spontaneous reaction at all temperature is (a) ∆H < 0 and ∆S > 0(b) ∆H < 0 and ∆S < 0(c) ∆S > 0 and ∆S = 0 (d) ∆H < 0 and ∆S > 0
Answer» (a) ∆H < 0 and ∆S > 0

Discussion

Explore topic-wise InterviewSolutions in .

Which one of the following gases possesses the largest internal energyA. 2 moles of helium occupying `1m^(2) at 300 k`B. 56 kg nitrogen at `107Nm^(2)` and 300 kC. 8 grams of oxygen at 8 atm and 300 kD. `6xx10^(26)` moleculer of oxygen occupying `40m^(2)` at 900 k

If the work done by a system on its surroundings is 100 J and the increase in the internal energy of the system is 100 cal, what must be the heat supplied to the system? (Given : J = 4.186J/cal)

Which one of the following gases possesses the largest internal energyA. 2 moles of helium occupying 1 `m^(3)` at 300 KB. 56 kg of nitrogen at `10^(7) Nm^(-2)` and 300 KC. 8 grams of oxygen at 8 atm and 300 KD. `6 xx 10^(26)` molecules of argon occupying `40 m^(3)` at 900 K

When air of the atmosphere rises up, it cools. Why?

What is the value of specific heat of water in S.I. units? Does it vary with temperature?

Define Radiation.

State second law of thermodynamics.

What is the value of work done in cyclic process?

Unit mass of liquid of volume `V_(1)` completely turns into a gas of volume `V_(2)` at constant atmospheric pressure P and temperature T. The latent heat of vaporization is "L". Then the change in internal energy of the gas isA. ZeroB. `P(V_(1)-V_(2))`C. `L-P(V_(2)-V_(1))`D. L

A gas expands `0.25m^(2)` at constant perssure `10^(3)N//m^(2),` the work done isA. 2.5 ergsB. 250 jC. 250 WD. 340 N

Consider a Carnot’s cycle operating between T1 = 500 K and T2 = 300 K producing 1 KJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.

The correct thermodynamics conditions for the spontaneous reaction at all temperature isA. `Delta H lt 0` and `Delta S gt 0`B. `DeltaH lt 0` and `DeltaS lt 0`C. `Deltah lt 0` and `DeltaS =0`D. `DeltaH gt 0` and `Delta S lt 0`

What is an adiabatic process ?

For a sample of perfect gas when its pressure is changed isothermally from `p_(i)` to `p_(f)` , the entropy change is given byA. `DeltaS= nRln((p_(f))/( p_(i)))`B. `DeltaS= nRln((p_(i))/( p_(f)))`C. `DeltaS= nRTln((p_(f))/( p_(i)))`D. `DeltaS= nRTln((p_(i))/( p_(f)))`

Why internal energy is a state function but work is not ?

Which of the following reaction is associated with an increase in entropy?A. `N_(2(g))+3H_(2(g)) rarr 2NH_(3(g))`B. `2H_(2(g))+O_(2(g)) rarr 2H_(2)O_((l))`C. `H_(2(g))+I_(2(g)) rarr 2HI_((g))`D. `C("graphite")+H_(2)O(g) rarr CO_((g))+H_(2(g))`

The value of `DeltaS` is negative for the processA. Burning of rocket FuelB. Dissolution of sugarC. Sublimation of IodineD. Freezing of water

`1` mole of an ideal gas at `25^(@)` C is subjected to expand reversibly and adiabatically to ten times of its initial volume. Calculate the change in entropy during expansion (in `J K^(-1) mol^(-1)`)A. `19.15`B. `-19.15`C. `4.7`D. zero

An insulated container is divided into two equal portions. One portion contains as ideal gas at pressure `P` and tenperature `T`. The other portion is a perfect vaccum. If a hole is opened between the two portions, claculate a. the change in internal energy. b. the change in temperature.

`1 mol` of an ideal gas at `25^(@)C` is subjected to expand reversibly `10` times of its initial volume. Calculate the change in entropy of expansions.

A system has internal energy equal to `U_(1), 450J` of heat is taken out of it and `600J` of work is done on it. The final enegry of the system will beA. `(U_(1) +150)`B. `(U +1050)`C. `(U_(1) - 150)`D. None of these

The molar heat capacity for a gas at constant `T` and `P` isA. `(3)/(2)R`B. `(5)/(2)R`C. Dependent on the atomicity of the gasD. Infinity

The bond enthalpies of `C-C`,`C=C` and `C=C` bonds are 348, 610 and 835 kj`//` mole respectively at 298 K and 1 bar . Calculate enthalpy of polymerisation per mole of butyne at 298 K and 1 bar :A. `-123 kJ`B. `-132 kJ`C. `-139 kJ`D. `-37 kJ`

The following is (are) endothermic reaction(s):A. Combustion of mehtaneB. Decomposition of waterC. Dehydrogenation of ethane to ethyleneD. Conversion of graphite to diamond

Which of the reaction defines molar `DeltaH_(f)^(0)`?A. `CaO(s) + CO_(2)(g) rarr CaCO_(3)(s)`B. `(1)/(2)Br_(2)(l) + (1)/(2)H_(2)(g) rarr HBr(g)`C. `N_(2)(g) + 2H_(2)(g) + (3)/(2)O_(2)(g) rarr NH_(4)NO_(3)(s)`D. `(1)/(2)I_(2)(s) + (1)/(2)H_(2)(g) rarr HI(g)`

Ethanol can undergo decompostion to form two sets of products. If the molar ratio of `C_(4)H_(4) "to" CH_(3)CHO` is 8:1 in a set of product gases, then the energy involved in the decomposition of 1 mole of ethanol is: A. 65.98 KJB. 48.137 KJC. 48.46 KJD. 57.22 KJ

In thermo dynamics a process is called reversible when :A. surroundings and system change into each otherB. there is no boundary between system and surroundingsC. the surroundings are always in equilibrium with systemD. the system chnges into surrounding spontanously

What is `DeltaU` when 2.0 mole of liquid water vaporise at `100^(@)C ` ? The heat of vaporistaion , `DeltaH` vap. of water at `100^(@)C` is `40.66 kJ mol^(-1)`.

The enhtaply change for the reaction of 50ml of ethylene with 50.0 mL if `H_(2)` at 1.5 atm pressure is `DeltaH=-0.31 KJ` . What is the `DeltaE`?A. `-0.3024`B. `0.6048`C. `0.1.2`D. None

Spontaneous adsorption of a gas on a solid surface is exothermic process becauseA. enthalpy of the system increase.B. enthropy increases.C. entropy decreasesD. free energy change increase.

Sponetaneous adsorption of a gas on solid surface is an exothermic process becauseA. `DeltaH` increases for system.B. `DeltaS` increases for gas.C. `DeltaS` decreases for gas.D. `DeltaG` increases for gas.

Calculate the entropy change in surroundings when 1 mol-1 of H2O(I) is formed under standard conditions. Given ∆H = – 286 kJ mol-1

Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions, ΔfHΘ = -286 kJ mol-1

In a reversible process, the entropy of Universe is ……(a) greater than zero (b) less than zero(c) equal to zero (d) remains constant

Neither `q` nor `w` is a state function but `q +w` is a state function. Explain why?

Give an example of an isolated system.

Neither q nor W is a state function but q + W is a state function. Why?

What will happen to internal energy if work is done by the system?

What is Gibb‘s Helmhaltz equation?

Predict the sign of S for the following reactions. CaCO3(s) + CO2(g) heat→ CaO(s) + CO2(g)

For a reaction to occur spontaneouslyA. `DeltaS` must be negativeB. `(-DeltaH+T DeltaS)` must be positiveC. `DeltaH + TDeltaS` must be negativeD. `DeltaH` must be negative

State the first law of thermodynamic.

The correct thermodynamic conditions for the spontaneous reaction at all temperature is (a) ∆H &lt; 0 and ∆S &gt; 0(b) ∆H &lt; 0 and ∆S &lt; 0(c) ∆S &gt; 0 and ∆S = 0 (d) ∆H &lt; 0 and ∆S &gt; 0

The correct thermodynamic conditions for the spontaneous reaction at all temperature is (a) ∆H < 0 and ∆S > 0(b) ∆H < 0 and ∆S < 0(c) ∆S > 0 and ∆S = 0 (d) ∆H < 0 and ∆S > 0