4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 19 InterviewSolution

901.	Which has maximum internal energy at `290K` ?A. Neon gasB. Nitrogen gasC. Ozone gasD. All have equal value
Answer» Correct Answer - C

Discussion

902.	`1 mol` of `NH_(3)` gas at `27^(@)C` is expanded under adiabatic condition to make volume `8` times `(gamma=1.33)`. Final temperature and work done, respectively, areA. `150K, 900 cal`B. `150K, 400cal`C. `250K, 100cal`D. `200K, 800cal`
Answer» `T_(2)= T_(1) ((V_(1))/(V_(2)))^(gamma-1) = 150K` `w =- C_(V) DeltaT =- C_(V) (T_(2) -T_(1))` `=- 3 xx 2xx (150 -300)` `= 900 kcal`

Discussion

903.	Which has maximum internal energy at `290K` ?A. Neon gasB. Nitrogen gasC. Ozone gasD. Equal for (a),(b),(c )
Answer» Correct Answer - C

Discussion

904.	1 mole of `NH_(3)` gas at `27^(@)` C is expanded under reversible adiabatic conditions to make volume 8 times (`gamma=1.33`). Final temperature and work done respectively are :A. 150 K, 900 calB. 150 K, 400 calC. 250 K, 1000 calD. 200 K, 800 cal
Answer» Correct Answer - A

Discussion

905.	Which of the following interactions operate during to propcess of solvation?A. Solvent - soluteB. Solvent - solventC. Solute-soluteD. All of these
Answer» Correct Answer - A Solvation is the process in which solvent molecules surround and interact with solute inons or molecules.

Discussion

906.	The enthalpies of combustion of carbon and carbon monoxide are `-390 kJ mol^(-1)` and `-278 kJ mol^(-1)` respectively. The enthalpy of formation of carbon monoxide isA. `668 kJ mol^(-1)`B. `112 kJ mol^(-1)`C. `-112 kJ mol^(-1)`D. `-668 kJ mol^(-1)`
Answer» Correct Answer - C `DeltaH=H_(P)-H_(R)`

Discussion

907.	Lattice enthalpy is the enthalpy change for completely separating one mole of a solid ionic compound into its constituent ions in the _______phase.A. solidB. gasC. liquidD. solution
Answer» Correct Answer - B `M^(+) X^(-) overset(Delta_("Lattice") H^(@)) rarr M^(+) (g) + X^(-) (g)`

Discussion

908.	The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure lattice enthalpy of `NaCl(s)`.
Answer» Lattice energy cannot be determined directly by experiment. It is determined indirectly by Born-Haber cycle

Discussion

909.	What is meant by reversible process? Explain why the efficiency of a reversible engine is maximum?
Answer» The process which can be retraced at any stage of its operation by reversing the boundary conditions is called a reversible process. The efficiency of a reversible engine is maximum, because in such a device, no dissipation of energy take place.

Discussion

910.	A heat engine coverts disordered mechanical motion into ordered mechanical motion. Comment.
Answer» As is known, a heat engine converts heat energy into mechanical energy. Now, heat energy is due to disorderd mechanical motion of the molocules. And mechanical energy is often due to orderd mechanical motion. Therefore, the statement is true.

Discussion

911.	Prove that cp is greater than cv.
Answer» Out of the two specific heats of a gas, C_{p >}C_v This can be understood as follows: In case C_v volume of the gas is kept constant and heat is required only for raising the temperature of gas through 1°C or 1K. No heat whatsoever is spent in expansion of the gas. In case of C_P as pressure of the gas is kept constant and the gas could expand on heating. ∴ Some heat is spent in expansion of the gas against external pressure. Hence, more heat is required to raise the temperature of 1 gm mole of the gas through 1°C or 1K when heated at constant pressure. Hence, C_{p >}C_v .

911.

Prove that cp is greater than cv.

Answer»

Out of the two specific heats of a gas,

C_{p >}C_v

This can be understood as follows:

In case C_v volume of the gas is kept constant and heat is required only for raising the temperature of gas through 1°C or 1K. No heat whatsoever is spent in expansion of the gas.

In case of C_P as pressure of the gas is kept constant and the gas could expand on heating.

∴ Some heat is spent in expansion of the gas against external pressure.

Hence, more heat is required to raise the temperature of 1 gm mole of the gas through 1°C or 1K when heated at constant pressure. Hence, C_{p >}C_v .

Discussion

912.	Derive the equation of adiabatic changes
Answer» Consider 1 g mole of n ideal gas enclosed in a cylinder fitted with a perfectly friction less piston. Let P, V, T be initial pressure, volume and temperature. Suppose a small amount of heat dQ spent in two ways: Increasing the temperature, of gas by small range dT at constant volume = c_vdT Expansion of gas by small volume dV portion of heat spend = PdV dQ= C_vdT + PdV In adiabatic change dQ = 0 C_vdT + PdV = 0 …(1) According to standard gas equation PV = RT Differentiate both sides, PdV + VdP = RdT dT = \(\frac{PdV+VdP}{R}\) Putting eq. (1), \(\frac{C_v(PdV+VdP)}{R}\) + PdV = 0 (C_v + R)PdV + C_vVdP = 0 …(2) But C_p − C_v = R From (2) dividing both sides by C_vPV \(\frac{C_PdV}{C_vPV}+\frac{C_vVdV}{C_vPV}=0\) \(\frac{\gamma dV}{v}+\frac{dP}{P}\) γ =\(\frac{C_p}{C_v}\) Integrating both sides. Γ∫ \(\frac{dV}{V}\) + ∫ \(\frac{dP}{P}\) = C γloge V + logeP = C log_eV^γ + log_eP = C Log_kPV^γ = C PV^γ = K This is the required relation

912.

Derive the equation of adiabatic changes

Answer»

Consider 1 g mole of n ideal gas enclosed in a cylinder fitted with a perfectly friction less piston. Let P, V, T be initial pressure, volume and temperature.

Suppose a small amount of heat dQ spent in two ways:

Increasing the temperature, of gas by small range dT at constant volume = c_vdT

Expansion of gas by small volume dV portion of heat spend = PdV

dQ= C_vdT + PdV

In adiabatic change

dQ = 0 C_vdT + PdV = 0 …(1)

According to standard gas equation

PV = RT

Differentiate both sides,

PdV + VdP = RdT

dT = \(\frac{PdV+VdP}{R}\)

Putting eq. (1),

\(\frac{C_v(PdV+VdP)}{R}\) + PdV = 0

(C_v + R)PdV + C_vVdP = 0 …(2)

But C_p − C_v = R

From (2) dividing both sides by C_vPV \(\frac{C_PdV}{C_vPV}+\frac{C_vVdV}{C_vPV}=0\)

\(\frac{\gamma dV}{v}+\frac{dP}{P}\)

γ =\(\frac{C_p}{C_v}\)

Integrating both sides.

Γ∫ \(\frac{dV}{V}\) + ∫ \(\frac{dP}{P}\) = C

γloge V + logeP = C

log_eV^γ + log_eP = C

Log_kPV^γ = C

PV^γ = K This is the required relation

Discussion

913.	A geyser heats water flowing at a rate of 3.0 litres per minute from 27°C to 77°C. If geyser operates on a gas burner, what is the rate consumption of the fuel if the heat of combustion is 4. 0 × 104 J/g
Answer» Rate of flow of water = 3 litres/minute ∴ Mass of water flowing per minute, M = 3000g / minute Initial temperature, T₁ = 27°C Final temperature, T₂= 77°C Rise in temperature of the flowing water ∆ T = T₂ – T₁ = (77 – 27)° C = 50° C specific heat of water, c = 4200 J kg^-1 °C^-1 = 4.2 Jg^-1 °C^-1 ∴ Total heat supplied by geyser, ∆ Q = mc ∆T = 3000 × 4.2 × 50 = 630000 J/min = 6.3 × 10⁵ J/min Heat of combustion of geyser = 4 × 10⁴Jg^-1 ∴ Rate of consumption of fuel, r = \(\frac{6.3 \times 10^5}{4.0 \times 10^4}\) = 15. 75 g/min.

913.

A geyser heats water flowing at a rate of 3.0 litres per minute from 27°C to 77°C. If geyser operates on a gas burner, what is the rate consumption of the fuel if the heat of combustion is 4. 0 × 104 J/g

Answer»

Rate of flow of water = 3 litres/minute

∴ Mass of water flowing per minute,

M = 3000g / minute

Initial temperature, T₁ = 27°C

Final temperature, T₂= 77°C

Rise in temperature of the flowing water

∆ T = T₂ – T₁ = (77 – 27)° C

= 50° C

specific heat of water, c = 4200 J kg^-1 °C^-1

= 4.2 Jg^-1 °C^-1

∴ Total heat supplied by geyser,

∆ Q = mc ∆T

= 3000 × 4.2 × 50

= 630000 J/min

= 6.3 × 10⁵ J/min

Heat of combustion of geyser

= 4 × 10⁴Jg^-1

∴ Rate of consumption of fuel,

r = \(\frac{6.3 \times 10^5}{4.0 \times 10^4}\) = 15. 75 g/min.

Discussion

914.	An ideal monoatomic gas following the process `(P)/(V^(2))=" constant will have heat capacity " (11R)/(X)`. Find the value of X.
Answer» Correct Answer - 6

Discussion

915.	The heat of combustion of C(graphite) and CO(g) are` -390 kJ//"mole and " -280 kJ//"mole"` respectively. `CO(g)+Cl_(2)(g)to COCl_(2)(g)Delta H=+208 kJ` Calculate heat of formation `("in kJ"//"mole")` of `COCl_(2)(g)` (in kJ/mole)
Answer» Correct Answer - 98

Discussion

916.	Will the heat release in the following two reactions be equal? Give reasons in support of your answer:(i) H2 (g) + 1/2 O2 (g) → H2O (g)(ii) H2 (g) + 1/2 O2 (g) → H2O (l)
Answer» No, The heats released in the two reactions are not equal. The heat released in any reaction depends upon the reactants, products and their physical states. Here in reaction (i) the water produced is in the gaseous state whereas in reaction (ii) liquid water is formed. As we know, that when water condenses to form water, heat equal to the latent heat of vaporisation is released. Thus, more heat is released in reaction (ii).

Discussion

917.	Show that for an ideal gas CP - CV = R
Answer» When a gas is heated under constant pressure conditions, the heat is required for raising the temperature of the gas and also for doing mechanical work against the external pressure during expansion. Consider one mole of gas to expand from volume V₁ to V₂ at a constant pressure p, when the temperature is raised from T K to (T + 1) K. Then by definition C_P - C_V = work done by the gas due to expansion. Work done by the gas during expansion = pΔV = p(V₂ - V₁) Therefore, C_P - C_V = p(V₂ - V₁) For one mole of the gas, using gas equation, one can write pV₂ = R (T + 1) and pV₁ = RT or, pV₂ - pV₁ = R(T + 1) - RT = R(T + 1) - RT = R or, p(V₂ - V₁) = R Therefore, C_P - C_V = R

917.

Show that for an ideal gas CP - CV = R

Answer»

When a gas is heated under constant pressure conditions, the heat is required for raising the temperature of the gas and also for doing mechanical work against the external pressure during expansion.

Consider one mole of gas to expand from volume V₁ to V₂ at a constant pressure p, when the temperature is raised from T K to (T + 1) K. Then by definition

C_P - C_V = work done by the gas due to expansion. Work done by the gas during expansion

= pΔV = p(V₂ - V₁)

Therefore, C_P - C_V = p(V₂ - V₁)

For one mole of the gas, using gas equation, one can write

pV₂ = R (T + 1)

and pV₁ = RT

or, pV₂ - pV₁ = R(T + 1) - RT

= R(T + 1) - RT = R

or, p(V₂ - V₁) = R

Therefore, C_P - C_V = R

Discussion

918.	Given the bond energies of `H - H` and `Cl - Cl` are `430 kJ mol^(-1)` and `240 kJ mol^(-1)`, respectively, and `Delta_(f) H^(@)` for `HCl` is `- 90 kJ mol^(-1)`. Bond enthalpy of `HCl` isA. 290 `KJ "mol"^(-1)`B. `380 KJ "mol"^(-1)`C. `425 KJ "mol"^(-1)`D. `245 KJ "mol"^(-1)`
Answer» Correct Answer - B `DeltaH_("reaction")=Delta_(H-H) + DeltaH_(Cl-Cl)-2DeltaH_(HCl)` or `DeltaH_(Cl-Cl) =(430+ 240 -(-90))/(2)` `=(760)/(2)=380 KJ "mol"^(-1)`

Discussion

919.	Which of the following are not state functions ? (I)q`+` w ,(II) q, (III) w, (IV) H-TSA. (I) and (IV)B. (II), (III) and (IV)C. (I),(II) and (III)D. (II) and (III)
Answer» Correct Answer - d `q+w= DeltaU` and`DeltaU` is a state function `=H-TS = G `and G is a state function.

Discussion

920.	How many types of Enthalpies of reactions?
Answer» Different types of Enthalpies of reactions: (i) Enthalpy of combustion (_cH), (ii) Enthalpy of formation (_fH) (iii) Enthalpy of neutralization (iv) Enthalpy of solution (v) Enthalpy of atomization (_aH), (vi) Enthalpy of Ionisation (_iH) (vii) Enthalpy of Hydration (_hyol.H) (viii) Enthalpy of fusion (_fus.H) (ix) Enthalpy of vaporization (_vap.H) (x) Enthalpy of sublimation (_sub.H)

920.

How many types of Enthalpies of reactions?

Answer»

Different types of Enthalpies of reactions:

(i) Enthalpy of combustion (_cH),

(ii) Enthalpy of formation (_fH)

(iii) Enthalpy of neutralization

(iv) Enthalpy of solution

(v) Enthalpy of atomization (_aH),

(vi) Enthalpy of Ionisation (_iH)

(vii) Enthalpy of Hydration (_hyol.H)

(viii) Enthalpy of fusion (_fus.H)

(ix) Enthalpy of vaporization (_vap.H)

(x) Enthalpy of sublimation (_sub.H)

Discussion

921.	What is the First Law of Thermodynamics ?
Answer» It is law of conservation energy. Energy can neither be created not destroyed, it may be converted from one from into another. Mathematically U = q + w, w = –p. V (work of expansion) U = q – p. V or q = U + p. V, q,w are not state function. But U is state function.

921.

What is the First Law of Thermodynamics ?

Answer»

It is law of conservation energy.

Energy can neither be created not destroyed, it may be converted from one from into another.

Mathematically U = q + w, w = –p. V (work of expansion)

U = q – p. V or q = U + p. V, q,w are not state function. But U is state function.

Discussion

922.	One mole of an ideal gas expands at a constant temperature of `300 K` from an initial volume of `10` litres to a final volume of `20` liters. The work done in expanding the gas is `(R=8.31 J// "mole"-K)` (in joules)A. `750`joulesB. `1728`joulesC. `1500`joulesD. `3456`joules
Answer» Correct Answer - B

Discussion

923.	Which of the following statements is correct for any thermodynamic systemA. The internal energy changes in all processesB. Internal energy and entropy are state functionsC. The change in entropy can never be zeroD. The work done in an adiabatic process is always zero
Answer» Correct Answer - B

Discussion

924.	Which one is not correct for a cyclic process as shown in the figure ? A. `dU = 0`B. `q =- w`C. w = `314 J`D. w = `31.4 J`
Answer» For a cyclic process, `dU = 0` `:. Q = DeltaU +(-w) rArr q =- w` Also, `w =` Area covered by sphere `= pir^(2) = pi xx [((V_(2)-V_(1)))/(2)]^(2)` `= (pixx(20)^(2))/(2^(2))` `= 100 pi = 100 xx 3.14 = 314 J`

Discussion

925.	A gas occupies `2L` at `STP`. It is provided `300J` heat so that it volume becomes `2.5L` at `1atm`. Caluclate the change in its internal enegry.
Answer» Work done `=- P xx DeltaV = 1 xx (2.5 - 2.0)` `=- 0.5 L atm` Therefore, work is carried out at constant `P` and thus irreversible. `=- (0.5 xx 1.987xx4.184)/(0.0821)J =0.5 xx 101.328J` `=- 50.631J` From the first law of thermodynamics, `:. q = DeltaU - w` `300 = DeltaU +50.63:. DeltaU = 249.37J`

Discussion

926.	Predict whether it is possible or not to reduce magnesium oxide using carbon at `298K` according to the reaction. `MgO(s) +C(s) rarr Mg(s) +CO(g)` `Delta_(r)H^(Theta) = +491.18 kJ mol^(-1)` and `Delta_(r)S^(Theta) = 197.67 J K^(-1) mol^(-1)` If not at what temperature, the reaction becomes spontaneous.
Answer» `MgO_((s))+C_((s))toMg_((s))+CO_((g))` `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `=491.18-298xx[197.67xx10^(-3)]` `=432.27 kJ` Thus reaction is non-spontaneous at `298 K`. For spontaneous nature `DeltaG^(@)=-ve i.,e TDeltaS^(@)gtDeltaH^(@)` or `Txx[197.67xx10^(-3)]gt491.18` or `Tgt491.18/(197.67xx10^(-3))gt2484.8 K`

Discussion

927.	For the process `H_(2)O(l) rarr H_(2)O(g)` at `T=100 ^(@)C` and 1 atmosphere pressure, the correct choice isA. `DeltaS_("system") gt 0` and `DeltaS _("surroundings") gt 0`B. `DeltaS_("system") gt 0` and `DeltaS _("surroundings") lt 0`C. `DeltaS_("system") lt 0` and `DeltaS _("surroundings") gt 0`D. `DeltaS_("system") lt 0` and `DeltaS _("surroundings") lt 0`
Answer» Correct Answer - B At `T= 100^(@)C` and 1 atmosphere presssure (i.e.,boilingpoint ) ,thereexists the equilibrim `H_(2)O(l) hArr H_( 2)O (g)`, For this conditions,`DeltaS_("total") = 0` Hence,`DeltaS_("system") + Delta_("surroundings") = 0` or `DeltaS_("system") = - DeltaS_("surroundings")` But as `H_(2)O(l)` changes into `H_(2)O(g)`, `DeltaS_("system") = + ve , i.e., gt 0` `:. DeltaS_("surroundings")` willbe`lt 0`.

Discussion

928.	Extensive properties depend on the quantity of matter but intensive properties do not. Explain whether the following properties are extensive or intensive. Mass,internal energy , pressure , heat capacity , molar heat capacity , density, mole fraction, specific heat, temperature and molarity.
Answer» Extensive properties . Mass , internal energy , heat capacity. Intensive properties.Pressure, molar heat capacity, density, mole fraction , specific heat, temperature and molarity. Note. Mole fraction or molarity of a solution is same whether we take a small amount of the soution or large amount of the solution . Otherwise too. ratio of two extensive properties is always intensive, i.e., `("Extensive")/(Extensive")= ` Intensive e.g., Mole fraction `= ("Molar of the component") /( "Total no. of moles")=( "Extensive")/( "Extensive") ` and Molarity`= ( "Moles")/( "Volume") = ( "Extensive")/("Extensive")`

Discussion

929.	In which of the following sets, all properties belong to same category (all extensive or all intensive) ?A. Mass, volume, pressureB. Temperature, pressure, volumeC. Heat capacity, density, entropyD. Enthalpy, internal energy, volume.
Answer» Correct Answer - D

Discussion

930.	Pressure of a liquid is linear function of volume `P=a+bV.` Calculate work done for change in state form `(1"bar",2L)` to `(2"bar",5L)`:A. `+810J`B. `-450J`C. `-810J`D. `+450J`
Answer» Correct Answer - C

Discussion

931.	If `H_(2)S+3/2 O_(2) rarr H_(2)O+SO_(2)`, `DeltaH=-136.7` Kcal `H_(2)+1/2 O_(2) rarr H_(2)O, DeltaH=-68.4` Kcal `S+O_(2) rarr SO_(2), DeltaH=-71` Kcal What is `DeltaH_(f)` of `H_(2)S`A. 2.7 KcalB. `-2.7` KcalC. 5.4 KcalD. `-5.4` Kcal
Answer» Correct Answer - B `H_(2)+S rarr H_(2)S` `DeltaH=H_(R)-H_(P)`

Discussion

932.	The internal energy change when a system goes fromk state `A` to `B` is `40 kJ mol^(-1)`. If the system goes from `A` to `B` by a reversible path and returns to state `A` by an irreversible path, what would be the net change in internal energy?A. `lt40 k J`B. ZeroC. `40 kJ`D. `gt40 k J`
Answer» Correct Answer - b In a cyclic process net change in internal energy is zero.

Discussion

933.	The internal energy change when a system goes fromk state `A` to `B` is `40 kJ mol^(-1)`. If the system goes from `A` to `B` by a reversible path and returns to state `A` by an irreversible path, what would be the net change in internal energy?A. zeroB. `40 kJ`C. `gt 40 kJ`D. `lt 40 kJ`
Answer» Correct Answer - A Internal energy is a state function. Thus, `Delta U = U_(f) - U_(i)` Since the system returns to its initial state, `U_(f) = U_(i)`. Hence, `Delta U = 0` It does not matter what path is followed.

Discussion

934.	In an adiabatic expansion of an ideal gas -A. `W=-Delta U`B. `W= Delta U`C. `Delta U=0`D. `W=0`
Answer» Correct Answer - B

Discussion

935.	The temperature of the system decreases in anA. Adiabatic compressionB. Isothermal compressionC. Isothermal expansionD. Adiabatic expansion
Answer» Correct Answer - D When a real gas is forced through a prorus plug into a region of low pressure, it is found that due to expansion, the gas on the side of low pressure gets cooled. The phenomenon of producing lowering of temperature when a gas is made to expand adiabatically from a region of high pressure into a region of low pressure is known as Joule-Thomson effect.

Discussion

936.	Calculate the amount of heat that must be absorbed by `5g` of ice at `120^(@)C` to convert it to water at `2^(@)C`. Use the specfic heat of ice `(2.09 Jg^(-1) 1^(@)C^(-1))`, enthalpy of fusion of ice `(334 J g^(-1))`, and specific heat of water `(4.18 J g^(-1) .^(@)C^(-1))` for calculation Strategu : We must calculate the amount of heat absorbed during three steps. ltbrrgt (i) Wariming `50 g` of ice from `-12^(@)C` to its melting point `0^(@)C`. This requires the specific heat of ice. Melting the ice with no change in temperature . This needs the entholpy of fusion of ice at `0^(@)C`. Warming the resulting liquid from `0^(@)C` to `20^(@)C`. This requires specific heat of water `underset("Ice at" - 12^(@)C)(50 g) overset(Step 1)rarr underset("Ice at" 0^(@)C)(50 g) overset(Setp 2)rarr underset("Water at" 0^(@)C)(50g) overset(Setp 3)rarr underset("Water at" 20^(@)C)(50 g)`
Answer» Steip`1 : q = m Delta T` `= (50 g) ((2.90 J)/(g .^(@)C)) [0^(@)C - (-12^(@)C)]` `1.25 xx 10^(3) J` Step `2 : q = m Delta_(fus) H^(@)` `= (50 g) (334 J g^(-1))` `= 1.67 xx 10^(4) J` Setp `3 : q = mc Delta T` `= (50 g) ((4.18 J)/(g .^(@)C)) (20^(@)C - 0^(@)C)` `4.18 xx 10^(4) J` `= 22.4 kJ`

Discussion

937.	Which of the following is not correct?A. Dissolution of `NH_(4) Cl` in excess of water is an endothermic process.B. Neutralization is always exothermic.C. The absolute value of enthalpy `(H)` can be determined experimentally.D. The heat of reaction at constant volume is denoted by `Delta U`
Answer» Correct Answer - C Defining equation of enthalpy `(H)` is `H = U + pV` We cannot determine experimentally the absolute value of internal energy because many energy contributions are relative and approximate. Thus, the absoulte value of enthalpy `(H)` cannot be determined experimentally.

Discussion

938.	Find the value of `Delta_(f) H^(@)` for the reaction `N_(2) O_(4) (g) + 3 CO(g) rarr N_(2) O (g) + 3 CO_(2) (g)` Standard enthalpies of formation of `CO(g), CI_(2) (g), N_(2) O (g)`, and `N_(2) O_(4) (g)` are `- 110, - 393, 81`, and `9.7 kJ mol^(-1)`, respectively. Strategy : The standard enthalpy change of a reaction is equal to the sum of the standard molar enthalpie of formation of the products each miltiplied by its stiochiometric coefficient in the balanced equation, minus the corresponding sum of the standard molar enthalpies of formation of the reactants
Answer» According to Eq. `Delta_(1) H^(@) = sum_(i) a_(i) Delta_(f) H^(@)` (products) `- sum_(i) b_(i) Delta_(f) H^(@)` (reactants) `[Delta_(f) H^(@) (N_(2) O, g) + 3 Delta_(f) H^(@) (CO_(2), g)]` `- [Delta_(f) H^(@) (N_(2) O_(4), g) + 3 Delta_(f) H^(@) (CO, g)]` `= [(1 mol) (81 kJ mol^(-1)) + (3 mol) (- 393 kJ mol^(-1))]` `[(9 mol) (9.7 kJ mol^(-1)) + (3 mol) (- 110 kJ mol^(-1))]` `- 777.7 kJ`

Discussion

939.	A reaction occurs spontaneously if:A. `T Delta S lt Delta H` and both `Delta H` and `Delta S` are `+ ve`B. `T Delta S gt Delta H` and both `Delta H` and `Delta S` are `+ ve`C. `T Delta S = Delta H` and both `Delta H` and `Delta S` is `+ ve`D. `T Delta S lt Delta H` and `Delta H` is `+ ve` and `Delta S` is `-ve`
Answer» Correct Answer - B According to thermodynamics, `Delta G = Delta H - T Delta S` If both `Delta H` and `Delta S` are positive, then `Delta G` will be negative only when the `T Delta S` term is greater in magnitude than `Delta H`.

Discussion

940.	How much heat is released when `850 kg` of ammonia is produced according to the following thermochemical equation? `N_(2) (g) + 3 H_(2) (g) rarr 2 NH_(3) (g)`, `Delta_(1) H^(@) = -91.8 kJ` Strategy : To do the calculation, we must convert grams of `NH_(3)` to moles of `NH_(3)`, and then to kilojoules of heat. Grams of `NH_(3) rarr ` Moles of `NH_(3) rarr` Kilojoules of heat
Answer» Step 1: Calculating moles of `NH_(3)` `n_(NH_(3)) = ("Mass"_(NH_(3)))/("Molar mass"_(NH_(3))` ` = (850 xx 10^(3) g)/(17 g NH_(3) mol^(-1))` `5 xx 10^(4) mol` Step 2: 2 mol `NH_(3)` formation releases `91.8 kJ`. This, the formation of `5 xx 10^(4) mol NH_(3)` will release `(91.8 Kj)/(2 mol) xx 5 xx 10^(4) mol = 2.30 xx 10^(6) kJ`

Discussion

941.	The equilibrium constant at 25°C for the process. CO3+(aq) + 6NH3(aq) ⇌ [CO(NH3)6]3(aq) is 2.5 × 106. Calculate the value of ΔG° at 25°C. (R = 8.314 JK1 mol-1). In which direction is the reaction spontaneous under standard conditions?
Answer» ΔG° = -2.303 RT log K = -2.303 × 3.134 × 298 log (2.5 x 10⁶) = -5705.8 [0.3980 + 6.0000] = -5705.8 × 6.3980 = -36.505 k/mol. The reaction is spontaneous in forward direction under standard conditions.

941.

The equilibrium constant at 25°C for the process. CO3+(aq) + 6NH3(aq) ⇌ [CO(NH3)6]3(aq) is 2.5 × 106. Calculate the value of ΔG° at 25°C. (R = 8.314 JK1 mol-1). In which direction is the reaction spontaneous under standard conditions?

Answer»

ΔG° = -2.303 RT log K

= -2.303 × 3.134 × 298 log (2.5 x 10⁶)

= -5705.8 [0.3980 + 6.0000]

= -5705.8 × 6.3980 = -36.505 k/mol.

The reaction is spontaneous in forward direction under standard conditions.

Discussion

942.	The thermochemical equation for the combustion of ethylene gas, `C_(2)H_(4)`, is `C_(2)H_(4)(g) rarr 2CO_(2)(g) +2H_(2)O(l), DeltaH^(Theta) =- 337 kcal` Assuming `70%` efficiency, calculate the weight of water at `20^(@)C` that can be converted into system at `100^(@)C` by buring `1m^(3)` of `C_(2)H_(4)` gas measured at `STP`. The heat of vaporisation of water at `20^(@)C` and `100^(@)C` are `1.00 kcal kg^(-1)` and `540 kcal kg^(-1)` respectively.
Answer» Number of moles in `1m^(3)` of ethylene `= 44.6` mol ltbr `DeltaH` for `1m^(3)` of ethylene `(44.6 mol` of tthylene) `= n(C_(2)H_(4)) xx DeltaH (1 "mole")` `=- 1.50 xx 104 kcal` Therefore, the useful heat `= 1.05 xx 10^(4) cal` For the overall procee, consider two stages: `H_(2)O(l) 20^(@)C rarr H_(2)O(g) 100^(@)C`, `DeltaH = C_(P) DeltaT = (1.00 kcal kg^(-1),K) (80K) = 80 kcal kg^(-1)` `H_(2)O(l) 100^(@)C rarr H_(2)O(g) 100^(@)C, DeltaH = 540 kcal kg^(-1)` `:. Delta_("total")H = 620 kcal kg^(-1)` Therefore, weight of water converted into steam `=("Amount of heat avaiable")/("Heat required/kg") = (1.05 xx 10^(4))/(620) = 16.9 kg`

Discussion

943.	What is the change in entropy when `2.5` mole of water is heated from `27^(@)C` to `87^(@)C`? Assume that the heat capacity is constant `(C_(p))_(m)(H_(2)O)=4.2J//g=k,ln(1.2)=0.18)`A. 16.4 `JK^(-1)`B. 34.02 `JK^(-1)`C. 2.89 `JK^(-1)`D. 18.2 `JK^(-1)`
Answer» Correct Answer - B

Discussion

944.	Statement-1 : In a cyclic process, both `Delta H and Delta U` are zero. And Statement-2 : `Delta H and Delta U` are path dependent functions.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 - C

Discussion

945.	Identify the correct statement regarding a spontaneous process:A. For a spontaneous process in an isolated system, the change in entropy is positiveB. Endothermic processes are never spontaneousC. Exthermic processes are always spontaneousD. Lowering of energy in the reaction process is the only criterion for spontaneity
Answer» Correct Answer - A

Discussion

946.	A system absorbs reversibly 600 J of heat and performs 250 J of work. The increase in the internal energy of system is :-A. 850 JB. 250 JC. 600 JD. 350 J
Answer» Correct Answer - D

Discussion

947.	The search of a perpetual motion machine has kept inventors occupied for centuries comment on this statement.
Answer» For centuries, scientists have been trying to fabticate machines which can generate perpetual motion. The statement is perfectly true. However, their attempts have not met with success so far. Theoretically, there can be generally be three types of perpetual motion machines: (i) A machanism which would produce machines with efficency more than 100%. This would violate first law of thermodynamics. (ii) A device that would extract heat from a source and convert it completly into other forms of energy. For example, an ocean liner that uses from the ocean in populsion. This would violate the second law of thermodynamics. (iii) A meachanism by which energy dissipation could be eliminated cent per cent. superconductors provided a genuine example of this type. When some current is made to flow through a superconductor, energy loss is zero and it would keep on flowing undiminished. But the current can do not work? so all the attempts are futile.

Discussion

948.	An ideal engine operates by taking in steam from a boiler at `327^(@)C` and rejecting heat to a sink at `27^(@)C`. The engine runs at `500 rpm` and heat taken is `600 k cal` in each revolution. Calculate(i) efficiency of engine (ii) work done in each cycle (iii) heat rejected in each revolution and (iv) power output of engine.
Answer» Correct Answer - `50% ; 1.26xx10^(6)J ; 300 K cal; 1.05xx10^(4)kW` Here, `327^(@)C= (327+273)K= 600K`, `T_(2)=27^(@)C= (27+273)= 300K` `n=500 rpm= (500)/(60)rps, Q_(1)= 600kcal` (i) `eta= 1-(T_(2))/(T_(1))= 1-(300)/(600)=0.5= 50%` (ii) `eta=W/(Q_(1))` `:. =etaQ_(1)= 0.5xx600= 300Kcal` `= 300xx4.2xx10^(3)J=1.26xx10^(6)J` (iii) `Q_(2)=Q_(1)-W=600 kcal-300kcal` `=300kcal`, (iv) `"Power output" = ("work")/("time")= (500)/(60)xx1.26xx10^(6)J//s` `=1.05xx10^(4)KW`

Discussion

949.	`1//R(R"is universal gas constant")` moles of an ideal gas `(gamma=1.5)` undergoes a cyclic process `(ABCDA)` as shown in figure. Assuming the gas to be ideal. If the net heat exchange is `10x` Joules, find the value of x ? `[In 2=0.7]`
Answer» For cyclic process `sumQ=sumW` Net change in heat energy `=W_(AB)+W_(BC)+W_(CD)+W_(DA)` `=nR(T_(B)-T_(A))+nRT_(B)ln ,(P_(B))/(P_(C))+nR(T_(D)-T_(C))+nRT_(D) , ln ,(P_(D))/(P_(A))` `=70 J` Answer is `x=7` .

Discussion

950.	For n moles of a gas, the difference between the molar heat capacity at constant pressure and that at constant volume is equal to `"…............."`.
Answer» `nR(C_(p) -C_(v)=nR)`

Discussion

Explore topic-wise InterviewSolutions in .

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