4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 70 InterviewSolution

3451.	One mole of anhydrous `MgCl_(2)` dissolves in water and librates `25 cal//mol` of heat. `Delta H_("hydration")` of `MgCl_(2)=30 cal//mol`. Heat of dissolution of `MgCl.H_(2)O`A. `+5` cal/molB. `-5` cal/molC. 55 cal/molD. `-55` cal/mol
Answer» Correct Answer - A `MgCl_(2).H_(2)O rarrMgCl_(2)(aq)Delta H_(1)= ?` `MgCl_(2)(aq)rarrMg^(2+)(aq)+2Cl^(-)(aq)Delta H_(2)= -30 cal//mol` `MgCl_(2)(s)rarrMg^(2+)(aq)+2Cl^(-1)(aq)Delta H_(3)= -25 cal//mol` `Delta H_(1)+Delta H_(2)=Delta H_(3)` `Delta H_(1)=Delta H_(3)-Delta H_(2)= -25+30= +5 cal//mol`

Discussion

3452.	`Delta H^(@)` for the reaction `X_((g))+Y_((g))hArrZ_((g))` is `-4.6 kcal`, the valueof `DeltaU^(@)` of th e reaction at `227^(@)C` is `(R=2 cal.mol^(-1)K^(-1))`:A. `-3.6 kcal`B. `-5.6 kcal`C. `-4.6 kcl`D. `-2.6 kcal
Answer» Correct Answer - A `DeltaH^(o)=Delta U^(o)+Deltan_(g)RT` `=4.6=Delta U^(@)-(1xx2xx500)/(1000)rArrDelta U^(o)= -3.6 kCal`

Discussion

3453.	Consider the following spontaneous reaction `3X_(2)(g)rarr2X_(3)(g)`. What are the sign of `DeltaH`, `DeltaS` and `DeltaG` for the reaction ?A. `+ve, +ve, +ve`B. `+ve, -ve, -ve`C. `-ve, +ve, -ve`D. `-ve, -ve, -ve`
Answer» Correct Answer - D For spontaneous reaction `DeltaG=-ve` `DeltaS=-ve(Deltan_(g)=-ve)` `DeltaH=-ve` `DeltaG=DeltaH-TDeltaS`

Discussion

3454.	A system is provided `50` joule of heat and the change in internal energy during the process is `60 J`. Magnitude of work done on the system is:
Answer» Correct Answer - 10

Discussion

3455.	A system works in a cyclic process. It absorbs `20` calories of heat and rejects `60 J` of heat during the process. The magnitude of work done `(J)` is `[ 1 "calorie"=4.2 J]`:
Answer» Correct Answer - 24

Discussion

3456.	The standard enthalpies fo formation of `CO_(2) (g), H_(2) O (1)`, and glucose (s) at `25^(@)C` are `- 400 kJ mol^(-1), - 300 kJ mol^(-)`, and `- 1300 kJ mol^(-1)`, respectively. The standard enthalply of combustion per gram of glucose at `25^(@)C` isA. `+ 16.11 kJ`B. `- 16.11 kJ`C. `+ 2900 kJ`D. `- 2900 kJ`
Answer» Correct Answer - B Thermochemical equation for the combusion of glucose is `C_(6) H_(12) O_(6) (s) + 6O_(2) (g) rarr 6 CO_(2) (g) + 6 H_(2) + 6 H_(2) O (1)` According to thermodynamics, `Delta_(r) H^(@) = sum a_(i) Delta_(f) H^(@)` (products) `- sum b_(i) Delta_(f) H^(@)` (reactants) `Delta_(C ) H^(@) = [(6 mol) Delta_(f) H^(@) (CO_(2), g) + (6 mol) Delta_(f) H^(@) (H_(2) O, 1)] - [(1 mol) Delta_(f) H^(@) (C_(6) H_(12) O_(6), s) + (6 mol) Delta_(f) H^(@) (O_(2), g)]` `= [6(-400)+6(-300)]-(-1300)` `= - 2900 kJ mol^(-1)` ` = (- 2900 kJ mol^(-1))/(180 g mol^(-1))` (molar mass of glucose = `180 g mol^(-1))` `= 16.11 kJ g^(-1)` Note that `Delta_(f) H^(@) (O_(2)) = 0` Here students often make the mistake of not calculating the standard enthalpy of combustion per gram. In the examination hall, they just read the question as ..... standard enthalpy of combustion of glucose...

Discussion

3457.	The standard enthalpies fo formation of `CO_(2) (g), H_(2) O (1)`, and glucose (s) at `25^(@)C` are `- 400 kJ mol^(-1), - 300 kJ mol^(-)`, and `- 1300 kJ mol^(-1)`, respectively. The standard enthalply of combustion per gram of glucose at `25^(@)C` isA. `+2900` KJB. `-2900` KJC. `-16.11` KJD. `+16.11 `KJ
Answer» Correct Answer - C `C_(6)H_(12)O_(6)(g) rarr 6CO_(2)(g) + 6H_(2)O(l)` `Delta_(c)H = 6 xx Delta_(f)H(CO_(2)) + 6 Delta_(f)H (H_(2)O) - Delta_(f)H(C_(6)H_(12)O_(6)) - 6Delta_(f)(O^(2),g)` `=6 xx (-400-300) - (-1300)-0` `=-4200 + 1300` =-2900KJ/mol For one gram of glucose, enthalphy of combustion `=-(2900)/(180) =-16.11 kJ//g`

Discussion

3458.	The accompanying diagram represents a reversible cannot cycle for an ideal gas: How many heat is rejected at the lower temperature, 200K, during the isothermal compression?A. 150 KJB. 30KJC. 120KJD. data are not sufficient to calculate exact value
Answer» Correct Answer - B `-q_(c) = q_(n)((T_(c))/(T_(h))) = 150 xx (200)/(1000)` Heat rejected =30 KJ

Discussion

3459.	One mole of an ideal monoatomic gas at temperature `T` and volume `1L` expands to `2L` against a constant external pressure of one atm under adiabatic conditions, then final temperature of gas will be:A. `T+(2)/(3xx0.0821)`B. `T-(2)/(3xx0.0821)`C. `(T)/(2^(5//3-1))`D. `(T)/(2^(5//3+1))`
Answer» Correct Answer - B `Delta U=W` `nCV(T_(2)-T)= -Pxx(V_(2)-V_(1))` `(3)/(2)R(T_(2)-T)= -1 " " :. T_(2)=T-(2)/(3xx0.0821)`

Discussion

3460.	A carnot engine takes in `1000 K cal` of heat from a reservoir at `627^(@)C` and exhausts heat to sink at `27^(@)C`. What is its efficiency? What is useful work done//cycle by the engine.
Answer» Correct Answer - `66.67% ; 2.8xx10^(6)J`

Discussion

3461.	A carnot engine working between `300 K and 600 K` has work output of `800 J` per cycle. What is amount of heat energy supplied to the engine from source per cycle?A. `800 J`B. `1600 J`C. `3200 J`D. `6400 J`
Answer» Correct Answer - B Here, `T_(1)=300K,T_(2)=600K,W= 800J` As `eta=1 -(T_(2))/(T_(1))= W/(Q_(1)) or 1-(300)/(600)= (800)/(Q_(1))` `Q_(1)= 1600J`

Discussion

3462.	A Carnot engine working between K 300 and 600 K has work output of 800 J per cycle. What is amount of heat energy supplied to the engine from source per cycleA. 1800 j cycleB. 1000 j cycleC. 2000 j cycleD. 1600 j cycle
Answer» Correct Answer - D

Discussion

3463.	One mole of a diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperature at A,B and C are 400K, 800K and 600K respectively. Choose the correct statement: A. change in internal energy in the process AB `=350R`B. change in internal energy in the process BC is `-500R`C. change in internal energy in the whole cyclic process is `250R`D. change in internal energy in the process CA is `700R`
Answer» Correct Answer - B `DeltaU_(AB)=nC_(v)(T_(B)-T_(A))` `=1xx(5R)/2(800-400)=1000R` `DeltaU_(BC)= n C_(v)(T_(c)-T_(B))` `= 1xx(5R)/2(600-800)= -500R` `DeltaU_(total)=0 ( :. "Process is cyclic")` `DeltaU_(CA)= n C_(v)(T_(A)-T_(C))` `= 1xx(5R)/2(400-600)= -500R`

Discussion

3464.	Figure shows a cycle `ABCDA` undergone by `2` moles of an ideal diatomic gas. The curve `AB` is a rectangular hyperbola and `T_(1)=300K` and `T_(2)=500K`. Determine the work done by the gas in the process `ArarrB`. A. `-3.320kJ`B. `4.326kJ`C. `2.326kJ`D. `3.326kJ`
Answer» Correct Answer - A Evidently, for the process `ArarrB` `Vprop(1)/(T) ` or `VT`=constant (say,K) or `TdV+VdT=0`or `dV=-(VdT)/(T)` (a) Now, work done by the gas in a process is given by `W=SigmaPdV` or `W_(AB)=Sigma(nRTdV)/(V)=Sigma(-nRdT)` `W_(AB)=overset(500)underset(300)(int)-nRdT=-(2 mol) (8.314 J//mol-K)` `[500-300]K=-3.326kJ`

Discussion

3465.	One mole of monoatomic ideal gas follows a proces `AB`, as shown. The specific heat of the process is `(13R)/(6)`. Find the value of x on P-axis. A. `4P_(0)`B. `5P_(0)`C. `6P_(0)`D. `8P_(0)`
Answer» Correct Answer - C `C=(13R)/(6) C_(v)=(3)/(2)R (DeltaW)/(DeltaT)=C-C_(v)=(2R)/(3)` `DeltaT=(P_(0)V_(0))/(R)(5n-3) DeltaW=2P_(0)V_(0) (5n+3)` `:.n=6`

Discussion

3466.	An ideal gas heat engine operates in a Carnot cycle between `227^(@)C and 127^(@)C`. It absorbs `6K cal.` of heat at higher temperature. The amount of heat in `k cal` rejected to sink isA. `4.8`B. `2.4`C. `1.2D. `6.0`
Answer» Correct Answer - A Here, `T_(1)= 227^(@)C = 227+ 273= 500K` `T_(2)= 127^(@)C= 127+273= 400K` `Q_(1)= 6k cal Q_(2)=?` `(Q_(2))/(Q_(1))= (T_(2))/(T_(1))= (400)/(500)= 4/5` `Q_(2)= 4/5Q_(1)= 4/5xx6 kcal= 4.8Kcal`

Discussion

3467.	An ideal gas heat engine operates in Carnot cycle between `227^@C` and `127^@C`. It absorbs `6.0 xx 10^4 cal` of heat at high temperature. Amount of heat converted to work is :A. `2.4xx10^(4) cal`B. `4.8xx10^(4) cal`C. `1.2xx10^(4) cal`D. `6.0xx10^(4) cal`
Answer» Correct Answer - C `eta = (T_(2)-T_(1))/(T_(2))=(500-400)/500=1/5` `:. W=etaxxQ=1/5xx6.0xx10^(4)=1.2xx10^(4)cal`

Discussion

3468.	An ideal gas heat engine operates in Carnot cycle between `227^(@)C` and `127^(@)C`. It absorbs `6 x 10^(4) cals` of heat at higher temperature. Amount of heat converted to work isA. `2.4xx10^(4)`calB. `6xx10^(4)`calC. `1.2xx10^(4)`calD. `4.8xx10^(4)`cal
Answer» Correct Answer - A

Discussion

3469.	An ideal gas heat engine operates in Carnot cycle between `227^(@)C` and `127^(@)C`. It absorbs `6 x 10^(4) cals` of heat at higher temperature. Amount of heat converted to work isA. `4.8xx10^(4)cals`B. `2.4xx10^(4)cals`C. `1.2xx10^(4)cals`D. `6xx10^(4)cals`
Answer» Correct Answer - C Here, `T_(1)= 227^(@)C= 227+273= 500K` `T_(2)= 127^(@)C= 127+273= 400K` `Q_(1)= 6xx10^(4)cals. W=?` As `(Q_(2))/(Q_(1))= (T_(2))/(T_(1))` `:. Q_(2)= (T_(2))/(T_(1))xxQ_(1)= (400)/(500)xx6xx10^(4)` `=4.8xx10^(4)cals` `W= Q_(1)-Q_(2)= 6xx10^(4)-4.8xx10^(4)` `=1.2xx10^(4)cals.`

Discussion

3470.	Can the Carnot engine be realised in practice?
Answer» No, because carnot engine is an ideal heat engine whose conditions are not realized in practice.

Discussion

3471.	Why cannot the Carnot’s engine be realised in practice?
Answer» Because of the following reasons (i) The main difficulty is that the cylinder should come in contact with the source,sink and stand again and again over a complete cycle which is very difficult to achieve in practice. (ii) The working substance should be an ideal gas however no gas fulfils the ideal gas behaviour. (iii) A cylinder with a perfectly frictionless piston cannot be realised

3471.

Why cannot the Carnot’s engine be realised in practice?

Answer»

Because of the following reasons

(i) The main difficulty is that the cylinder should come in contact with the source,sink and stand again and again over a complete cycle which is very difficult to achieve in practice.

(ii) The working substance should be an ideal gas however no gas fulfils the ideal gas behaviour.

(iii) A cylinder with a perfectly frictionless piston cannot be realised

Discussion

3472.	At 300 K, 4 gm calcium is dissolved in hydrochloric acid in an open vessel at the atmoshphere pressure 0.821 atm. Calculate work done by the system.
Answer» Correct Answer - `-2.463` atm litre

Discussion

3473.	When `Fe_(S)` is dissovled in aqueous hydrochloric acid in a closed vessel the work done is __________
Answer» Correct Answer - zero

Discussion

3474.	Calculate the fall in temperature of helium initially at `15^(@)C`, when it is suddenly expanded to `8 times` its original volume `(gamma=5//3)`.
Answer» Here, `T_(1)=273+15=288 K` , `T_(2)=?, V_(2)=8V_(1), gamma=5//3` As expansion is sudden //adiabatic `:. T_(2)V_(2)^(gamma-1)= T_(1)V_(1)^(gamma-1)` `:. T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1) = 288((V_(1))/(8V_(1)))^((5/3-1))` or `logT_(2)= log288+2/3log(1/8)` `log288+2/3[log1-log8]` `logT_(2)= 2.4594+2/3[0-0.90031]` `1.8573` `T_(2)= antilog 1.8573= 71.99K` `:.` fall in temperature of helium `T_(1)-T_(2)=288-71.99= 216.01K` `=216.01^(@)C` `( :. "in magnitude", 1^(@)C=1K)`

Discussion

3475.	The following thermochemical equations represent combustion of ammonia and hydrogen: 4NH3(g) + 3O2(g) → 6H2O(l) + 2N2(g); ΔH = -1516 kJ 2H2(g) + O2(g) → 2H2O(l); ΔH = -572kJ Calculate enthalpy of formation of ammonia.
Answer» Required equation is 1/2N_2(g) + 3/2H_2(g) → NH_3(g) ΔH = ? 4NH_3(g) + 3O_2(g) → 2N_2(g) + 6H₂O_2(l)ΔH = -1516 kJ …(1) 2H_2(g) + O_2(g) → 2H₂O_(l) ΔH = -572 kJ …(2) Reserving equation (1) and dividing by 4 we get: 1/2N_2(g) + 2H₂O_2(l) → NH_3(g)+ 3/4O_2(g) ΔH = +379 kJ …(3) Multiplying equation (2), by 3/4 we get, 3/2H_2(g) + 3/4O₂ → 3/2H₂O_(l) ΔH = -572 × 3/4 = -429kJ ....(4) Adding (3) and (4) we get , 1/2N_2(g) + 3/2H_2(g) → NH_3(g) ;ΔH = +379 – 429 = -50 kJ/mol

3475.

The following thermochemical equations represent combustion of ammonia and hydrogen: 4NH3(g) + 3O2(g) → 6H2O(l) + 2N2(g); ΔH = -1516 kJ 2H2(g) + O2(g) → 2H2O(l); ΔH = -572kJ Calculate enthalpy of formation of ammonia.

Answer»

Required equation is 1/2N_2(g) + 3/2H_2(g) → NH_3(g) ΔH = ?

4NH_3(g) + 3O_2(g) → 2N_2(g) + 6H₂O_2(l)ΔH = -1516 kJ …(1)

2H_2(g) + O_2(g) → 2H₂O_(l) ΔH = -572 kJ …(2)

Reserving equation (1) and dividing by 4 we get:

1/2N_2(g) + 2H₂O_2(l) → NH_3(g)+ 3/4O_2(g) ΔH = +379 kJ …(3)

Multiplying equation (2), by 3/4 we get,

3/2H_2(g) + 3/4O₂ → 3/2H₂O_(l) ΔH = -572 × 3/4 = -429kJ ....(4)

Adding (3) and (4) we get ,

1/2N_2(g) + 3/2H_2(g) → NH_3(g) ;ΔH = +379 – 429 = -50 kJ/mol

Discussion

3476.	Bond energies of `N-=N, H-H` and `N-H` bonds are 945,463 & 391 kJ `mol^(-1)` respectively, the enthalpy of the following reactions is : `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`
Answer» Correct Answer - `+93 KJ` `DeltaH = 6xxDeltaH_(N-H) - DeltaH_(N-N) - 3DeltaH_(H-H) = 93 KJ`

Discussion

3477.	The absolute enthalpy of neutralization of the reaction, `MgO(s)+2HCl(aq.)+H_(2)O(l)` will beA. less than `- 57.33 kJ mol^(-1)`B. `- 57.33 kJ mol^(-1)`C. greater than `- 57.33 kJ mol^(-1)`D. `57.33 kJ mol^(-1)`
Answer» Correct Answer - C Note that in the reactions 1 mol of `H_(2) O` is formed according to the reaction `O^(2-) + 2H^(+) rarr H_(2)O` Since `O^(2-)` dinegative, its interaction with `H^(+)` releases relatively more energy. Thus, the absoulte enthalpy of neutralization of `MgO` by `HCl` is greater than `- 57.33 kJ mol^(-1)`

Discussion

3478.	The heat required to raise the temperature of a body by `1^(@)C` is calledA. Specific heatB. Heat capacityC. water equivalentD. Heat energy
Answer» Correct Answer - B

Discussion

3479.	Heat capacity isA. `(dQ)/(dT)`B. `dQxxdT`C. `sum Q. 1/(dT)`D. `dQ-dT`
Answer» Correct Answer - A

Discussion

3480.	Which is an extensive property of the system?A. No. of molesB. ViscosityC. TemperatureD. Refractive index
Answer» Correct Answer - A

Discussion

3481.	Which one of the following pairs represents the intensive properties?A. Specific heat and TemperatureB. Entropy and densityC. Enthalpy and molefractionD. Heat and temperature
Answer» Correct Answer - A

Discussion

3482.	In which of the following sets, all properties belong to same category (all extensive or all intensive) ?A. Mass, Volume, specific heatB. Temperature, Concentration, VolumeC. Heat capacity, Concentration, EntropyD. Enthalpy, Entropy, Volume
Answer» Correct Answer - D

Discussion

3483.	For the water gas reaction: `C(s) +H_(2)O(g) hArr CO(g) +H_(2)(g)` the standard Gibbs enegry for the reaction at `1000K` is `-8.1 kJ mol^(-1)`. Calculate its equilibrium constant.
Answer» `-DeltaG^(@)=2.303RTlogK_(C)` `+8.1xx10^(3)=2.303xx8.314xx1000 logK_(C)` `K_(C)=2.648mol litre^(-1)`

Discussion

3484.	For the water gas reaction, `C(s) +H_(2)O(g) hArr CO(g)+H_(2)(g)` the standard Gobbs free energy of reaction (at `1000K)` is `-8.1 kJ mol^(-1)`. Calculate its equilibrium constant.
Answer» We know, `K - antilog ((-DeltaG^(Theta))/(2.3030 RT)) …(i)` Given that, `DeltaG^(Theta) =- 8.1 kJ mol^(-1)` `R = 8.314 xx 10^(-3) k J K^(-1)mol^(-1)` `T = 1000K` Substituting these values in equation (i), we get `K = antilog [(+(8.1))/(2.303 xx8.314xx 10^(-3)xx1000)]` `K = 2.65`

Discussion

3485.	Which of the following are extensive properties ?A. Elevation in boiling pointB. Boiling pointC. emf of cellD. `E^(Theta)` of cell
Answer» Mass-dependent properties are extensive.

Discussion

3486.	Which of the following are intensive properties?A. Heat capacityB. Refractive indexC. Specific volumeD. Entropy
Answer» Mass-independent properties are intensive.

Discussion

3487.	`DeltaG^(Theta)` tells us: a. Whether a change is feasible or not. b. How far a reaction will proceed. c. About emegry of activation.
Answer» `DeltaG` tells about feasibility of a reaction.

Discussion

3488.	For the water gas reaction: `C(s) +H_(2)O(g) hArr CO(g) +H_(2)(g)` the standard Gibbs enegry for the reaction at `1000K` is `-8.1 kJ mol^(-1)`. Calculate its equilibriu constant.
Answer» We know that: `Delta_(r)G^(Theta) = - 2.303 RT log K` or `logK = (-Delta_(r)G)/(2.303 RT)` `Delta_(r)G^(Theta) =- 8.1 kJ mol^(-1), T = 1000K`, `R = 8.314 xx 10^(-3) kJ mol^(-1)K^(-1)` `:. Log K =- (8.1)/(2.303 xx 8.314 xx 10^(-3) xx 1000) = 0.423` or `K = 2.64`

Discussion

3489.	The standard Gibbs energy change value `(Deltat_(r)G^(Theta))` at `1773K` are given for the following reactions: `4Fe +3O_(2) rarr 2Fe_(2)O_(3), Delta_(r)G^(Theta) = - 1487 kJ mol^(-1)` `4AI +3O_(2) rarr 2AI_(2)O_(3),Delta_(r)G^(Theta) =- 22500 kJ mol^(-1)` `2CO +O_(2) rarr 2CO_(2),Delta_(r)G^(Theta) =- 515 kJ mol^(-1)` Find out the possibility of reducing `Fe_(2)O_(3)` and `AI_(2)O_(3)` with `CO` at this temperature.
Answer» `2AI_(2)O_(3)rarr 4AI +3O_(2), Delta_(r)G^(Theta) = 22500 kJ mol^(-1)` Since `Delta_(r)G^(Theta)` is `+ve` decomposition is non-spontaneous. Let us consider another reaction, `2CO +O_(2) rarr 2CO_(2), Delta_(r)G^(Theta) = - 515 kJ mol^(-1)` We wish to consider whther reduction of `AI_(2)O_(3)` with `CO` is feasible or not.

Discussion

3490.	Which of the following are not state functions? (I) `q+w` (II)`q` (III) `w` (IV) `H-TS`A. I and IVB. II, III and IVC. I, II and IIID. II and III
Answer» Correct Answer - D The thermodynamic parameters which depend only upon the initial and final states of system, are called state functions, such as enthalpy (H=q+W) , Gibbs free energy (G=H-TS) , etc. While those parameters which depend on the path by which those parameters whch depend on the path by which the process is performed rather than on the initial and final states, are called path functions, such as work done , heat, etc.

Discussion

3491.	A container of volume `1m^(3)` is divided into two equal compartments by a partition. One of these compartments contains an ideal gas at 300 K . The other compartment is vacuum. The whole system is thermally isolated from its surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would beA. 300 kB. 239 kC. 200 kD. 100 k
Answer» Correct Answer - A

Discussion

3492.	110 J of heat is added to a gaseous system, whose internal energy change is 40 j. then the amount of external work done isA. `150j`B. `70j`C. `110j`D. `40j`
Answer» Correct Answer - B

Discussion

3493.	Ten litres of water are boiled at 100°C, at a pressure of 1.013 × 105 Pa, and converted into steam. The specific latent heat of vaporization of water is 539 cal/g. Find (a) the heat supplied to the system (b) the work done by the system (c) the change in the internal energy of the system. 1 cm3 of water on conversion into steam, occupies 1671 cm3 (J = 4.186 J/cal)
Answer» Data : P = 1.013 × 10⁵ Pa, V (water) = 10 L = 10^-3 × 10³ m , V(steam) = 1671 × 10^-3 × 10³ m , L = 539 cal/g = 539 × 10³ \(\frac{cal}{kg}\) = 539 × 10³ × 4.186 \(\frac{J}{kg}\) as J = 4.186 J/cal, mass of the water (M) = volume × density = 10 × 10^-3 m³ × 10³ kg/m³ = 10 kg (a) Q = ML = (10) (5.39 × 4.186 × 10⁵) J = 2.256 × 10⁷ J This is the heat supplied to the system (b) W = P∆V = (1.013 × 10⁵) (1671 – 1) × 10^-2J = (1.013) (1670) × 10³ J = 1.692 × 10⁶ J This is the work done by the system. (c) ∆ U = Q – W = 22.56 × 10⁶ – 1.692 × 10⁶ = 2.0868 × 107J This is the change (increase) in the internal energy of the system.

3493.

Ten litres of water are boiled at 100°C, at a pressure of 1.013 × 105 Pa, and converted into steam. The specific latent heat of vaporization of water is 539 cal/g. Find (a) the heat supplied to the system (b) the work done by the system (c) the change in the internal energy of the system. 1 cm3 of water on conversion into steam, occupies 1671 cm3 (J = 4.186 J/cal)

Answer»

Data : P = 1.013 × 10⁵ Pa, V (water) = 10 L = 10^-3 × 10³ m , V(steam) = 1671 × 10^-3 × 10³ m ,

L = 539 cal/g = 539 × 10³ \(\frac{cal}{kg}\) = 539 × 10³ × 4.186 \(\frac{J}{kg}\) as J = 4.186 J/cal, mass of the water (M) = volume × density = 10 × 10^-3 m³ × 10³ kg/m³ = 10 kg

(a) Q = ML = (10) (5.39 × 4.186 × 10⁵) J = 2.256 × 10⁷ J

This is the heat supplied to the system

(b) W = P∆V = (1.013 × 10⁵) (1671 – 1) × 10^-2J = (1.013) (1670) × 10³ J = 1.692 × 10⁶ J

This is the work done by the system.

(c) ∆ U = Q – W = 22.56 × 10⁶ – 1.692 × 10⁶ = 2.0868 × 107J

This is the change (increase) in the internal energy of the system.

Discussion

3494.	For the percipitation reaction of `Ag^(o+)` ions with `NaCI`, which of the following statements is true?A. `DeltaH=0`B. `Delta G=0`C. `Delta G=` -veD. `Delta G=Delta H`
Answer» Correct Answer - C

Discussion

3495.	Enthalpy of solution of `NaCl(s)` is`+ 4kJ mol^(-1)` and enthalpy of hydration is -784kJ `mol^(-1)`. Lattice enthalpy of NaCl(s) will be `"…..............." kJ mol^(-1)`
Answer» `+ 788kJ mol^(-1)( Delta_("sol") H = Delta_("lattice")H + Delta_(hyd) H ) `

Discussion

3496.	Lattice energy of `NaCl_((s))` is `-788kJ mol^(-1)` and enthalpy of hydration is `-784kJ mol^(-1)`. Calculate the heat of solution of `NaCl_((s))`.A. `-4kJ`B. `4KJ`C. `-6KJ`D. `2KJ`
Answer» Correct Answer - B heat of solution = L.E+H.E

Discussion

3497.	The value of ∆H for cooling 2 moles of an ideal mono atomic gas from 125°C to 25°C at constant pressure will be [given CP = \(\frac{5}{2}\) R] ……(a) -250 R (b) -500 R (c) 500 R (d) +250 R
Answer» (b) -500 R T_i = 125°C = 398K T_f = 25°C = 298 K ∆H = nC_p (T_f – T_i ) ∆H = 2 x \(\frac{5}{2}\) R(298 – 398) ∆H = -500 R

3497.

The value of ∆H for cooling 2 moles of an ideal mono atomic gas from 125°C to 25°C at constant pressure will be [given CP = \(\frac{5}{2}\) R] ……(a) -250 R (b) -500 R (c) 500 R (d) +250 R

Answer»

(b) -500 R

T_i = 125°C = 398K

T_f = 25°C = 298 K

∆H = nC_p (T_f – T_i )

∆H = 2 x \(\frac{5}{2}\) R(298 – 398)

∆H = -500 R

Discussion

3498.	Calculate the work done when `56g` of iron reacts with hydrochloric acid in `(a)` a closed vessel of fixed volume and `(b)`an open beaker at `25^(@)C`.
Answer» a. Vessel is of fixed volume, hence `DeltaV = 0`. No work is done. b. The gas driver back the atmosphere hence. `W =- P_(ex) DeltaV` Also, `DeltaV = V_("final") - V_("initial") = V_("final") ( :. V_("initial") = 0)` `:. DeltaV = (nRT)/(P_(ex))` or `W =- P_(ex) =- nRT` where `n` is the number of mole of `H_(2)`gas obtained from `n` mole of `Fe(s)`. `Fe(s) +2HCI(aq) rarr FeCI_(2)(aq) +H_(2)(g)` 1mol, 1mol `:. n = (56)/(56) = 1mol` `:. w =- 1 xx 8.314 xx 298 =- 2477.57J` The raaction mixture in the given system does `2.477kJ` of work driving back to atmosphere.

Discussion

3499.	When ice melts, the change in internal energy is greater than the heat supplied. Why ?
Answer» When ice melts, volume of water formed is less than that of ice. So, surroundings (environment) does work on the system (ice). And by first law, ∆Q = ∆ W + ∆U ∆U = ∆Q - ∆ W (∆ W= negative as work is done on the system) ∆U > ∆Q

3499.

When ice melts, the change in internal energy is greater than the heat supplied. Why ?

Answer»

When ice melts, volume of water formed is less than that of ice. So, surroundings (environment) does work on the system (ice).

And by first law,

∆Q = ∆ W + ∆U

∆U = ∆Q - ∆ W (∆ W= negative as work is done on the system)

∆U > ∆Q

Discussion

3500.	Can the specific heat of a gas be infinity ?
Answer» Yes, the specific heat of a gas be infinity.

Discussion

Explore topic-wise InterviewSolutions in .

One mole of anhydrous `MgCl_(2)` dissolves in water and librates `25 cal//mol` of heat. `Delta H_("hydration")` of `MgCl_(2)=30 cal//mol`. Heat of dissolution of `MgCl.H_(2)O`A. `+5` cal/molB. `-5` cal/molC. 55 cal/molD. `-55` cal/mol

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A carnot engine working between `300 K and 600 K` has work output of `800 J` per cycle. What is amount of heat energy supplied to the engine from source per cycle?A. `800 J`B. `1600 J`C. `3200 J`D. `6400 J`

A Carnot engine working between K 300 and 600 K has work output of 800 J per cycle. What is amount of heat energy supplied to the engine from source per cycleA. 1800 j cycleB. 1000 j cycleC. 2000 j cycleD. 1600 j cycle

Figure shows a cycle `ABCDA` undergone by `2` moles of an ideal diatomic gas. The curve `AB` is a rectangular hyperbola and `T_(1)=300K` and `T_(2)=500K`. Determine the work done by the gas in the process `ArarrB`. A. `-3.320kJ`B. `4.326kJ`C. `2.326kJ`D. `3.326kJ`

One mole of monoatomic ideal gas follows a proces `AB`, as shown. The specific heat of the process is `(13R)/(6)`. Find the value of x on P-axis. A. `4P_(0)`B. `5P_(0)`C. `6P_(0)`D. `8P_(0)`

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An ideal gas heat engine operates in Carnot cycle between `227^@C` and `127^@C`. It absorbs `6.0 xx 10^4 cal` of heat at high temperature. Amount of heat converted to work is :A. `2.4xx10^(4) cal`B. `4.8xx10^(4) cal`C. `1.2xx10^(4) cal`D. `6.0xx10^(4) cal`

An ideal gas heat engine operates in Carnot cycle between `227^(@)C` and `127^(@)C`. It absorbs `6 x 10^(4) cals` of heat at higher temperature. Amount of heat converted to work isA. `2.4xx10^(4)`calB. `6xx10^(4)`calC. `1.2xx10^(4)`calD. `4.8xx10^(4)`cal

An ideal gas heat engine operates in Carnot cycle between `227^(@)C` and `127^(@)C`. It absorbs `6 x 10^(4) cals` of heat at higher temperature. Amount of heat converted to work isA. `4.8xx10^(4)cals`B. `2.4xx10^(4)cals`C. `1.2xx10^(4)cals`D. `6xx10^(4)cals`

Can the Carnot engine be realised in practice?

Why cannot the Carnot’s engine be realised in practice?

At 300 K, 4 gm calcium is dissolved in hydrochloric acid in an open vessel at the atmoshphere pressure 0.821 atm. Calculate work done by the system.

When `Fe_(S)` is dissovled in aqueous hydrochloric acid in a closed vessel the work done is __________

Calculate the fall in temperature of helium initially at `15^(@)C`, when it is suddenly expanded to `8 times` its original volume `(gamma=5//3)`.

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Bond energies of `N-=N, H-H` and `N-H` bonds are 945,463 & 391 kJ `mol^(-1)` respectively, the enthalpy of the following reactions is : `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`

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Which is an extensive property of the system?A. No. of molesB. ViscosityC. TemperatureD. Refractive index

Which one of the following pairs represents the intensive properties?A. Specific heat and TemperatureB. Entropy and densityC. Enthalpy and molefractionD. Heat and temperature

In which of the following sets, all properties belong to same category (all extensive or all intensive) ?A. Mass, Volume, specific heatB. Temperature, Concentration, VolumeC. Heat capacity, Concentration, EntropyD. Enthalpy, Entropy, Volume

For the water gas reaction: `C(s) +H_(2)O(g) hArr CO(g) +H_(2)(g)` the standard Gibbs enegry for the reaction at `1000K` is `-8.1 kJ mol^(-1)`. Calculate its equilibrium constant.

For the water gas reaction, `C(s) +H_(2)O(g) hArr CO(g)+H_(2)(g)` the standard Gobbs free energy of reaction (at `1000K)` is `-8.1 kJ mol^(-1)`. Calculate its equilibrium constant.

Which of the following are extensive properties ?A. Elevation in boiling pointB. Boiling pointC. emf of cellD. `E^(Theta)` of cell

Which of the following are intensive properties?A. Heat capacityB. Refractive indexC. Specific volumeD. Entropy

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For the water gas reaction: `C(s) +H_(2)O(g) hArr CO(g) +H_(2)(g)` the standard Gibbs enegry for the reaction at `1000K` is `-8.1 kJ mol^(-1)`. Calculate its equilibriu constant.

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For the percipitation reaction of `Ag^(o+)` ions with `NaCI`, which of the following statements is true?A. `DeltaH=0`B. `Delta G=0`C. `Delta G=` -veD. `Delta G=Delta H`

Enthalpy of solution of `NaCl(s)` is`+ 4kJ mol^(-1)` and enthalpy of hydration is -784kJ `mol^(-1)`. Lattice enthalpy of NaCl(s) will be `"…..............." kJ mol^(-1)`

Lattice energy of `NaCl_((s))` is `-788kJ mol^(-1)` and enthalpy of hydration is `-784kJ mol^(-1)`. Calculate the heat of solution of `NaCl_((s))`.A. `-4kJ`B. `4KJ`C. `-6KJ`D. `2KJ`

The value of ∆H for cooling 2 moles of an ideal mono atomic gas from 125°C to 25°C at constant pressure will be [given CP = \(\frac{5}{2}\) R] ……(a) -250 R (b) -500 R (c) 500 R (d) +250 R

Calculate the work done when `56g` of iron reacts with hydrochloric acid in `(a)` a closed vessel of fixed volume and `(b)`an open beaker at `25^(@)C`.

When ice melts, the change in internal energy is greater than the heat supplied. Why ?

Can the specific heat of a gas be infinity ?