4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 69 InterviewSolution

3401.	Which one of the following statement is false :A. work is a state functionB. temperetaue is a state functionC. change in the state is completely defined when the initial and final states are specified.D. work appears at the boundary of the system.
Answer» Correct Answer - A

Discussion

3402.	Select the option which is correct regarding true//false of the given stateements: Statements-1: Thermal decomposition of solid `CaCO_(3)` is expected to be spontaneous as high temperature and non-spontaneous at low temperatures. Statements-2: If the rate law for an acid catalysed reaction is `r=K(Reactant)^(1)[Catalyst]^(@)` then rate will be independent of presence / absence of catalyst. Statement-3: When water freezes in a glass beaker, volumes of the system increases at constant temperature and hence entropy will also increses.A. All statement are correctB. Only statemen-3 is incorrectC. only statement-1 is correctD. Only statement-2 is incorrect
Answer» Correct Answer - C

Discussion

3403.	What will be the standard internal energy change for the reaction at 298 K ? `OF_(2(g))+H_(2)O_((g))+2HF_((g)),DeltaH^(@)=-310kJ`A. `-312.5Kj`B. `-125.03kJ`C. `-310 kJ`D. `-156kJ`
Answer» Correct Answer - A `DeltaH=DeltaU+Deltan_(g)RT` `DeltaH=-310xx10^(3)J, Deltan_(g)=3-2=1, R="8.314 J K"^(-1)"mol"^(-1),` T = 298 K `DeltaU=-310 xx10^(3)-(1xx 8.314xx298)` `=-312.477xx106(3)J=-312.5kJ`

Discussion

3404.	Given the following equations and `DeltaH^(@)` values, determine the enthalpy of reaction at 298 K for the reaction : `C_(2)H_(4)(g)+6F_(2)(g) to 2HF_(4)(g)+4HF(g)` `H_(2)(g)+F_(2)(g)to2HF(g)+2HF(g)," "DeltaH_(1)^(@)=-537" kJ"` `C(s)+2F_(2)(g)toCF_(4)(g)," "DeltaH_(2)^(@)=-680" kJ"` `2C(s)+2H_(2)(g)toC_(2)H_(4)(g)," "DeltaH_(3)^(@)=52" kJ"`A. `-1165`B. `-2382`C. `+1165`D. `+2382`
Answer» Correct Answer - B

Discussion

3405.	Given the following equations and `DeltaH^(@)` values, determine the enthalpy of reaction at 298 K for the reaction : `C_(2)H_(4)(g)+6F_(2)(g) to 2HF_(4)(g)+4HF(g)` `H_(2)(g)+F_(2)(g)to2HF(g)+2HF(g)," "DeltaH_(1)^(@)=-537" kJ"` `C(s)+2F_(2)(g)toCF_(4)(g)," "DeltaH_(2)^(@)=-680" kJ"` `2C(s)+2H_(2)(g)toC_(2)H_(4)(g)," "DeltaH_(3)^(@)=52" kJ"`A. `-1165`B. `-2486`C. `+1165`D. `+2486`
Answer» Correct Answer - B `DeltaH^(@)=2xxDeltaH_(1)^(@)+2xxDeltaH_(2)^(@)-DeltaH_(3)^(@)`

Discussion

3406.	Find bond enthalpy of S-S bond from the following data: `C_(2)H_(5)-S-C_(2)H_(5)," "DeltaH_(f)^(@)=-147.2kJ" "mol^(-1)` `C_(2)H_(5)-S-S-C_(2)H_(5)," "DeltaH_(f)^(@)=-201.9kJ" "mol^(-1)` `S(g)," "DeltaH_(f)^(@)=222.8kJ" "mol^(-1)`A. `-168.1 KJ//"mol"`B. `+168.1 KJ//"mol"`C. `-277.5 KJ//"mol"`D. `+277.5 KJ//"mol"`
Answer» Correct Answer - D `C_(2)H_(5)-S-C_(2)H_(2)+Srarr C_(2)H_(5)-S-S-C_(2)H_(5)` `DeltaH_(f)^(@)(C_(2)H_(5)-S-C_(2)H_(5))=DeltaH_(f)^(@)(C_(2)H_(5)_S-S-C_(2)H_(5))-DeltaH_(S-S)-DeltaH_("sub")S.` `DeltaH_(S-S)=277.5 KJ//"mol"`

Discussion

3407.	If `900J//g `of heat is exchanged at boiling point of water then water is the increase in entropy.A. 43.4 J/K-moleB. 87.2 J/K moleC. 900 JK/molD. Zero
Answer» Correct Answer - A `Delta S? = (Delta H)/(T)` `Delta H = 900 J//g` `Delta H = 900xx18 J//mol` and boiling point of water = 373 K `therefore Delta S = (900xx18)/(373)=43.4 J//K-mol`

Discussion

3408.	a monoatomic gas `(Cv=(3)/(2)R)` is allowed to expand adiabaticaly and reversibly from initial volume of 8 L to 300 K to a volume `V_(2)` at 250 K. `V_(2)` is :A. 10.5 LB. 23 LC. 8.5 LD. 50.5 L
Answer» Correct Answer - A

Discussion

3409.	5 mol of an ideal gas expand reversibly from a volume of `8 dm^(3)` to `80 dm^(3)` at an temperature of `27^(@)C`. Calculate the change in entropy.A. `41.57 JK^(-1)`B. `-95.73 JK^(-1)`C. `95.73 JK^(-1)`D. `-41.57 JK^(-1)`
Answer» Correct Answer - C `Delta S = 2.303 nR log ((V_(2))/(V_(1)))` `Delta S = 2.303xx5xx8.314 log ((80)/(8))`

Discussion

3410.	`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. `19.15`B. `-19.15`C. `4.7`D. zero
Answer» Correct Answer - D

Discussion

3411.	A gas `(C_(v.m) = (5)/(2)R)` behaving ideally is allowed to expand reversibly and adiabatically from `1` litre to `32` litre. Its initial temperature is `327^(@)C`. The molar enthalpy change (in `J//mol`) for the process is :A. `-1125R`B. `-675`C. `-1575R`D. None of these
Answer» Correct Answer - C `(T_(2))/(T_(1))=((V_(1))/(V_(2)))^(gamma-1)` `T_(2)=T_(1).((1)/(32))^((7)/(5)-1)=600.((1)/(2^(5)))^((2)/(5))` `=600(0.5)^(2)=150K` `DeltaH_(m)=(7)/(2)Rxx(150-600)=-1575R`

Discussion

3412.	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. `16.6J//K`B. `9J//K`C. `34.46J//K`D. `1.89J//K`
Answer» Correct Answer - C

Discussion

3413.	What is the change in entropy when `2.5` mole of water is heated from `27^(@)` to `87^(@) C` ? Assume that the heat capacity is constant . `(C_(p.m)(H_(2)O) = 4.2 J//g-k In (1.2)= 0.18)`A. `16.6` J/KB. `9` J/KC. `34.02` J/KD. `1.89` J/k
Answer» Correct Answer - C

Discussion

3414.	For the reaction, `2HgO_((s)) rarr 2Hg_((l))+O_(2(g))`A. `Delta H gt 0` & `Delta S lt 0`B. `Delta H gt 0` & `Delta S gt 0`C. `Delta H lt 0` & `Delta S lt 0`D. `Delta H lt 0` & `Delta S gt 0`
Answer» Correct Answer - B

Discussion

3415.	If a steam engine delivers `6.0xx10^(8)J` of work per minute and absorbs `5.4xx10^(9)J` of heat per minute from its boiler then the efficiency of the engine isA. 0.11B. 0.12C. 0.13D. 0.14
Answer» Correct Answer - A Efficiency of engine =(working out put(W))/(heat input(Q)) `=(6.0xx106(8))/(5.4xx10^(9))=0.11=0.11xx100%=11%`

Discussion

3416.	An engine has an efficiency of 0.25 when temperature of sink is reduced by `58^(@)C` , If its efficency is doubled, then the temperature of the source isA. `150 ^(@)C`B. `222 ^(@)C`C. `242 ^(@)C`D. `232 ^(@)C`
Answer» Correct Answer - D Here `eta_(1)=1-(T_(2))/(T_(1))` or 0.25 =1 -`(T_(2))/(T_(1)) rArr (1)/(4) =1-(T_(2))/(T_(1))` `(T_(2))/(t_(1))=1-(1)/(4)=(3)/(4)` According to question, `eta_(2)=2eta_(1)` and `T_(2)=T_(2)-58^(@)C` `therefore 2xx(1)/(4)=1(T_(2)-58^(@)C)/(T_(1)) rArr (1-(1))/(2)=(T_(2)-58^(@)C)/(T_(1))` `(1)/(2)=(T_(2))/(T_(1))-(58^(@))/(t_(1))rArr(3)/(4)-(1)/(2)=(58)/(T_(1)) rArr T_(1)=232^(@)C`

Discussion

3417.	Prove that for an adiabatic process TVγ-1 constant.
Answer» We know that for an adiabatic process PV^γ = constant ……(1) For an ideal gas, \(\frac{PV}{T}\) = constant, C …….(2) ∴ P = \(\frac{C.T}{V}=\) ........(3) using equation (3) in (1), \((\frac{C.T}{T})V^γ\) = constant. i.e. T.V^γ-1 = constant.

3417.

Prove that for an adiabatic process TVγ-1 constant.

Answer»

We know that for an adiabatic process

PV^γ = constant ……(1)

For an ideal gas, \(\frac{PV}{T}\) = constant, C …….(2)

∴ P = \(\frac{C.T}{V}=\) ........(3)

using equation (3) in (1),

\((\frac{C.T}{T})V^γ\) = constant.

i.e. T.V^γ-1 = constant.

Discussion

3418.	Assetion : A refrigerator transfers heat from a lower temperature to a higher temperature. Reason: Heat cannot flow from a lower temperature to a higher temperature.A. If both assetion and reason are true and reaasons is the correct expanation of assetionB. If both assetion and reason are tur but reason is not the correct explanation of assetionC. If assertion is true but reason is falseD. If both assertion and reason are false.
Answer» Correct Answer - B Refrigerator uses the electric energy to transfer heat from a lower temperature to a higher temperature.

Discussion

3419.	What is a heat engine? Define efficiency of a heat engine.
Answer» Heat engine is a device that performs the conversion of heat energy to mechanical work through cyclic process. The efficiency of a heat engine is defined as the ratio of work done by the heat engine to heat absorbed per cycle. If a heat engine absorbs Q₁ amount of energy from the source and dissipates Q₂ amount of energy to sink, the efficiency η, is given by, \(η = \frac{Q_1-Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}.\)

3419.

What is a heat engine? Define efficiency of a heat engine.

Answer»

Heat engine is a device that performs the conversion of heat energy to mechanical work through cyclic process. The efficiency of a heat engine is defined as the ratio of work done by the heat engine to heat absorbed per cycle. If a heat engine absorbs Q₁ amount of energy from the source and dissipates Q₂ amount of energy to sink, the efficiency η, is given by,

\(η = \frac{Q_1-Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}.\)

Discussion

3420.	In a refrigerator one removes heat from a lower temperature and deposits to the surrounding at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power, and heat is transferred from −3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50 % of a perfect engine.
Answer» Given : T₁ = 27°C = 27 + 273 = 300 K T₂ = −3°C = −3 + 273 = 270 K Efficiency, η = 1 – \(\frac{T_2}{T_1}\) = 1 - \(\frac{270}{300}\) = 1 – 0.9 = 0.1 = \(\frac{1}{10}\) Efficiency of refrigerator is 50% of percentage engine. η' = 50% of η = 0.5 ×\(\frac{1}{10}\)=\(\frac{1}{20}\) ∴ Coefficient of performance, \(β=\frac{Q_2}{W}=\frac{1-η'}{η'}\) \(β=\frac{1-\frac{1}{20}}{\frac{1}{20}}=\frac{1-0.05}{0.05}\) = \(\frac{0.95}{0.05}\) = 19 Q₂= 19% of work done by motor on refrigerator = 19 × 1 = 19 K J/S

3420.

In a refrigerator one removes heat from a lower temperature and deposits to the surrounding at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power, and heat is transferred from −3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50 % of a perfect engine.

Answer»

Given :

T₁ = 27°C = 27 + 273 = 300 K

T₂ = −3°C = −3 + 273 = 270 K

Efficiency,

η = 1 – \(\frac{T_2}{T_1}\)

= 1 - \(\frac{270}{300}\)

= 1 – 0.9

= 0.1 = \(\frac{1}{10}\)

Efficiency of refrigerator is 50% of percentage engine.

η' = 50% of η = 0.5 ×\(\frac{1}{10}\)=\(\frac{1}{20}\)

∴ Coefficient of performance,

\(β=\frac{Q_2}{W}=\frac{1-η'}{η'}\)

\(β=\frac{1-\frac{1}{20}}{\frac{1}{20}}=\frac{1-0.05}{0.05}\)

= \(\frac{0.95}{0.05}\) = 19

Q₂= 19% of work done by motor on refrigerator

= 19 × 1

= 19 K J/S

Discussion

3421.	One mole of carbon undergoes incomplete combustion to produce carbon monoxide, Calculate `(Delta H - Delta U)` for the formation of CO at 298 K. Given R = 8.314 `JK^(-1) mol^(-1)`.
Answer» 1238.786 J `mol^(-1)`

Discussion

3422.	Why `Delta H_(f)^(@)` of elements is zero?
Answer» It is because elements occur naturally but cannot be prepared.

Discussion

3423.	The standard heat of formation of at 298 K for `CCl_(4)(g), H_(2)O(g), CO_(2)(g) and HCI(g)` are -25.5, -57.8, -94.1 and -22.1 kcal `mol^(-1)` respectively. Calculate `Delta_(r)H^(Theta)` for the reaction r`CCl_(4)(g)+2H_(2)O(g) rarr CO_(2)(g)+4HCl(g)` Hint : `Delta_(r)H^(Theta) = sum Delta_(f)H_("Products")^(Theta)- sum Delta_(f)H_("Reactants")^(Theta)`
Answer» Correct Answer - 41.4 kcal.

Discussion

3424.	At 298 K, the heats of formation fo `C Cl_(4) (g), H_(2)O(g), CO_(2)(g)` and `HCl(g)` are `-25, -57, -94` and `-22` K.cal `mol^(-1)` respectively. The `DeltaH` of the reaction: `C Cl_(4)(g) +2H_(2)O rarr CO_(2) (g) +4HCl(g)`A. `-43` K. calB. `+72` K.calC. `14` K.calD. `-125` K.cal
Answer» Correct Answer - A `DeltaH=H_(P)-H_(R)`

Discussion

3425.	Work is a state function which is expressed in joule. work appears only at the boundary of the system.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE
Answer» Correct Answer - D

Discussion

3426.	Which of the following statement is true? Worik appears at the boundary of the system. (ii) Change in the state is completely defined when the initial and fianl states are specified. (iii) Temperature is a state funciton. (iv) Work is a state function.A. (i),(ii),(iii)B. (i),(ii),(iii),(iv)C. (ii),(iii),(iv)D. (ii),(iii)
Answer» Correct Answer - A Work is not a state function because system does not process any work. It can only exchange work with its surroundings. It is a path function.

Discussion

3427.	Which statement is correct?A. `"In a coffe-cup calorimeter", q=DeltaH`B. `"In a coffe-cup calorimeter", w=0`C. `"In bomb calorimeter", q=DeltaS`D. `"In bomb calorimeter", wgt0`
Answer» Correct Answer - A

Discussion

3428.	`DeltaH_(f)^(@)` of water is `-285.5 KJ mol^(-1)`. If enthalpy of neutraliztion of monoacidic strong base is `-57.3KJ mol^(-1)` then `DeltaH_(f)^(@)` of `OH^(-)` ion will be :A. `-285.5KJmol^(-1)`B. `285.5KJmol^(-1)`C. `114.5KJmol^(-1)`D. `-114.5KJmol^(-1)`
Answer» Correct Answer - a

Discussion

3429.	Which one of the following is a state property/function?A. HeatB. WorkC. Internal energyD. Potential enegry
Answer» Internal enegry is state funciton.

Discussion

3430.	The enthalpy of neutralisation of a strong acid by a string base is `-57.32 kJ mol^(-1)`. The enthalpy of formation of water is `-285.84 kJ mol^(-1)`. The enthalpy of formation of hydroxy`1` ion isA. `+228.52 kJ mol^(-1)`B. `-114.26 kJ mol^(-1)`C. `-228.52 kJ mol^(-1)`D. `+114.2 kJ mol^(-1)`
Answer» The process of neutralisation is `H^(o+)(aq) + overset(Theta)OH (aq) rarr H_(2)O(l),DeltaH^(Theta) =- 57.32 kJ mol^(-1)` `DeltaH_(reaction)^(Theta)=sum` Heat of formation of products `-sum`Heat of fromation of reactants `= Delta_(f)H_(H_(2)O(l))^(Theta)-(Delta_(f)H_(H^(o+)(aq))^(Theta) +Delta_(f)H_(OH_((aq))^(Theta))^(Theta))` `-57.32 =- 285.84 +57.32 =- 228.52 kJ` `x =- 285.84 + 57.32 =- 228.52 kJ` Thus, the entalpy of formation of hydroxy`1` ion is `-228.52 kJ`.

Discussion

3431.	A process in which pressure remians constant is calledA. Isochroic processB. Isothermal processC. Adiabatic processD. Isobaric process
Answer» Isobaric process `(DeltaP = 0)`.

Discussion

3432.	Which of the following acid will release maximum amount of heat when completely neutralised by strong base `NaOH`?A. `1 M HCI`B. `1M HNO_(3)`C. `1M HCIO_(4)`D. `1M H_(2)SO_(4)`
Answer» Ionisation of `H_(2)SO_(4)` given double amount of `H^(o+)` ions as compared to other acids. `H_(2)SO_(4)rarr 2H^(o+) +SO_(4)^(2-)`

Discussion

3433.	An ideal gas is expanded adiabatically at an initial temperature of 300 K so that its volume is doubled. The final temperature of the hydrogen gas is `lambda=1.40)`A. 227.36 kB. 500.30 kC. 454.76 kD. `-47^(@)`
Answer» Correct Answer - A

Discussion

3434.	One mole of an ideal gas `(C_(v,m)=(5)/(2)R)` at 300 K and 5 atm is expanded adiabatically to a final pressure of 2 atm against a constant pressure of 2 atm. Final temperature of the gas :A. 270 KB. 273 KC. 248.5 KD. 200 K
Answer» Correct Answer - C

Discussion

3435.	Which of the following pairs of a chemical reaction is certaion to result a spontaneous reaction ?A. endothermic and decreasing disorderB. exothermic and increasing disorderC. endothermic and increasing disorderD. exothermic and decreasing disorder
Answer» Correct Answer - B

Discussion

3436.	The change in enthalpy of a substqance when one mole of the substance is completely burnt in excess of air is calledA. Heat of sublimationB. Heat of combustionC. Heat of fusionD. Both (1) & (2)
Answer» Correct Answer - B

Discussion

3437.	A system absorbs 300 cal of heat. The work done by the system is 200 cal. `Delta U` for the above change isA. 100 calB. 500 calC. `-500` calD. `-100` cal
Answer» Correct Answer - A

Discussion

3438.	`1 gH_(2)` gas `STP` is expanded so that the volume is doubled. Hence, work done isA. `22.4 L-atm`B. `5.6 L-atm`C. `11.2 L-atm`D. `44.8 L-atm`
Answer» `V_(1) (volume of 1g H_(2)) = 11.2 L at STP` `V_(2)(volume of 1g H_(2)) = 22.4 L` ltbr. `:. W =P DeltaV = 11.2 L atm`

Discussion

3439.	Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction ?A. Exothermic and decreasing disoderB. Endothermic and increasing disorderC. Exothermic and increasing disorderD. Endothermic and decreasing disorder
Answer» Correct Answer - C If reaction is exothermic , therefore `DeltaH` is negative and on increasing disorder, `DeltaS` is positive thus, at these condition, `DeltaG` is negative according to following equation `DeltaG=DeltaH-TDeltaS` `DeltaG` = negative, and for spontaneous reaction `DeltaG` must be negative.

Discussion

3440.	`DeltaU=q+w`, is mathematical expression forA. first law of thermodynamicsB. second law of thermodynamicsC. third law of thermodynamicsD. zeroth law of thermodynamics
Answer» Correct Answer - A

Discussion

3441.	In an isothermal processA. q = 0 and `Delta U` = 0B. q `ne` 0 and `Delta U` = 0C. q = 0 and `Delta U ne` 0D. `q ne 0 and Delta U ne 0`
Answer» Correct Answer - B

Discussion

3442.	16 g of `O_(2)` gas at STP is expanded so that volume is doubled. Hence, work done isA. `-22.4` L atmB. `-44.8` L atmC. `-11.2` L atmD. `-5.6` L atm
Answer» Correct Answer - C

Discussion

3443.	What will be the work done when one mole of a gas expands isothermally from 15 L to 50 L against a constant pressure of 1 atm at `25^(@)C`?A. `-3542 cal`B. `-843.3cal`C. `-718cal`D. `-60.23cal`
Answer» Correct Answer - C

Discussion

3444.	Considering entropy `(S)` as a thermodynamics parameter, the criterion for the spontaneity of any process isA. `DeltaS_("system") + DeltaS_("surrounding") gt 0 `B. `DeltaS_("system") - DeltaS_("surrounding") gt0`C. `DeltaS_("system") gt 0`D. `DeltaS_("surrounding") gt 0`
Answer» Correct Answer - A For spontaneous process, `DeltaS` must be positive . In reversible process `DeltaS_("system") + DeltaS_("surrounding")=0` Hence, system is present in equilibrium . (i.e., it is not spontaeous process) While in irreversible process `DeltaS_("system") + DeltaS_("surrounding") gt 0` Hence, in the process `DeltaS` is positive.

Discussion

3445.	A reaction occurs spontanecously if `:-`A. `TDeltaS lt DeltaH` and both `DeltaH and DeltaS` are +veB. `TDeltaS gt DeltaH` and both `DeltaH and DeltaS` are +veC. `TDeltaS = DeltaH` and both `DeltaH and DeltaS` are +veD. `TDeltaS gt DeltaH and DeltaH` is +ve and `DeltaS` is -ve
Answer» Correct Answer - B The spontaneity of a reaction is based upon the negative value of `DeltaG and DeltaG` is based upon T, `DeltaS and DeltaH` according to following equation (Gibbs Helmholtz equation) `DeltaG=DeltaH-TDeltaS` If the magnitude of `DeltaH-TDeltaS` If the magnitude of `DeltaH-TDeltaS` is negative, then the reaction is spontaneous when `TDeltaS gt DeltaH` or we can say that `DeltaH` and `DeltaS` are positive , then `DeltaG` is negative.

Discussion

3446.	The work done during the expanision of a gas from a volume of `4 dm^(3)` to `6 dm^(3)` against a constant external pressure of 3 atm is (1 L atm = 101.32 J)A. `-6J`B. `-608J`C. `+304J`D. `-304J`
Answer» Correct Answer - B

Discussion

3447.	The work done during the expansion of a gas from a volume of 3 `dm^(3)` against a constant external pressure of 3 atm is (1 L atm = 101.3 J)A. `-608` JB. `+608` JC. `-304` JD. `+304` J
Answer» Correct Answer - A

Discussion

3448.	The work done during the expanision of a gas from a volume of `4 dm^(3)` to `6 dm^(3)` against a constant external pressure of 3 atm is (1 L atm = 101.32 J)A. `-6 J`B. `-608 J`C. `+304 J`D. `-304 J`
Answer» Correct Answer - B Work done (W) `=-P_("ext") (V_(2)-V_(1))` `=-3xx(6-4)=-6 L atm` `=-6xx101.32` J `(because` 1 L atm =101.32 J`)` `=-607.92 ~~-608 J`

Discussion

3449.	The work done during the expanision of a gas from a volume of `4 dm^(3)` to `6 dm^(3)` against a constant external pressure of 3 atm is (1 L atm = 101.32 J)A. `- 6 J`B. `- 608 J`C. `+ 304 J`D. `- 304 J`
Answer» Correct Answer - B Work done by the system during an irreversible expansion againt a constant atmospheric pressure is given by `W = - P_(ext) Deltav = - P_(ext) (V_(f) - V_(i))` `= - 3 atm (6 L - 4 L)` `= - 6 L atm` ` = - L atm ((101.32 J)/(1 L atm))` `= - 607.92 kJ`

Discussion

3450.	The ratio of`C_(p)` value of a triatomic gas to the `C_(v)` value of a monoatomic gas is
Answer» Correct Answer - 3

Discussion

Explore topic-wise InterviewSolutions in .

Which one of the following statement is false :A. work is a state functionB. temperetaue is a state functionC. change in the state is completely defined when the initial and final states are specified.D. work appears at the boundary of the system.

What will be the standard internal energy change for the reaction at 298 K ? `OF_(2(g))+H_(2)O_((g))+2HF_((g)),DeltaH^(@)=-310kJ`A. `-312.5Kj`B. `-125.03kJ`C. `-310 kJ`D. `-156kJ`

If `900J//g `of heat is exchanged at boiling point of water then water is the increase in entropy.A. 43.4 J/K-moleB. 87.2 J/K moleC. 900 JK/molD. Zero

a monoatomic gas `(Cv=(3)/(2)R)` is allowed to expand adiabaticaly and reversibly from initial volume of 8 L to 300 K to a volume `V_(2)` at 250 K. `V_(2)` is :A. 10.5 LB. 23 LC. 8.5 LD. 50.5 L

5 mol of an ideal gas expand reversibly from a volume of `8 dm^(3)` to `80 dm^(3)` at an temperature of `27^(@)C`. Calculate the change in entropy.A. `41.57 JK^(-1)`B. `-95.73 JK^(-1)`C. `95.73 JK^(-1)`D. `-41.57 JK^(-1)`

`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. `19.15`B. `-19.15`C. `4.7`D. zero

A gas `(C_(v.m) = (5)/(2)R)` behaving ideally is allowed to expand reversibly and adiabatically from `1` litre to `32` litre. Its initial temperature is `327^(@)C`. The molar enthalpy change (in `J//mol`) for the process is :A. `-1125R`B. `-675`C. `-1575R`D. None of these

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. `16.6J//K`B. `9J//K`C. `34.46J//K`D. `1.89J//K`

What is the change in entropy when `2.5` mole of water is heated from `27^(@)` to `87^(@) C` ? Assume that the heat capacity is constant . `(C_(p.m)(H_(2)O) = 4.2 J//g-k In (1.2)= 0.18)`A. `16.6` J/KB. `9` J/KC. `34.02` J/KD. `1.89` J/k

For the reaction, `2HgO_((s)) rarr 2Hg_((l))+O_(2(g))`A. `Delta H gt 0` & `Delta S lt 0`B. `Delta H gt 0` & `Delta S gt 0`C. `Delta H lt 0` & `Delta S lt 0`D. `Delta H lt 0` & `Delta S gt 0`

If a steam engine delivers `6.0xx10^(8)J` of work per minute and absorbs `5.4xx10^(9)J` of heat per minute from its boiler then the efficiency of the engine isA. 0.11B. 0.12C. 0.13D. 0.14

An engine has an efficiency of 0.25 when temperature of sink is reduced by `58^(@)C` , If its efficency is doubled, then the temperature of the source isA. `150 ^(@)C`B. `222 ^(@)C`C. `242 ^(@)C`D. `232 ^(@)C`

Prove that for an adiabatic process TVγ-1 constant.

What is a heat engine? Define efficiency of a heat engine.

One mole of carbon undergoes incomplete combustion to produce carbon monoxide, Calculate `(Delta H - Delta U)` for the formation of CO at 298 K. Given R = 8.314 `JK^(-1) mol^(-1)`.

Why `Delta H_(f)^(@)` of elements is zero?

At 298 K, the heats of formation fo `C Cl_(4) (g), H_(2)O(g), CO_(2)(g)` and `HCl(g)` are `-25, -57, -94` and `-22` K.cal `mol^(-1)` respectively. The `DeltaH` of the reaction: `C Cl_(4)(g) +2H_(2)O rarr CO_(2) (g) +4HCl(g)`A. `-43` K. calB. `+72` K.calC. `14` K.calD. `-125` K.cal

Which statement is correct?A. `"In a coffe-cup calorimeter", q=DeltaH`B. `"In a coffe-cup calorimeter", w=0`C. `"In bomb calorimeter", q=DeltaS`D. `"In bomb calorimeter", wgt0`

`DeltaH_(f)^(@)` of water is `-285.5 KJ mol^(-1)`. If enthalpy of neutraliztion of monoacidic strong base is `-57.3KJ mol^(-1)` then `DeltaH_(f)^(@)` of `OH^(-)` ion will be :A. `-285.5KJmol^(-1)`B. `285.5KJmol^(-1)`C. `114.5KJmol^(-1)`D. `-114.5KJmol^(-1)`

Which one of the following is a state property/function?A. HeatB. WorkC. Internal energyD. Potential enegry

The enthalpy of neutralisation of a strong acid by a string base is `-57.32 kJ mol^(-1)`. The enthalpy of formation of water is `-285.84 kJ mol^(-1)`. The enthalpy of formation of hydroxy`1` ion isA. `+228.52 kJ mol^(-1)`B. `-114.26 kJ mol^(-1)`C. `-228.52 kJ mol^(-1)`D. `+114.2 kJ mol^(-1)`

A process in which pressure remians constant is calledA. Isochroic processB. Isothermal processC. Adiabatic processD. Isobaric process

Which of the following acid will release maximum amount of heat when completely neutralised by strong base `NaOH`?A. `1 M HCI`B. `1M HNO_(3)`C. `1M HCIO_(4)`D. `1M H_(2)SO_(4)`

An ideal gas is expanded adiabatically at an initial temperature of 300 K so that its volume is doubled. The final temperature of the hydrogen gas is `lambda=1.40)`A. 227.36 kB. 500.30 kC. 454.76 kD. `-47^(@)`

One mole of an ideal gas `(C_(v,m)=(5)/(2)R)` at 300 K and 5 atm is expanded adiabatically to a final pressure of 2 atm against a constant pressure of 2 atm. Final temperature of the gas :A. 270 KB. 273 KC. 248.5 KD. 200 K

Which of the following pairs of a chemical reaction is certaion to result a spontaneous reaction ?A. endothermic and decreasing disorderB. exothermic and increasing disorderC. endothermic and increasing disorderD. exothermic and decreasing disorder

The change in enthalpy of a substqance when one mole of the substance is completely burnt in excess of air is calledA. Heat of sublimationB. Heat of combustionC. Heat of fusionD. Both (1) & (2)

A system absorbs 300 cal of heat. The work done by the system is 200 cal. `Delta U` for the above change isA. 100 calB. 500 calC. `-500` calD. `-100` cal

`1 gH_(2)` gas `STP` is expanded so that the volume is doubled. Hence, work done isA. `22.4 L-atm`B. `5.6 L-atm`C. `11.2 L-atm`D. `44.8 L-atm`

Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction ?A. Exothermic and decreasing disoderB. Endothermic and increasing disorderC. Exothermic and increasing disorderD. Endothermic and decreasing disorder

`DeltaU=q+w`, is mathematical expression forA. first law of thermodynamicsB. second law of thermodynamicsC. third law of thermodynamicsD. zeroth law of thermodynamics

In an isothermal processA. q = 0 and `Delta U` = 0B. q `ne` 0 and `Delta U` = 0C. q = 0 and `Delta U ne` 0D. `q ne 0 and Delta U ne 0`

16 g of `O_(2)` gas at STP is expanded so that volume is doubled. Hence, work done isA. `-22.4` L atmB. `-44.8` L atmC. `-11.2` L atmD. `-5.6` L atm

What will be the work done when one mole of a gas expands isothermally from 15 L to 50 L against a constant pressure of 1 atm at `25^(@)C`?A. `-3542 cal`B. `-843.3cal`C. `-718cal`D. `-60.23cal`

Considering entropy `(S)` as a thermodynamics parameter, the criterion for the spontaneity of any process isA. `DeltaS_("system") + DeltaS_("surrounding") gt 0 `B. `DeltaS_("system") - DeltaS_("surrounding") gt0`C. `DeltaS_("system") gt 0`D. `DeltaS_("surrounding") gt 0`

A reaction occurs spontanecously if `:-`A. `TDeltaS lt DeltaH` and both `DeltaH and DeltaS` are +veB. `TDeltaS gt DeltaH` and both `DeltaH and DeltaS` are +veC. `TDeltaS = DeltaH` and both `DeltaH and DeltaS` are +veD. `TDeltaS gt DeltaH and DeltaH` is +ve and `DeltaS` is -ve

The work done during the expanision of a gas from a volume of `4 dm^(3)` to `6 dm^(3)` against a constant external pressure of 3 atm is (1 L atm = 101.32 J)A. `-6J`B. `-608J`C. `+304J`D. `-304J`

The work done during the expansion of a gas from a volume of 3 `dm^(3)` against a constant external pressure of 3 atm is (1 L atm = 101.3 J)A. `-608` JB. `+608` JC. `-304` JD. `+304` J

The work done during the expanision of a gas from a volume of `4 dm^(3)` to `6 dm^(3)` against a constant external pressure of 3 atm is (1 L atm = 101.32 J)A. `-6 J`B. `-608 J`C. `+304 J`D. `-304 J`

The work done during the expanision of a gas from a volume of `4 dm^(3)` to `6 dm^(3)` against a constant external pressure of 3 atm is (1 L atm = 101.32 J)A. `- 6 J`B. `- 608 J`C. `+ 304 J`D. `- 304 J`

The ratio of`C_(p)` value of a triatomic gas to the `C_(v)` value of a monoatomic gas is