540 + Interview Questions in CHEMICAL-THERMODYNAMICS IN GENERAL AWARENESS Page 3 InterviewSolution

101.	If 100 calorie of heat are added to the same system as in example 1 and a work of 50 calorie is done on the system, calculate the energy change of the system.
Answer» Heat absorbed q=100cal Work done on the system, `w=+50cal` ltBrgt Applying the first law of thermodynamics, `DeltaD=q+w=(100+50)=150`calorie In the above examples, the final state is same but the paths adopted are different. Thus, the change in energy of the system depends onthe intial and finial states but does not depend on the path by which the final state has reached, q and w are, therefore, not state functions but `DeltaU` is a state functions.

Discussion

102.	If 500 calorie of heat energy are added to a system and the system does 350 calorie of work on the surroundings, what is the energy change of the system?
Answer» Heat absorbed, q=500cal Work done on the system `w=+50cal` Applying the first law of htermodynamics `DeltaU=q+w=(100+50)=150` calorie In the above two examples, the final state is same but the paths adopted are different. Thus, the change in energy of the system depends on the initial and final states but does not depend on the path by which the final state has reached, q and w are , therefore, not state functions but `DeltaU` is a state functins.

Discussion

103.	For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant Kin terms of change in entropy is described byA. With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive.B. With increase in temperature, the value of K for endothemic reaction increases because unfavourable change in entropy of the surrounding decreasesC. With increases in temperature, the value of K for exothermic reaction decreases because because favourable change in entropy of the surroundings decreasesD. With increases in temperature, the value of K for endothemic reaction increases because the entropy change of the system negative.
Answer» Correct Answer - B::C For endothermic reactions: `DeltaH=+ve` therefore, equilibrium constant increases with temperature. `log((K_(1))/(K_(2)))=(DeltaH)/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]` `DeltaS_(surr)=-(q_("system"))/(T)` `DeltaS_(surr) lt `0 for endothermic process thus decrease in entropy of surroudings is unfavourable. For exothermic reaction, equilibrium constant will decreases with rise in temperature `DeltaS_(surr) lt0` (for exothermic process which is favourable for surroundings).

Discussion

104.	Entropy of vaporisation of water at `100^(@)C`, if molar heat of vaporisation is `9710 cal mol^(-1)` will beA. `20 cal mol^(-1)K^(-1)`B. `26 cal mol^(-1)K^(-1)`C. `24 cal mol^(-1)K^(-1)`D. `28 cal mol^(-1)K^(-1)`
Answer» Correct Answer - D `DeltaS_(vap)=DeltaH_(vap)//T_(b)=9710//373` `=26.032 cal K^(-1) mol^(-1)`

Discussion

105.	A gas expands from `3 dm^(3)` to `5 dm^(3)` against a constant pressure of 3 atm. The work done during expansion is used to heat 10 mol of water at a temperature of 290 K. Calculate final temperature of water. Specific heat of water `=4.184 J g^(-1)K^(-1)`
Answer» Work done `=PxxdV=3.0xx(5.0-3.0)` `=6.0litre-atm=6.0xx101.3J` `=607.8J` ltBrgt let `DeltaT` be the change in temperature Heat absorbed `=mxxsxxDeltaT` `=10.0xx18xx4.184xxDeltaT` Given, `PxxdV=mxxsxxDeltaT` or `DeltaT=(PxxdV)/(mxxs)=(607.8)/(10.0xx18.0xx4.184)=0.807` Final temperature `=290+0.807=290.807K`

Discussion

106.	A sample of a gas contracts `200cm^(3)` by an average of .5 atmosphere while 8.5 J heat flows out into the surroundings what is the change in energy of the system? (1 litre-atm=101.3J)
Answer» Energy of the system increases by 1.63J

Discussion

107.	A chemical process is carried out in a thermostat maintained at `25^(@)C`.The process may be termed asA. isobaric processB. isoentropic processC. adiabatic processD. isothermal process.
Answer» Correct Answer - D Because the temperature is constant.

Discussion

108.	An isolated system comprises the liquid in equilibrium with vapours. At this stage, the molar entropy of the vapour isA. less than that of liquidB. more than that of liquidC. equal to zeroD. equal to that of liquid
Answer» Correct Answer - D In isolated systems at equilibrium `DeltaS=0`. i.e., `(S_(P)-S_(R ))=0` , `S_(P)=S_(R )`

Discussion

109.	A sample of a gas expands from `200cm^(3)` to `500cm^(3)` against an average pressure of 750 torr while 1.5 J heat flows into the system. What is the change in energy of the system? (1 litre-atm=101.3J)
Answer» Energy of the system is decreased by 28.5 J, `DeltaE=-28.5J`

Discussion

110.	A cup of tea placed in the room eventually acquires a room temperature by losing heat. The process may be considered close toA. Cyclic processB. Reversible processC. Isothermal processD. None of these
Answer» Correct Answer - B Question based on facts/definitions

Discussion

111.	When a solid changes into liquid, the entropyA. becomes zeroB. becomes minimumC. increasesD. remains constant
Answer» Correct Answer - C Because in liquids, molecules have greater randomness.

Discussion

112.	Calculate the pressure-volume work by the system when the gas expands from 1.0 litre to 2.0 litre against a constant external pressure of 10 atmospheres. Express the answer in calorie and joule.
Answer» `-10` litre-atm, -242.2 cal, -1013.28J

Discussion

113.	Energy changes accompanying the chemical reactions can take placeA. in the form of heat onlyB. in the form of heat as well as light onlyC. in the form of light onlyD. in any form depending upon the nature of the system.
Answer» Correct Answer - D Question based on facts/definitions

Discussion

114.	A gas expands from a volume of `1m^(3)` to a volume of `2m^(3)` against an external pressure of `10^(5)Nm^(-2)`. The work done of the gas will beA. `10^(5)kJ`B. `10^(2)kJ`C. `10^(2)J`D. `10^(3)J`
Answer» Correct Answer - B `W=PDeltaV=-10^(5)Nm^(-2)xx(2-1)m^(-2)` `=10^(5)Nm=10^(5)J=10^(2)kJ`.

Discussion

115.	In endothermic reactionsA. the heat is given to surroundingsB. `DeltaH=-ve`C. `DeltaH=+ve`D. `H_(R) gt H_(P)`
Answer» Correct Answer - C For endothermic reactions, heat is absorbed. `DeltaH=+ve` [`:. H_(P) gt H_(R )`]

Discussion

116.	An ideal gas expands from `10^(-3) m^(3)` to `10^(-2) m^(3)` at 300 K against a constant pressure of `10^(5) Nm^(-2)`. The workdone isA. `900KJ`B. `-900KJ`C. `270KJ`D. `-900J`
Answer» Correct Answer - A Work done during expansion (or by the system) `=-PxxDeltaV=-1xx10^(5)[10^(-2)-10^(-3)]` `=-1xx10^(5)xx9xx10^(-3)=-900J`

Discussion

117.	An ideal gas expands from `10^(-3) m^(3)` to `10^(-2) m^(3)` at 300 K against a constant pressure of `10^(5) Nm^(-2)`. The workdone isA. `-0.9 kJ`B. `-900 kJ`C. `270 kJ`D. `-270 kJ`
Answer» Correct Answer - A `W=-PDeltaV` `P=10^(5)Nm^(-2)=1"atm"`, `DeltaV=10-1=9 "Litres"[1m^(3)=10^(3)L]` `W=-(1xx9xx101)/(1000)[1 L-"atom"=101 J]` `=0.9 kJ`

Discussion

118.	The heat absorbed in a reaction at constant temperature and constant volume isA. `DeltaE`B. `DeltaH`C. `-DeltaA`D. `-DeltaG`
Answer» Correct Answer - A `q_(v)=DeltaE`, at constant temperature.

Discussion

119.	What per cent `T_(1)` is of `T_(2)` for a heat engine whose efficiency is `10%`A. `92%`B. `10%`C. `90%`D. `100%`
Answer» Correct Answer - C `eta=1-(T_(1))/(T_(2))` `(T_(1))/(T_(2))=1-eta=1-0.10=0.90=90%` `T_(1)=90% T_(2)`

Discussion

120.	One mole of an ideal gas at `300K` is expanded isothermally from an inital volume of 1 litre to 10 litres. The `DeltaE` for this process is `(R=2cal mol^(-1)K^(-1))`A. `163.7 cal`B. `1381.1 cal`C. `9` lt atmD. zero
Answer» Correct Answer - D For isothermal process `dT=0`, thus `DeltaE=0`.

Discussion

121.	A gas expands isothermally and reversibly. The work done by the gas isA. ZeroB. MaximumC. MinimumD. Cannot be determined
Answer» Correct Answer - B Work done during isothermal reversible expansion of a gas is maximum.

Discussion

122.	One mole of an ideal gas at `300K` is expanded isothermally from an inital volume of 1 letre to 10 litres. The `DeltaE` for this process is `(R=2cal mol^(-1)K^(-1))`A. `163.7cal`B. zeroC. `138.1cal`D. `9L` atm
Answer» Correct Answer - D For isothermal process `DeltaE=0` because `DeltaT=0`

Discussion

123.	The work done by the system in a cyclic process involving one mole of an ideal monoatomic gas is `-50 kJ//cycle`. The heat absorbed by the system per cycle isA. ZeroB. `50 kJ`C. `-50 kJ`D. `250 kJ`
Answer» Correct Answer - B For a cyclic process, the work done is equal to heat absorbed `q=DeltaE-W` `:. DeltaE=0` Thus, `q=-W=50 kJ`

Discussion

124.	The maximum efficiency of a steam engine operating between `110^(@)C` and `25^(@)C`A. `20%`B. `22.2%`C. `25%`D. `30%`
Answer» Correct Answer - B `eta=(T_(2)-T_(1))/(T_(2))=(383-298)/(383)=0.222` or `eta=22.2%`

Discussion

125.	Work done on a system when one mole of an an ideal gas at 500 K is compressed isothermally and reversibly to 1/10th of its original volume. (R = 2cal)A. `500 kcal`B. `1.51 kcal`C. `-23.03 kcal`D. `2.303 kcal`
Answer» Correct Answer - D `W=-2.303 nRT log (V_(2))/(V_(1))` `=-2.303xx1xx2xx500 log (1)/(10)` `=(-2.303xx1000xlog10^(-1))/(1000)kJ` `=+2.303 kJ`

Discussion

126.	A gas is allowed to expand at constant temperature from a volume of `1.0 L` to `10.0 L` against an external pressure of `0.50` atm. If the gas absorbs `250 J` of heat from the surroundings, what are the values of `q` and `DeltaE` ? (Given `1 L "atm" =101 J`)A. `{:(q,w,DeltaE),(250J,-455J,-205J):}`B. `{:(q,w,DeltaE),(-250J,-455J,-710J):}`C. `{:(q,w,DeltaE),(250J,455J,710J):}`D. `{:(q,w,DeltaE),(250J,455J,205J):}`
Answer» Correct Answer - A `q="heat absorbed"=+250 J` `W=-PDeltaV=-0.5"atm"xx(10-1)` `=-4.5 L-"atm"` `=-4.5xx101J=-454.5 J` Now `q=DeltaE-W` `DeltaE=q+W=250-454.5` `=-204.5 J`

Discussion

127.	The heat of combusion of benzne determined in a bomb calorimeter is `-870Kcal mol^(-1)` at `298 K`. The value of `DeltaE` for the reaction isA. `-1740 kcal mol^(-1)`B. `+870 kcal mol^(-1)`C. `-870 kcal mol^(-1)`D. `+1740 kcal mol^(-1)`
Answer» Correct Answer - C The heat change determined in a bomb valorimeter is `DeltaE` value. Thus `DeltaE=-870kcal mol^(-1)`

Discussion

128.	The work done by the system in the conversion of `1 mol` of water at `100 ^(@)C` and `760` torr to steam is `-3.1 kJ`. Calculate the `DeltaE` for the conversion (Latent heat of vaporisation of water is `40.65 kJ mol^(-1)`)A. `43.75 kJ`B. `101.35 kJ`C. `37.55 kJ`D. `-40.65 kJ`
Answer» Correct Answer - C According to first law, `DeltaE=q+W` `:. DeltaE=40.65-3.1=37.55 kJ`

Discussion

129.	`H_(2)O(l)toH_(2)O(g)` , `DeltaH=+43.7 kJ` `H_(2)O(s)toH_(2)O(l)` , `DeltaH=+6.05kJ` The value of `DeltaH_((sublimation))` of ice isA. `49.75 kJ mol^(-1)`B. `37.65 kJ mol^(-1)`C. `43.7 kJ mol^(-1)`D. none of these
Answer» Correct Answer - A Add the two equations to get `H_(2)O(s) to H_(2)O(g)`

Discussion

130.	One mole of a gas absorbs 200 J of heat at constant volume. Its temperature rises from 298 K to 308 K. The change in internal energy is :-A. `200 J`B. `-200 J`C. `200xx(308)/(298)J`D. `200xx(298)/(308)J`
Answer» Correct Answer - A `DeltaE=q` at constant volume and temperature.

Discussion

131.	The heat of reaction at constant volume `(DeltaE)` and that at constant pressure `(DeltaH)` are related asA. `DeltaE=DeltaH+DeltanRT`B. `DeltaH=DeltaE-DeltanRT`C. `DeltaH=DeltaE+DeltanRT`D. `DeltaH=DeltaE+nRT`
Answer» Correct Answer - C `DeltaH=DeltaE+Deltan_(g)RT`.

Discussion

132.	The enthalpy of the reaction : `H_(2)O_(2)(l)toH_(2)O(l)+(1)/(2)O_(2)(g)` is `-23.5 kcal mol^(-1)` and the enthalpy of formation of `H_(2)O(l)` is `-68.3 kcal mol^(-1)`. The enthalpy of formation of `H_(2)O_(2)(l)` isA. `-44.8 kcal mol^(-1)`B. `44.8 kcal mol^(-1)`C. `-91.8 kcal mol^(-1)`D. `91.8 kcal mol^(-1)`
Answer» Correct Answer - A `H_(2)+O_(2)toH_(2)O_(2)` is `DeltaH_(f)` of `H_(2)O_(2)`. To get `H_(f)` for this reaction subtract equation `(i)` from `DeltaH_(f)` of `H_(2)O`.

Discussion

133.	Under which of the following condition is the relation `DeltaH = DeltaU +P DeltaV` valid for a closed system atA. Constant pressureB. Constant temperatureC. Constant temperature and pressureD. Constant temperature, pressure and composition.
Answer» Correct Answer - C This equation is valid for a closed system at constant temperature and pressure.

Discussion

134.	At `27^(@)C` for reaction, `C_(6)H_(6)(l)+(15)/(2)O_(2)(g)to6CO_(2)(g)+3H_(2)O(l)` proceeds spontaneously because the magnitude ofA. `DeltaH = T Delta S`B. `DeltaH gt T Delta S`C. `DeltaH lt T Delta S`D. `DeltaH gt 0` and `T Delta S lt 0`
Answer» Correct Answer - B In this reaction, `DeltaS=-ve` (because products have lesser gaseous molecules) and `DeltaH=-ve`. Thus, the reaction is spontaneous only if `DeltaH gt T DeltaS`

Discussion

135.	The heat change for the following reaction at `298 K` and constant pressure is `+7.3 kcal` `A_(@)B(s)to2A(s)+(1)/(2)B_(2)(g)` ,` DeltaH=+7.3 kcal` The heat change at constant volume would beA. `+7.3kcal`B. More than `7.3kcal`C. Less than `7.3 kcal`D. Zero
Answer» Correct Answer - C `DeltaH=DeltaE+Deltan_(g)RT` or `DeltaE=DeltaH-Deltan_(g)RT` `:. DeltaE=+7.3-(1)/(2)xx2xx298` `=7300 cal-298` `=7002 cal=7.00 kcal`

Discussion

136.	Enthalpy of formation of compound isA. always positiveB. always negativeC. can be either negative or zeroD. can be positive or negative
Answer» Correct Answer - D Enthalpy of formation of a compound may be `+ve` or `-ve`

Discussion

137.	For the gaseous reaction: `N_(2)O_(4) rarr 2NO_(2)`A. `DeltaHgtDeltaE`B. `DeltaHltDeltaE`C. `DeltaH=DeltaE`D. `DeltaH=0`, but `DeltaE` may have any value.
Answer» Correct Answer - A `Deltan_(g)=2-1=1` Thus `DeltaH gt Delta E`.

Discussion

138.	For a reaction `A(g)+3B(g)to2C(g)` , `DeltaH^(@)=-24kJ` The value of `DeltaG^(@)` is `-9 kJ`. The standard entropy change of reaction isA. `5 JK^(-1)`B. `50 JK^(-1)`C. `500 JK^(-1)`D. `0.5 JK^(-1)`
Answer» Correct Answer - B Apply `DeltaS^(@)=(DeltaH^(@)-DeltaG^(@))/(T)` Here, `T=298 K` (Standard temperature)

Discussion

139.	A hypothetical reaction, `A rarr 2B`, proceeds via following sequence of steps `{:(ArarrC,,DeltaH = q_(1),,,),(CrarrD,,DeltaH = q_(2),,,),((1)/(2)D rarrB,,DeltaH = q_(3),,,):}` The heat of reaction isA. `q_(1)-q_(2)+2q_(3)`B. `q_(1)+q_(2)-2q_(3)`C. `q_(1)+q_(2)+2q_(3)`D. `q_(1)+2q_(2)-2q_(3)`
Answer» Correct Answer - C Multiply eqn. `(iii)` by `2` and add all the three.

Discussion

140.	Which of the following equations express `DeltaH` as heat of combustion ?A. `H_(2)(g)+I_(2)to2HI(g)`B. `CaCO_(3)(s)toCaO(s)+CO_(2)(g)`C. `CH_(4)(g)+2O_(2)(g)toCO_(2)(g)+2H_(2)O(l)`D. `2KCIO_(3)(s)to2KCI(s)+3O_(2)(g)`
Answer» Correct Answer - C According to definition of heat of combustion.

Discussion

141.	`N_(2)(g) +2O_(2)(g) rarr 2NO_(2)(g) +X kJ` `2NO(g)+O_(2)(g) rarr 2NO_(2) +Y kJ` The ethalpy of formation of `NO` isA. `(2X-2Y)`B. `X-Y`C. `(1)/(2)(Y-X)`D. `(1)/(2)(x-y)`
Answer» Correct Answer - D Multiply both the reactions by `(1)/(2)` and subtract first equation from `2nd`.

Discussion

142.	The enthalpy of formation of ammonia is `- 46.2 mol^(-1)`. The enthalpy change for the reaction `2NH_(3) rarr N_(2) + 3 H_(2)` isA. `42.0 kJ`B. `64.0 kJ`C. `80.0 kJ`D. `92.0 kJ`
Answer» Correct Answer - D `underset(2"moles")(2NH_(3))toN_(2)+3H_(2)` The enthalpy of formation of ammonia `=-46kJ mol^(-1)` `DeltaH"of given reaction " =-2xx(-46)=92kJ`

Discussion

143.	Given that `(1)/(2)S_(8)(s)+6O_(2)(g)to4SO_(3)(g)`, `DeltaH^(@)=-1590 kJ` The standard enthalpy of formation of `SO_(3)` isA. `-1590 kJ mol^(-1)`B. `-397.5 kJ mol^(-1)`C. `-3.975 kJ mol^(-1)`D. `+397.5 kJ mol^(-1)`
Answer» Correct Answer - B `DeltaH_(f)(SO_(3))=` heat of formation of `1` mole `=(-1590)/(4)-397.5 kJ mol^(-1)`

Discussion

144.	`HA+OH^(-) to H_(2)O+A^(-)+q_(1) kJ` `H^(+)+OH^(-) to H_(2)O+ q_(2) kJ` The enthalpy of dissociation of `HA` isA. `(q_(1)+q_(2))`B. `(q_(1)-q_(2))`C. `(q_(2)-q_(1))`D. `-(q_(1)+q_(2))`
Answer» Correct Answer - B Subtract `2nd` equation from first.

Discussion

145.	The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :A. `+46.0 kJ`B. `+92.0 kJ`C. `-23 kJ`D. `-92 kJ`
Answer» Correct Answer - B The given reaction involve decomposition of `2` mol of ammonia `DeltaH_(f)(NH_(3))=-46.0`, `DeltaH_("decomposition")"of" NH_(3)=+46` `:. DeltaH "of given reaction"=2xx46=92 kJ`

Discussion

146.	The enthalpy of formation of ammonia is `-46.0 kJ mol^(-1)`. The enthalpy for the reaction `2N_(2)(g)+6H_(2)(g)to4NH_(3)(g)`A. `-46.0 kJ`B. `46.0 kJ`C. `184.0 kJ`D. `-184.0kJ`
Answer» Correct Answer - D `DeltaH_(f)` refers to the formation of one mole of `NH_(3)`. `:. DeltaH_(f)` for `4 "moles"=4xx-46=-184 kJ`

Discussion

147.	Heat of neutralisation of `NaOH` and `HCl` is `-57.46 kJ //` equivalent. The heat of ionisation of water in `kJ //mol` is `:`A. `-57.3 kJ mol^(-1)`B. `-114.6 kJ mol^(-1)`C. `+57.3 kJ mol^(-1)`D. `+114.6 kJ mol^(-1)`
Answer» Correct Answer - C `NaOH+HCltoNaCl+H_(2)O` or `H^(+)+OH^(-)toH_(2)O` , `DeltaH=-57.3 kJ` ` :. H_(2)O to H^(+)+OH^(-)` , `DeltaH=+57.3 kJ` This is according to Laplace law.

Discussion

148.	For a given reaction`DeltaH=35.5kJmol^(-1)`and`DeltaS=83.6JK^(-1)mol^(-1)`.The reaction is spontaneous at :(Assume that `DeltaH`and`DeltaS` do no vary with temperature)A. `Tgt425K`B. All temperaturesC. `Tgt298K`D. `T lt 425K`
Answer» Correct Answer - D `DeltaG=DeltaH-TDeltaS` For equilibrium `DeltaG=0` `DeltaH-TDeltaS` `T_(eq).=(DeltaH)/(DeltaS)=(35.5xx1000)/(83.6)=425K` Since the reaction is endothemic it will be spontaneous at `Tgt 425K` . Option (1)

Discussion

149.	When 100 mL each of HCl and NaOH solutions are mixed, 5.71 kJ of heat was evolved. What is the molarity of two solution? The heat of neutralisation of HCl is 57.1 kJ
Answer» Correct Answer - 1M

Discussion

150.	Enthalpy of neutralisation of `NaOH` with `H_(2)SO_(4)` is `-57.3kJ mol^(-1)` and with ethanoic acid `-55.2 kJ mol^(-1)`. Which of the following is the best explanation of this difference?A. Ethanoic acid is weak acid and thus requires less `NaOH` for neutralisationB. Ethanoic acid is only partly ionized, neutralisation is therefore incompleteC. Ethanoic acid is monobasic while `H_(2)SO_(4)` is dibasicD. Some heat is used to ionize the ethanoic acid completely.
Answer» Correct Answer - D Enthanoic acid is weak acid and it requires so heat for its ionization.

Discussion

Explore topic-wise InterviewSolutions in .

If 100 calorie of heat are added to the same system as in example 1 and a work of 50 calorie is done on the system, calculate the energy change of the system.

If 500 calorie of heat energy are added to a system and the system does 350 calorie of work on the surroundings, what is the energy change of the system?

Entropy of vaporisation of water at `100^(@)C`, if molar heat of vaporisation is `9710 cal mol^(-1)` will beA. `20 cal mol^(-1)K^(-1)`B. `26 cal mol^(-1)K^(-1)`C. `24 cal mol^(-1)K^(-1)`D. `28 cal mol^(-1)K^(-1)`

A gas expands from `3 dm^(3)` to `5 dm^(3)` against a constant pressure of 3 atm. The work done during expansion is used to heat 10 mol of water at a temperature of 290 K. Calculate final temperature of water. Specific heat of water `=4.184 J g^(-1)K^(-1)`

A sample of a gas contracts `200cm^(3)` by an average of .5 atmosphere while 8.5 J heat flows out into the surroundings what is the change in energy of the system? (1 litre-atm=101.3J)

A chemical process is carried out in a thermostat maintained at `25^(@)C`.The process may be termed asA. isobaric processB. isoentropic processC. adiabatic processD. isothermal process.

An isolated system comprises the liquid in equilibrium with vapours. At this stage, the molar entropy of the vapour isA. less than that of liquidB. more than that of liquidC. equal to zeroD. equal to that of liquid

A sample of a gas expands from `200cm^(3)` to `500cm^(3)` against an average pressure of 750 torr while 1.5 J heat flows into the system. What is the change in energy of the system? (1 litre-atm=101.3J)

A cup of tea placed in the room eventually acquires a room temperature by losing heat. The process may be considered close toA. Cyclic processB. Reversible processC. Isothermal processD. None of these

When a solid changes into liquid, the entropyA. becomes zeroB. becomes minimumC. increasesD. remains constant

Calculate the pressure-volume work by the system when the gas expands from 1.0 litre to 2.0 litre against a constant external pressure of 10 atmospheres. Express the answer in calorie and joule.

Energy changes accompanying the chemical reactions can take placeA. in the form of heat onlyB. in the form of heat as well as light onlyC. in the form of light onlyD. in any form depending upon the nature of the system.

A gas expands from a volume of `1m^(3)` to a volume of `2m^(3)` against an external pressure of `10^(5)Nm^(-2)`. The work done of the gas will beA. `10^(5)kJ`B. `10^(2)kJ`C. `10^(2)J`D. `10^(3)J`

In endothermic reactionsA. the heat is given to surroundingsB. `DeltaH=-ve`C. `DeltaH=+ve`D. `H_(R) gt H_(P)`

An ideal gas expands from `10^(-3) m^(3)` to `10^(-2) m^(3)` at 300 K against a constant pressure of `10^(5) Nm^(-2)`. The workdone isA. `900KJ`B. `-900KJ`C. `270KJ`D. `-900J`

An ideal gas expands from `10^(-3) m^(3)` to `10^(-2) m^(3)` at 300 K against a constant pressure of `10^(5) Nm^(-2)`. The workdone isA. `-0.9 kJ`B. `-900 kJ`C. `270 kJ`D. `-270 kJ`

The heat absorbed in a reaction at constant temperature and constant volume isA. `DeltaE`B. `DeltaH`C. `-DeltaA`D. `-DeltaG`

What per cent `T_(1)` is of `T_(2)` for a heat engine whose efficiency is `10%`A. `92%`B. `10%`C. `90%`D. `100%`

One mole of an ideal gas at `300K` is expanded isothermally from an inital volume of 1 litre to 10 litres. The `DeltaE` for this process is `(R=2cal mol^(-1)K^(-1))`A. `163.7 cal`B. `1381.1 cal`C. `9` lt atmD. zero

A gas expands isothermally and reversibly. The work done by the gas isA. ZeroB. MaximumC. MinimumD. Cannot be determined

One mole of an ideal gas at `300K` is expanded isothermally from an inital volume of 1 letre to 10 litres. The `DeltaE` for this process is `(R=2cal mol^(-1)K^(-1))`A. `163.7cal`B. zeroC. `138.1cal`D. `9L` atm

The work done by the system in a cyclic process involving one mole of an ideal monoatomic gas is `-50 kJ//cycle`. The heat absorbed by the system per cycle isA. ZeroB. `50 kJ`C. `-50 kJ`D. `250 kJ`

The maximum efficiency of a steam engine operating between `110^(@)C` and `25^(@)C`A. `20%`B. `22.2%`C. `25%`D. `30%`

Work done on a system when one mole of an an ideal gas at 500 K is compressed isothermally and reversibly to 1/10th of its original volume. (R = 2cal)A. `500 kcal`B. `1.51 kcal`C. `-23.03 kcal`D. `2.303 kcal`

The heat of combusion of benzne determined in a bomb calorimeter is `-870Kcal mol^(-1)` at `298 K`. The value of `DeltaE` for the reaction isA. `-1740 kcal mol^(-1)`B. `+870 kcal mol^(-1)`C. `-870 kcal mol^(-1)`D. `+1740 kcal mol^(-1)`

The work done by the system in the conversion of `1 mol` of water at `100 ^(@)C` and `760` torr to steam is `-3.1 kJ`. Calculate the `DeltaE` for the conversion (Latent heat of vaporisation of water is `40.65 kJ mol^(-1)`)A. `43.75 kJ`B. `101.35 kJ`C. `37.55 kJ`D. `-40.65 kJ`

`H_(2)O(l)toH_(2)O(g)` , `DeltaH=+43.7 kJ` `H_(2)O(s)toH_(2)O(l)` , `DeltaH=+6.05kJ` The value of `DeltaH_((sublimation))` of ice isA. `49.75 kJ mol^(-1)`B. `37.65 kJ mol^(-1)`C. `43.7 kJ mol^(-1)`D. none of these

One mole of a gas absorbs 200 J of heat at constant volume. Its temperature rises from 298 K to 308 K. The change in internal energy is :-A. `200 J`B. `-200 J`C. `200xx(308)/(298)J`D. `200xx(298)/(308)J`

The heat of reaction at constant volume `(DeltaE)` and that at constant pressure `(DeltaH)` are related asA. `DeltaE=DeltaH+DeltanRT`B. `DeltaH=DeltaE-DeltanRT`C. `DeltaH=DeltaE+DeltanRT`D. `DeltaH=DeltaE+nRT`

Under which of the following condition is the relation `DeltaH = DeltaU +P DeltaV` valid for a closed system atA. Constant pressureB. Constant temperatureC. Constant temperature and pressureD. Constant temperature, pressure and composition.

At `27^(@)C` for reaction, `C_(6)H_(6)(l)+(15)/(2)O_(2)(g)to6CO_(2)(g)+3H_(2)O(l)` proceeds spontaneously because the magnitude ofA. `DeltaH = T Delta S`B. `DeltaH gt T Delta S`C. `DeltaH lt T Delta S`D. `DeltaH gt 0` and `T Delta S lt 0`

The heat change for the following reaction at `298 K` and constant pressure is `+7.3 kcal` `A_(@)B(s)to2A(s)+(1)/(2)B_(2)(g)` ,` DeltaH=+7.3 kcal` The heat change at constant volume would beA. `+7.3kcal`B. More than `7.3kcal`C. Less than `7.3 kcal`D. Zero

Enthalpy of formation of compound isA. always positiveB. always negativeC. can be either negative or zeroD. can be positive or negative

For the gaseous reaction: `N_(2)O_(4) rarr 2NO_(2)`A. `DeltaHgtDeltaE`B. `DeltaHltDeltaE`C. `DeltaH=DeltaE`D. `DeltaH=0`, but `DeltaE` may have any value.

For a reaction `A(g)+3B(g)to2C(g)` , `DeltaH^(@)=-24kJ` The value of `DeltaG^(@)` is `-9 kJ`. The standard entropy change of reaction isA. `5 JK^(-1)`B. `50 JK^(-1)`C. `500 JK^(-1)`D. `0.5 JK^(-1)`

A hypothetical reaction, `A rarr 2B`, proceeds via following sequence of steps `{:(ArarrC,,DeltaH = q_(1),,,),(CrarrD,,DeltaH = q_(2),,,),((1)/(2)D rarrB,,DeltaH = q_(3),,,):}` The heat of reaction isA. `q_(1)-q_(2)+2q_(3)`B. `q_(1)+q_(2)-2q_(3)`C. `q_(1)+q_(2)+2q_(3)`D. `q_(1)+2q_(2)-2q_(3)`

Which of the following equations express `DeltaH` as heat of combustion ?A. `H_(2)(g)+I_(2)to2HI(g)`B. `CaCO_(3)(s)toCaO(s)+CO_(2)(g)`C. `CH_(4)(g)+2O_(2)(g)toCO_(2)(g)+2H_(2)O(l)`D. `2KCIO_(3)(s)to2KCI(s)+3O_(2)(g)`

`N_(2)(g) +2O_(2)(g) rarr 2NO_(2)(g) +X kJ` `2NO(g)+O_(2)(g) rarr 2NO_(2) +Y kJ` The ethalpy of formation of `NO` isA. `(2X-2Y)`B. `X-Y`C. `(1)/(2)(Y-X)`D. `(1)/(2)(x-y)`

The enthalpy of formation of ammonia is `- 46.2 mol^(-1)`. The enthalpy change for the reaction `2NH_(3) rarr N_(2) + 3 H_(2)` isA. `42.0 kJ`B. `64.0 kJ`C. `80.0 kJ`D. `92.0 kJ`

Given that `(1)/(2)S_(8)(s)+6O_(2)(g)to4SO_(3)(g)`, `DeltaH^(@)=-1590 kJ` The standard enthalpy of formation of `SO_(3)` isA. `-1590 kJ mol^(-1)`B. `-397.5 kJ mol^(-1)`C. `-3.975 kJ mol^(-1)`D. `+397.5 kJ mol^(-1)`

`HA+OH^(-) to H_(2)O+A^(-)+q_(1) kJ` `H^(+)+OH^(-) to H_(2)O+ q_(2) kJ` The enthalpy of dissociation of `HA` isA. `(q_(1)+q_(2))`B. `(q_(1)-q_(2))`C. `(q_(2)-q_(1))`D. `-(q_(1)+q_(2))`

The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :A. `+46.0 kJ`B. `+92.0 kJ`C. `-23 kJ`D. `-92 kJ`

The enthalpy of formation of ammonia is `-46.0 kJ mol^(-1)`. The enthalpy for the reaction `2N_(2)(g)+6H_(2)(g)to4NH_(3)(g)`A. `-46.0 kJ`B. `46.0 kJ`C. `184.0 kJ`D. `-184.0kJ`

Heat of neutralisation of `NaOH` and `HCl` is `-57.46 kJ //` equivalent. The heat of ionisation of water in `kJ //mol` is `:`A. `-57.3 kJ mol^(-1)`B. `-114.6 kJ mol^(-1)`C. `+57.3 kJ mol^(-1)`D. `+114.6 kJ mol^(-1)`

For a given reaction`DeltaH=35.5kJmol^(-1)`and`DeltaS=83.6JK^(-1)mol^(-1)`.The reaction is spontaneous at :(Assume that `DeltaH`and`DeltaS` do no vary with temperature)A. `Tgt425K`B. All temperaturesC. `Tgt298K`D. `T lt 425K`

When 100 mL each of HCl and NaOH solutions are mixed, 5.71 kJ of heat was evolved. What is the molarity of two solution? The heat of neutralisation of HCl is 57.1 kJ