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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 3201. |
`Delta H_(1)^(@)` for `CO_(2)(g)`, `CO(g)` and `H_(2)O(g)` are `-393.5,-110.5` and `-241.8 kJ mol^(-1)` respectively. Standard enthalpy change for the reaction `CO_(2)(g)+H_(2)(g) rarr CO(g) + H_(2)O(g)` isA. `524.1`B. `+41.2`C. `-262.5`D. `-41.2` |
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Answer» `CO_(2)(g)+H_(2)(g) rarr CO(g) +H_(2)O(g)` `DeltaH^(Theta) = {[Delta_(f)H^(Theta) CO, (g)+Delta_(f)H^(Theta)H_(2)O,(g)] -Delta_(f)H^(Theta)CO_(2),(g)}` `= [(-110.5)+(-241.8(-(-393.5+0))]` `= +41.2 kJ` `Delta_(f)H^(Theta) H_(2),(g) = 0` `Delta_(f)H^(Theta) = 0` in elementary state |
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| 3202. |
What are state variables ? |
| Answer» Property of a system which is dependent only on the state of system. It is independent of the path adopted to attain a particular state. e.g. pressure, temperature, volume. | |
| 3203. |
For the process to occur under adiabatic condition write the correct condition. |
| Answer» `q=0` , No heat is allowed to enter or leave the system under adiabatic conditions. | |
| 3204. |
Which of the following statements is/are correct?A. Water in a beaker be made to boil by placing it in a bath of boiling waterB. Water can be made to boil without heatingC. Two pieces of ice stick to each other if they are pressed against each other and released due to regelationD. Ice can be made to sublimate |
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Answer» Correct Answer - B::C::D The water in beaker will be heated to `100^(@)C` but will not boil as for boiling it requires latent heat of vaporisation. Water can be made to boil by lowering the external pressure to the extent that its b.pt becomes equal to room temperature. releasing pressure and resolidification on releasing pressure. ice can be sublimed by reducing external pressure to its vapour pressure. |
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| 3205. |
For liquid enthalpy of fusion is `1.435 kcal mol^(-1)` and molar entropy change is `5.26 cal mol^(-1) K^(-1)`. The melting point of the liquid is |
| Answer» `:. T=(DeltaH_("Fusion"))/(DeltaS_("Fusion"))=(1.435xx10^(3))/5.26=273 K=0^(@)C` | |
| 3206. |
Assertion (A): The endothermic reactions are favoured at lower temperature and the exothermic reactions are favoured at higher temperature. Reason (R ) : when a system in equilibrium is disturbed by changing the temperature, it will tend to adjust itself so as to overcome the effect of the change.A. both (A) and (R ) are correct, and (R ) is the correct explanation for (A).B. both (A) and (R ) are correct, but(R ) is not a correct explanation for (A).C. (A) is correct, but (R ) is incorrect.D. (A) is incorrect, but (R ) is correct. |
| Answer» Endothermic reaction requires heat energy to produced in forward direction. | |
| 3207. |
Consider the reaction at `300 K` `H_(2)(g)+Cl_(2)(g)rarr2HCl(g), " "DeltaH^(@)= -185kJ` If `2` mole of `H_(2)` completely react with `2` mole of `Cl_(2)` to form `HCl`. What is `DeltaU^(@)` for this reaction? |
| Answer» Correct Answer - D | |
| 3208. |
If the enthalpy change for the transition of liquid water to steam is `30 kJ mol^(-1)` at `27^(@)C`, the entropy change for the process would beA. `100 J mol^(-1)K^(-1)`B. `10J mol^(-1) K^(-1)`C. `1.0 J mol^(-1) K^(-1)`D. `0.1 J mol^(-1) K^(-1)` |
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Answer» Correct Answer - a `DeltaS_("vap") = (Delta_("vap") H)/(T)` |
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| 3209. |
The heat of combustion of ethanol determinal in a bomb calorimeter is -670.48 K. Cals `"mole"^(-1)` at `25^(@)C`. What is `Delta H` at `25^(@)C` for the reaction :-A. `-335.24` K. Cals.B. `-671.08` K. Cals.C. `-670.48` K. Cals.D. `+670.48` K. Cals. |
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Answer» Correct Answer - B `C_(2)H_(5)OH_((l))+3O_(2(g))rarr 2CO_(2(g))+3H_(2)O_((l))` In Bomb calorimeter `Delta E = 670.48 kCal mol^(-1)` `Delta H=Delta E + Delta n_(g)RT` `Delta H=-670.48-1xx2xx10^(-3)xx298` |
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| 3210. |
Heat evolved during the combustion of 32 gm methanol is a bomb calorimeter was determined to be 470 kcal/mol at `25^(@)`C. The value of `Delta`u of the reaction at the same temperature isA. `-335.24` kcalB. `-669.28`kcalC. `-470` kcal/molD. `-196.5 xx 10^(4)` J |
| Answer» Correct Answer - C::D | |
| 3211. |
Assertion : Water kept in an open vessel will quickly evaporate on the surface of the moon. Reason : The temperature at the surface of the moon is much higher than boiling point of the water.A. If both, Assertion and Reason are true and Reason is the correct explanation of the Asserrion.B. If both,Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - C Water would evaporate quickly because there is no atmosphere on moon. |
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| 3212. |
The enthalpy changes for the following process are listed below : `Cl_(2)(g)=2Cl(g)," "242.3" kJ"mol^(-1)` `I_(2)(g)=2I(g)," "151.0" kJ"mol^(-1)` `ICl(g)=2I(g)+Cl(g)," "211.3" kJ"mol^(-1)` `I_(2)(s)=I_(2)(g)," "62.76" kJ"mol^(-1)` Given that standard states for iodine and chlorine are `I_(2)(s)` and `Cl_(2)(g)`, the standerd enthalpy of formation for ICl(g) is :A. `-16.8 kJ mol^(-1)`B. `+16.8 kJ mol^(-1)`C. `+244.8 kJ mol^(-1)`D. `-14.6 kJ mol^(-1)` |
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Answer» Correct Answer - B `(1)/(2)I_(2(g))+(1)/(2)Cl_(2(g))rarr ICl_((g)), Delta H_(f)(ICl) = ?` `eq.(v)=(1)/(2)[eq(i)+eq(ii)+eq(iv)]-eq(iii)` `Delta H=(1)/(2)[Delta H_(1)+Delta H_(2)+Delta H_(4)]-Delta H_(3)` |
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| 3213. |
The enthalpy changes for the following process are listed below : `Cl_(2)(g)=2Cl(g)," "242.3" kJ"mol^(-1)` `I_(2)(g)=2I(g)," "151.0" kJ"mol^(-1)` `ICl(g)=2I(g)+Cl(g)," "211.3" kJ"mol^(-1)` `I_(2)(s)=I_(2)(g)," "62.76" kJ"mol^(-1)` Given that standard states for iodine and chlorine are `I_(2)(s)` and `Cl_(2)(g)`, the standerd enthalpy of formation for ICl(g) is :A. `+16.8 kJ mol^(-1)`B. `+244.8 kJ mol^(-1)`C. `-14.6 kJ mol^(-1)`D. `-16.8 kJ mol^(-1)` |
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Answer» Correct Answer - A `1/2 I_(2)+1/2 Cl_(2) rarr ICl, DeltaH=H_(R)-H_(P)` |
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| 3214. |
A 5 litre cylinder contained 10 moles of oxygen gas at `27^(@)C`. Due to sudden leakage through the hole, all the gas escaped into the atmosphere and the cylinder got empty. If the atmosphere pressure is 1.0 atm. Calculate the work done by the gas. Assuming gas to be ideal. |
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Answer» `V_("initial")=5` litre, `T=27^(@)C=27+273 = 300 K` `V_("final")=(nRT)/(P)=(10xx0.0821xx300)/(1)=246.3` litre `Delta V=V_("final")-V_("initial")=246.3-5=241.3` litre `W_("exp")=-P Delta V=-1xx241.3` litre `xx` atm. `=-24443.69J=24.4 kJ` i.e. work done by gas is 24.4 kJ |
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| 3215. |
Two liters of `N_(2)` at `0^(@)C` and `5` atm pressure is expanded isothermally against a constant external pressure of `1` atm untill the pressure of gas reaches `1` atm. Assuming gas to be ideal, claculate the work of expansion. |
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Answer» Given `P_(1)=5` atm, `V_(1)=2` litre, `P_(2)=1` atm, `V_(2)=?` We know that - if T is constant then `P_(1)V_(1)=P_(2)V_(2)` `rArr V_(2)=(P_(1)V_(1))/(P_(2))=(5xx2)/(1)=10` litre `W=-P_(ext)Delta V=-1(V_(2)-V_(1))=-1(10-2)=-8` litre atm `=-8xx101.3=-810.4` joule |
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| 3216. |
Calculate the work done during isothermal reversible expansion expansion of one mol of an ideal gas from 10 atm to 1 atm at 300K. |
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Answer» Number of moles of ideal gas (n) = 1, Initial pressure `(P_(1))=10` atm, Final pressure `(P_(2))=1` atm Constant absolute temp (T) = 300 K, Gas constant (R )= 2 cal/mol. `therefore` Work done ofr isothermal reversible expansion of an ideal gas `W=-2.303 "nRT log" (P_(1))/(P_(2))=-2.303xx1xx2xx300xx"log"_(10)(10)/(1)=-1381.8` calories |
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| 3217. |
Calculate the amount of work done in each of the following cases `:-` (i) One mole of an ideal gas contained in a bulbof 10 litre capacity at 1 bar is allowed to enter into an evacuated bulb of 100 litre capacity. (ii) One mole of a gas is allowed to expand from a volumeof 1 litre to a volume of 5 litres against the constant external pressure of 1 bar ( 1 litre bar = 100 J) Calculate the internal energy change `(Delta U ) ` in each case if the process were carrid out adiabatically. |
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Answer» (i) `x= -P_(ext)xxDeltaV` As expansion takes place into the evacuated bulb, i.e., agains vacuum, `P_(ext) = 0`. Hence, w`=` 0 For adiabatic process `, q=0 :. Delta U = q+w= 0+0=0` (ii) `Delta V = V_(2)-v_(1) =5-1 = 4 `litres `P= 1`bar` :. w= - P Delta V = -1 xx 4` litre bar `= - 4 xx 100J = -400J ( 1 L ` bar `= 100 J )` The negative sign implies that the work is done by the system. For adiabatic process, `q=0` . Hence `Delta U = q+w= 0 - 405.2 J = - 405 . 2 J`. |
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| 3218. |
Find the work done in each case : (a) When one mole of ideal gas in 10 liter container aat 1 atm, is allowed to enter a vaccuated bulb of capacity 100 liter. (b) When 1 mole of gas expands from 1 liter to 5 liter a against constant stamospherice pressue . |
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Answer» Correct Answer - (a) `W=-PDeltaV` But since gas emters the vacurm bulb and pressuer in vacuuns zero . This type of expansion is called free expansion is always zero . Note :- Work done in free expansion is always zero (b) `W=- PDeltaV= - 1 (5-1) =- 42 liter-atm. |
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| 3219. |
Calculate the amount of work done in each of the following cases : (i) One mole of an ideal gas contained in a bulb of 10 litre capacity at 1 atm is allowed to enter into an evacuated bulb of 100 litre capacity. (ii) One mole of a gas is allowed of 1 atm (1 litre atm = 101.3 J) |
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Answer» (i) `W=-P_(ext)xx Delta V` As expansion takes place into the evacuated bulb, i.e., against vacuum, `P_(ext)=0`. Hence W = 0. (ii) `Delta V=V_(2)-V_(1)=5-1=4` litres, P = 1 atm `therefore W = -P Delta V = -1xx4` litre atm `=-4xx101.3J=-405.2 J` Alternatively, using the SI units directly P = 1 atm = 101325 Pa, `Delta V=4L=4xx10^(-3)m^(3)` `therefore W =-P xx Delta V=-101325xx4xx10^(-3)J=-405.3 J` The negative sign implies that the work is done by the system. |
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| 3220. |
Greenhouse gas `CO_(2)` can be converted to `CO(g)` by the following reaction `CO_(2)(g)+H_(2)(g) rarr CO_(2)+H_(2)O(g)` , termed as water gas reaction. Calculate `DeltaH` at 1400 K using the given data for 1000K , assuming the `C_(p)^(@)` values remain constant in the given temoerature range. `DeltaH=35040 J"mol"^(-1), C_(p)^(@)(CO_(2))=(42.31 + 10.09 xx 1^(-3)T) J "mol"^(-1)K_(1)` `C_(P)^(@)(H_(2))=(27.40 + 3.20 xx 10^(-3)T) J"mol"^(-1)K_(1)` `C_(P)^(@)(CO)=(28.34+ 4.14 xx 10^(-3)T)J"mol"^(-1)K^(-1)` `C_(P)^(@)(H_(2)O)=(30.09 + 10.67 xx 10^(-3)T)J"mol^(-1)K^(-1)` |
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Answer» Correct Answer - `int_(H_(1))^(H_(2))dH=int_(T_(1))^(T_(2)) Cp.dT` `DeltaH_(2)-DeltaH_(1) = int_(T_(1))^(T_(2)) (28.35 + 4.14 xx 10^(-3)T + 30.09 + 10.67 xx 10^(-3)T)-(42.31 + 10.09 xx 10^(-3)T + 27.4 + 3.2xx 10^(-3)T)dT` `=int_(T_(1))^(T_(2)) (-11.28 + 1.52 xx 10^(-3)T)dT` `=-11.28 xx 400 + 1.52 xx ((1400^(2) - 1000^(2)))/(2) xx 10^(-3)` `DeltaH_(2) - 35040 =- 3782.4 " " therefore DeltaH_(2) = 31258 "Joule"` |
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| 3221. |
For a chemical reaction,`DeltaH` is negative and `DeltaS` is negative and `DeltaS` is negative . This reaction isA. spontaneous al all temperaturesB. nonspontaneous at all temperatureC. spontaneous only at high temperatureD. spontaneous only at low temperature |
| Answer» Correct Answer - A | |
| 3222. |
Greenhouse gas `CO_(2)` can be converted to `CO(g)` by the following reaction `CO_(2)(g)+H_(2)(g) rarr CO_(2)+H_(2)O(g)` , termed as water gas reaction. Calculate `DeltaG` for the reaction at 1000K `(DeltaH_(1000K)=35040 J "mol"^(-1) DeltaS_(1000K) =32.11 J "mol"^(-1)K^(1))`. |
| Answer» Correct Answer - `DeltaG=DeltaH-TDeltaS " " therefore DeltaG=35040- 1000 xx 32.11=2930 J` | |
| 3223. |
6.24 g of ethanol are vaporized by supplying 5.89 KJ of heat energy . What is the ebnthalpy of vapourisation of ethanol?A. 43.42 KJB. 47.0 KJC. 21.75 KJD. 435.0 KJ |
| Answer» Correct Answer - A | |
| 3224. |
One mole of solid iron was vaporized in an oven at `3500` K . If iron boils at `3133`K abd enthalpy of vaporization is `349 KJ mol^(-1)` , determine `DeltaS_("system"), DeltaS_("surrounding")` and `DeltaS_("universe")`. (Oven is considered as surroundings). |
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Answer» Correct Answer - `DeltaS_("system")=111.4 JK^(-1), " " DeltaS_("surr")=-99.71 JK^(-1), `DeltaS_("surr")=+11.69 JK^(-1)` At boiling point , `" "DeltaS_("system")=(DeltaHvap)/(t_(b)) = (349xx10^(33))/(3133) = 111.4 JK^(-1)` Heat change in surrounding `=-349 KJ` `DeltaS_("surr") =-(349xx1000)/(3500) =-99.71 JK^(-1)` `implies " "DeltaS_("univ") = DeltaS_("sys") + DeltaS_("surr") = 111.4=99.71 =+ 11.69 JK^(-1)` |
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| 3225. |
Calculate the enthalpy change accompanying the transforming of C (graphite) to C(diamond) . Given that the enthalpies of combustion of graphite and diamond are 393.5 and 395.4 kJ `mol^(-1)` respectively. |
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Answer» We are given (i) C (graphite) `+ O_(2)(g) rarr CO_(2)(g) , Delta _(c ) H^(@)= - 393.5 kJ mol^(-1)` (ii) C (diamond) `+ O_(2)(g) rarr CO_(2)(g), Delta _(c ) H^(@) = - 395.4 kJ mol^(-1)` We aim at C( graphite ) `rarr ` C ( diamond) `, Delta _("trans") H^(@) = ?` Subtracting eqn. (ii) from eqn. (i), we get or C(graphite)- C (diamond) `rarr 0 , Delta _(r) H^(@) = - 393.5 - ( - 395.4) = + 1.9 kJ` |
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| 3226. |
The enthalpies of combustion of graphite and diamond are 393.5 kJ and 395.4 kJ respectively. Calculate the enthalp change accompanying the transformation of 1 mole of graphite into diamond. |
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Answer» Correct Answer - `1.9 kJ mol^(-1)` Given `: (i) C(gr) + O_(2)(g) rarr CO(_(2)(g), DeltaH = - 393. 5 kJ mol^(-1)` (ii) `C(dia) + O_(2)(g) rarr CO_(2)(g) , Delta H = -395kJ mol^(-1)` Aim `: C (gr) rarr C (dia) , Delta H = ?` Eqn. (i) - Eqn. (ii) gives the required result. |
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| 3227. |
Calculate the heat change accompanying the transformation of C (graphite) to C (diamond). Given that the heats of combustion of graphite and diamond are 393.5 and 395.4 kJ mol-1 respectively. |
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Answer» (i) C (graphite) + O2 (g) → CO2 (g) ΔH = -393.5 kJ (ii) C (diamond) + O2 (g) → CO2 (g) ΔH = -395.4 kJ C (graphite) → C (diamond), ΔH = ? Subtracting equation (ii) from (i), we get C (graphite) - C (diamond) = 0 ΔH = -393.5 - (-395.4) = +1.9 kJ or, C (graphite) → C (diamond) ΔH = +1.9 kJ |
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| 3228. |
When a sample of gas expand from `4.0L` to `12.0L` against a constant pressure of `0.30 atm`, the work involved isA. `243.19 J`B. `-243.19 J`C. `234.19 J`D. `-234.19 J` |
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Answer» Correct Answer - B `W = -P Delta V` |
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| 3229. |
`1.0 mol` of an ideal gas initially present in a `2.0L` insulated cylinder at `300K` is allowed to expand against vacuum to `8.0L`. Determine `w, DeltaU, Delta_("total")S`, and `DeltaG`. |
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Answer» `w =- P_(ext)DeltaV = 0, q =0`(insulated cylinder) `rArr DeltaU = 0 = DeltaH` `rArr T_(f) = 300K` `Delta_(sys)S = nR"In"(V_(2))/(V_(1)) = 2RIn 2 = 11.52 J K^(-1)` and `Delta_(surr)S = 0 :. Q_(sys) = q_(surr) = 0` `rArr Delta_("total")S = 11.52JK^(-1)` `rArr DeltaG =- TDelta_("total")S =- 300 xx 11.52 =- 3456J` |
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| 3230. |
In an adiabatic process, which of the following is true? (a) q = w (b) q = 0 (c) ΔE = q (d) PΔV = 0 |
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Answer» Answer: (a) q = 0 |
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| 3231. |
Heat of neutralization is always (a) positive (b) negative (c) zero (d) positive or negative |
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Answer» (b) negative |
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| 3232. |
Define cyclic process. Give example. |
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Answer» When a system returns to its original state after completing a series of changes, then it is said that a cycle is completed. This process is known as a cyclic process. For a cyclic process dU = 0, dH = 0, dP = 0, dV= 0 and dT= 0 |
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| 3233. |
Match the List-I and List-II using the correct code given below the list.List-IList-IIA. Cyclic process1. ΔU = q - PΔVB. Adiabatic process2. ΔU = qvC. Isobaric process3. q = - wD. Isochoric process4. ΔU = wCode: (a) A - 4, B - 2, C - 3, D - 1(b) A - 3, B - 4, C - 1, D - 2(c) A - 2, B - 1, C - 4, D - 3(d) A - 1, B - 3, C - 2, D - 4 |
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Answer» (b) A - 3, B - 4, C - 1, D - 2 |
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| 3234. |
Which one of the following is the quantity of heat required to raise the temperature of 1 gm of water by 1 °C? (a) 1 Joule (b) 1 Calorie (c) 1 Kelvin (d) 1 Kilo joule |
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Answer» (b) 1 Calorie |
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| 3235. |
Which one of the following is ail intensive property? (a) Specific heat capacity (b) Mass (c) Enthalpy (d) Heat capacity |
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Answer» (a) Specific heat capacity |
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| 3236. |
Calculate the amount of heat required to raise the temperature of 5 g of iron from `25^(@)C "to" 75^(@)C`. The specific heat capacity of iron is 0.45 J/g. |
| Answer» Correct Answer - 112.5 J | |
| 3237. |
Heat liberated when 100 ml of in NaOH is neutralized by 300 ml of in HCl ……(a) 22.92 Ici (b) 17.19 kJ (c) 11.46 kJ (d) 5.73 Id |
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Answer» Base = V1 = 100ml N1 = 1N Acid = V2 = 300ml N2 = 1N Enthalpy of neutralization of 1000 ml = 57.3 kJ. ∴ Enthalpy of neutralization of 100 ml x 100 = 5.73KJ / 1000 x 100 = 5.73 kJ. |
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| 3238. |
Which of the following is not a thermodynamic function? (a) internal energy (b) enthalpy (c) entropy (d) frictional energy |
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Answer» (d) frictional energy |
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| 3239. |
For a cyclic process, the change in internal energy of the system is ……(a) always + ve (b) equal to zero (c) always – ve (d) none of the above |
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Answer» (b) equal to zero |
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| 3240. |
The heat required to raise the temperature of a body by 1 K is called ……(a) specific heat (b) thermal capacity (c) water equivalent (d) none of these |
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Answer» (b) thermal capacity |
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| 3241. |
Heat capacity is ……(a) dq / dT(b) dq.dT (c) ∑q. 1/dT(d) none of these |
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Answer» Answer: (a) dq/ dT |
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| 3242. |
Calculate the values of ∆U and ∆H for an ideal gas in terms of CP and CV. |
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Answer» Calculation of ∆U and ∆H: For one mole of an ideal gas, we have, CV = dU / dT dU = C.dT For a finite change, we have, ∆U = (U2 – U1) = CV (T2 – T1) and for n moles of an ideal gas we get ∆U = nCV (T2 – T1 ) …….(1) We know, ∆H = ∆(U + PV) ∆H = ∆U + ∆(PV) ∆H = ∆U + ∆RT [∴ PV= RT] ∆H = ∆U + R∆T ∆H = CV (T2 – T1) + R (T2 – T1) ∆H = (CV + R) (T2 – T1 ) ∆H = CP(T2 – T1) [∴C – C = R] For n moles of an ideal gas we get ∆H = nCP (T2 – T1) ……..(2) |
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| 3243. |
Prove that for an ideal gas, CP is greater than CV . |
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Answer» 1. It is clear that two heat capacities are not equal and CP is greater than CV by a factor which is related to the work done. 2. At a constant pressure, a part of heat absorbed by the system is used up in increasing the internal energy of the system and the other for doing work by the system. 3. At constant volume, the whole of heat absorbed is utilized in increasing the temperature of the system as there is no work done by the system. Thus CP is greater than CV . CP = dH / dT ; CV = dU / dT 4. By definition, H = U + PV for 1 mole of an ideal gas. H = U + RT By differentiating this equation with respect to temperature T. we get, dH / dT = dU / dT + R CP = CV + R CP – CV = R Thus for an ideal gas, CP is greater than CV by the gas constant R. |
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| 3244. |
For an ideal gas ……(a) Cp – Cv = O (b) Cp – Cv = R (c) Cv – Cp = R (d) Cv – Cp > R |
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Answer» (b) Cp – Cv = R |
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| 3245. |
Which of the following properties is not a fùnction of state? (a) Concentration (b) Internal energy (c) Enthalpy (d) Entropy |
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Answer» (a) Concentration |
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| 3246. |
Which of the following always has a negative value? (a) heat of reaction (b) heat of solution (c) heat of combustion (d) heat of formation |
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Answer» (c) beat of combustion |
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| 3247. |
The relation between Cp and Cv is……(a) Cp – Cv = R (b) Cp + Cv = R (c) – 285 KJ (d) R – Cv = Cp |
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Answer» Answer: (a) Cp – Cv = R |
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| 3248. |
Which of the following relation is true? (a) Cp > Cv (b) Cv > Cp (c) Cp = Cv (d) Cp = Cv = 0 |
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Answer» Answer: (a) Cp >Cv |
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| 3249. |
Calculate the entropy change in the melting of 1 kg of ice at `0^(@)C` in SI units. Heat of funsion of ice = 80 cal `g^(-1)`. |
| Answer» `Delta S_(fus)=(q_(fus))/(T)=(80xx1000)/(273) (cal)/(K)=(8000)/(273)xx4.2JK^(-1)=1230JK^(-1)` | |
| 3250. |
Calculate the change in entropy for the fusion of 1 mole of ice (water). The melting point of water is 273 K and molar enthalpy of funsion for water `=6.0 kJ mol^(-1)` |
| Answer» `Delta S_(fus)=(Delta H_(fus))/(T)=(6000J mol^(-1))/(273K)=21.97 JK^(-1)mol^(-1)` | |