Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

One mole of an organic compound requires 0.5 mole of oxygen to produce an acid. The compound may be

Answer»

Aldehyde
Ether
Ketone
Alcohol

ANSWER :A
2.

One mole of an organic compound A with the formula C_(3)H_(8)O reacts completely with two moles of HI ot form X and Y. When Y is boiled with aqueous alkali it forms Z. Z answers the iodoforms test. The compound A is

Answer»

Propan-2-ol
Propan 1-ol
Ethoxygethane
Methoxyethane

Solution :If A reacts with HI, it could be ether
`CH_(3)-O-C_(2)H_(5)underset("(excess)")overset(HI)to underset((X))(CH_(3)I)+underset((Y))(C_(2)H_(5)I)+H_(2)O`
`underset(C_(2)H_(5)Ioverset(aq. KOH)to underset((Z))(C_(2)H_(5)OH)`
`C_(2)H_(5)OH` gives IODOFORM test.
3.

One mole of an organic compound is found to require only 0.5 mol of oxygen to produce an acid. Which class of compounds does the starting material belong to ?

Answer»

Alcohol
Ether
Ketone
Aldehyde

Solution :`RCHO + (1)/(2)O_(2) RARR RCOOH`.
4.

One mole of an organic compound A with the formula C_(3)H_(8)O reacts completely with two moles of HI to from X and Y. When Y is boiled with aqueous alkali it forms Z.Z answers the iodoform test. The compound A is

Answer»

A) propan-2-ol
B) propane-1-ol
C) ethoxyethane
D) methoxyethane

Solution :If A reacts with HI, it could be ether
`CH_(3)-underset((A))(O)-C_(2)H_(5)underset("(excess)")overset(HI)(to)underset((X))CH_(3)I+C_(2)H_(5)I+H_(2)O`
`underset((Y))(C_(2)H_(5)I)overset("aq. KOH")(to)underset((Z))(C_(2)H_(5)OH)`
`C_(2)H_(5)OH` gives IODOFORM test .
5.

One mole of an organic compound (A) having M.F. C_(2)H_(6)O reacts with MeMgI to liberate one mole of methane. (A) reacts with CH_(3)COCl to yield a sweet smelling liquid (B). Identify (A) and (B).

Answer»

Solution :(i) Since compound (A) with M.F. `C_(2)H_(6)O` reacts with MeMgI to give one mole of `CH_(4)`, the compound (A) must be ETHYL alcohol.
`MeMgI-=CH_(3)MgI`
`underset("Ethyl alcohol")(CH_(3)CH_(2)OH)+CH_(3)MgI to CH_(3)+Mg(OCH_(2)CH_(3))I`
(II) Compound A, i.e., ethyl alcohol reacts with acetyl chloride `(CH_(3)COCl)` to form sweet smelling compound (B). therefore, compound (B) must be ethyl acetate.
`underset("Acetyl chloride")(CH_(3)COCl)+underset("Ethyl alcohol")(HOCH_(2)CH_(3)) to underset("Ethyl acetate (B)")(CH_(3)CO-OCH_(2)CH_(3)+HCl)`
6.

One mole of an organic amide (A) upon alkaline hydrolysis gives one mole of NH_3and one mole of monobasic acid of equivalent weight 74. What is the molecular formula of (A)?

Answer»

SOLUTION :`C_2H_5CONH_2`
7.

One mole of an non linear triatomic ideal gas is expanded adiabatically at 300 K from 16 atm to 1 atm. Find the values of DeltaS_("sys"),DeltaS_("surr")&DeltaS_("tota") under the following conditions. (i) Expansion is carried out reversibly. (ii) Expansion is carried out irreversibly (iii) Expansion is free.

Answer»

Solution :For non-linear tri-atomic IDEAL gas
`C_(v)=3R,C_(p)=4R`
(i) `DeltaS_("sys")=nC_(v)"ln"T_(2)/T_(1)+NR"ln"v_(2)/v_(1)=0`
q = 0
`DeltaS_("surr")=-DeltaS_("sys")=0`
`DeltaS_("total")=0`
(ii) First of all we will have to calculate the temperature of the gas after it has undergoes the said adiabatic reversible expansion we have q = 0
`DeltaU=q+w`
`nC_(v)(T_(2)-T_(1))=-P_("ext")(v_(2)-v_(1))`
`3R(T_(2)-300)=-1[(RT_(2))/p_(2)-(RT_(1))/2]=-R[T_(2)/1-300/16]`
`T_(2)=229.68K`
`DeltaS_("sys")=nC_(p)"ln"T_(2)/T_(1)+nR"ln"p_(1)/p_(2)`
`=4R"ln"229.68/300+R"ln"16/1=-1.068R+2.77`
`R=1.702R`
`DeltaS_("surr")=(-q_("irr"))/T=0`
`DeltaS_("total")=DeltaS_("sys")=1.702R`
(iii) In free adiabatic expansion we have
w = 0
`q=0" "DeltaT=0`
`DeltaS_("sys")=nR"ln"p_(1)/p_(2)=Rln16=2.77R`
`DeltaS_("surr")=(-q_("irr"))/T=0`
`DeltaS_("total")=DeltaS_("sys")=2.77R`
8.

One mole of an non linear triatomic ideal gas is compressed adiabatically at 300 K from 1 atom to 16 atm. Calculate Work done under the following conditions. (i) Expansion is carried out reversibly. (ii) Expansion is carried out irreversibly.

Answer»

SOLUTION :q = 0 Adiabatic PROCESS
(i) `W_("rev")=DeltaU=C_(V)(T_(2)-T_(1))`
`P^(1-gamma)T^(gamma)=K`
`T_(2)=300(1/16)^((1-4//3)/(4//3))=600K`
w = 3R (600 - 300) = 900 R
(ii) `W_("irr")=-P_("EXT")(V_(2)-V_(1))=-P_("ext")((nRT_(2))/P_(2)-(nRT_(1))/P_(1))`
`-P_("ext")((nRT_(2))/P_(2)-(nRT_(1))/P_(1))=C_(v)(T_(2)-T_(1))`
`-16((RT_(2))/16-(RT_(1))/1)=3R(T_(2)-T_(1))`
`-(T_(2)-16T_(1))=3R(T_(2)-T_(1))`
`-T_(2)+16T_(1)=3T_(2)-3T_(1)`
`4T_(2)=19T_(1)`
`T_(2)=19/4xx300=1425K`
`w=DeltaU=3R(1425-300)=3375R`
9.

One mole of an non linear triatomic ideal gas is expanded adiabatically at 300 K from 16 atm to 1 atm. Calculate Work done under the following conditions. (i) Expansion is carried out reversibly. (ii) Expansion is carried out irreversibly

Answer»

Solution :q = 0
`W=DeltaU=C_(v)(T_(2)-T_(1))`
`C_(v)` for triatomic NON linear GAS = 3R
(i) For rev. process.
`P_(1)^(1-GAMMA)T_(1)^(gamma)=P_(2)^(1-gamma)T_(2)^(gamma)""r=4/3`
`T_(2)=T_(1)(P_(1)/P_(2))^((1-gamma)/v)=300(16)(1-4/3)/(4//3)=300(2^(4))^(-1//4)=150K`
`DeltaU=w-3R(150-300)=-450R`
(II) n = 1
`-P_("ext")(V_(2)-V_(1))=C_(V)(T_(2)-T_(1))`
`-P_("ext")((RT_(2))/P_(2)-(RT_(1))/P_(1))=3R(T_(2)-T_(1))`
`-1((RT_(2))/1-(RT_(1))/16)=3R(T_(2)-T_(1))`
`-R/16(16T_(2)-T_(1))=3R(T_(2)-T_(1))`
`-16T_(2)+T_(1)=48T_(2)-48T_(1)`
`49T_(1)=64T_(2)`
`T_(2)=49/64xx300`
`T_(2)=229.69`
`W_("irr")+C_(v)(T_(2)-T_(1))+3R(229.69-300)=-210.93R`
10.

One moleof an non linear triatomic ideal gasis expanded abiabatically at 300 K form16 atmto1 atmltbr gtFind the value DeltaS_(sys), DeltaS_(surr) & DeltaS_("total")under the following conditions. (i) Expansion is carriedout reversibly (ii) Expansionis carriedout irreversibly (iii) Expansion is free.

Answer»


SOLUTION :N//A
11.

One mole of annon lineartriatomicideal gas is compressed adiabaticallyat 300 K from 1 atm to 16 atm. Calculate Work doneunder the following conditions.(i) Expansion is carriedout reversibly (ii) Expansion is carried outirreversibly.

Answer»


SOLUTION :N//A
12.

One moleof an idealmonoatomic gas is put throughtreverslible pathas show in figure.Fill in theblankin thetablegivenbelow:

Answer»


Solution :`"State-1": """P= 1 atm""""V= 22.4L""""T= 273K"`
`"State-2": """P= 1 atm""""V= 44.8L""""T= 546K"`
`"State-3": """P= 0.5 atm""""V= 44.8L""""T= 273K"`
Step -A : Isobaricprocess
`W= - PDeltaV = - nRDeltaT=- R(273)`
`DeltaH=nC_(p,m)DELTAT = (5)/(2)Rxx 273`
`DeltaE=nCv_(m)DeltaT = (3)/(2)Rxx 273`
`q_(p) = DeltaH =(5)/(2)R xx 273`
Step -B : ISOCHORIC process
`W=0`
`DeltaH=nC_(p,m)DeltaT = -(5)/(2)Rxx 273`
`DeltaE=nC_(v,m)DeltaT = -(3)/(2)Rxx 273`
`q_(p) = DeltaE =-(3)/(2)R xx 273`
Step-C : Isothermal process
`W= - nRT ln.((V_(2))/(V_(1)))= 273ln2`
`DeltaH=0`
`DeltaE = 0`
`q= - W "" =- 273 R ln 2`
13.

One mole of an ideal monoatomic gas mixed with 1 mole of an ideal diatomic gas. The molar specific heat of thr mixture at constant vlume is :

Answer»

3 cal
4 cal
8 cal
9 cal

Answer :B
14.

One mole of an ideal monoatomic gas is caused to go through the cycle shown in figure.Then the change in the internal energy of gas from a tob and b to c is respectively :

Answer»

`(9P_@V_@)/2,6 RT_@`
`(9P_@V_@)/2, 10 RT_@`
`(15P_@V_@)/2,6 RT_@`
`(15P_@V_@)/2, 10 RT_@`

SOLUTION :NA
15.

One mole of an ideal monoatomic gas is heated according to path AB and AC. If temperature of state B and C are equal then calculate (q_(AC))/(q_(AB)) xx 10.

Answer»


ANSWER :8
16.

Onemoleof an ideal monoatomic gas iscarried through the reversilycycle of thegiven figure consisiting of stepA,and C andinvolvingstate1,2 and 3. Fill in the blankspacein thethe tablegivenbelow assumingreversible steps.

Answer»


Solution :`1 to 2 rArr "" " Isochoric preocess"`
`2 to 3 rArr "" " Isothermala preocess"`
`3 to 1 rArr "" " Isobaric preocess"`
`"State - 1:"""P=(NRT)/(V)= 1 atm""V= 22.4 L ""T= 273K`
`"State - 2:"""T= 546K""V= 22.4 L ""P= 2atm`
`"State - 3:"""T= 546K""V= 44.8 L ""P= 1atm`
Step - A:Isochoric process
`""W = 0`
`""DeltaE= n C_(v.m)DELTAT= (3)/(2) R xx 273`
`""DeltaH= n C_(p.m)DeltaT= (5)/(2) R xx 273`
`""q_(v)= DeltaE= (5)/(2) R xx 273`
Step- B : Isothermal processltb`""DeltaE=0`
`""DeltaH=0`
`""W= - nRT ln .((V_(1))/(V_(2))) = - 1 xx R xx546 ln 2`
`""= - 546R ln 2`
`""q = - W =546R ln 2`
Step-C : Isobaricprocess
`""DeltaE = nC_(v.m)DeltaT= -(3)/(2) R xx 273`
`""DeltaH = nC_(p.m)DeltaT= -(5)/(2) R xx 273`
`""W =- PDeltaV= - nRT = R xx 273`
`""q_(p) = DeltaH = -(5)/(2) R xx 273`
17.

One mole ofan ideal monoatomic gasexpanded irreversibly in two stage expansion . "State"-1""("8.0 bar, 4.0 liters, 300K") "State"-2""("2.0 bar, 16 liters, 300K") "State"-3""("1.0 bar, 32 liters, 300K") Total heatabsored by thegas in theprocess is :

Answer»

<P>116 J
40 J
4000 J
None of these

Solution :`W= - P_(ext)[V_(2)-V_(1)]`
`W_(1)= - 2 [16-4] xx 100 ""= - 2400J`
`W_(2) = - [32 -16] = - 16 xx 100= - 1600 J`
""`-4000`
`q = - W = 4000 J`
18.

Onemole ofan idealmonoatomic gas (C_(V.M)= 1.5 R)is subjected to the followingsequence of steps : (a)The gasis heated reversiblyat constantpressure of 1 atm from 298 K to 373 K. (b) Next, thegas isheatedreversiblyand isothermallyto doubleits volume. (c)Finally , thegasis cooled reversiblyandadiabatically to 308 K . Calcuated q, w, DeltaH for the overall porcess.

Answer»


Solution :(a)`""W= intP_("ext").dV = - P_("ext). DeltaV`
`= - nRDeltaT`
`=- 1xx 8.14 XX (373 - 298)`
` = -623.55 J//mol`
`DeltaH = q = nC_(P)DeltaT`
`= - 2.5 xx 8.314 xx 75"" (C_(P)=2.5 R)`
`= 1558.8 J//mol`
`DeltaU = 1.5 xx 8.314 xx 75""(C_(V) = 1.5 R)`
` =935.5 J//mol`
(b)` W=-2.303RT log ((V_(1))/(V_(2)))`
` = - 2.303 xx 8.314 xx 373 log 2`
` = - 2149. 7 J//mol`
`DeltaU = 0 ,DeltaH = 0 ""q =2149.7 J//mol^(-1)`
(C)`W = Deltau = nC_(V)DeltaT""("For adiabaticprocess "q = 0, DeltaU ne W)`
`= 1 xx 1.5 xx 8.314 (308-373)`
`= - 810.62 J//mol`
`q= 0`
`DeltaH = nC_(P)DeltaT=- 1 xx 2.5 xx 8.314 (308-373)`
`= -1351 .03 J//mol`
For overall process `q_(net) = 3708.59 J//mol`
`W_("net") = - 3583. 88 J//mol""DeltaU_("net")= 124.71 J//mol`
`DeltaH_("net")= 207.85 J//mol`
19.

One mole of an ideal gas undergoes reversible isothermal expansion from an initial volume V_(1) to a final volume 10V_(1), and does 10 kJ of work. The initial pressure was 1xx10^(7) Pa. (a) Calculate V_(1) . (b) If there were two moles of the gas, what had been its temperature ?

Answer»

Solution :For reversible ISOTHERMAL expansion of an ideal gas,
(a) `W=-2.303nRTlog.(V_(2))/(V_(1))=-10,000("GIVEN")`
or `2.303xx1xx8.314xxTlog.(10V_(1))/(V_(1))=10,000`
`T=522.27K`
We have
`pV=nRT`
`10^(7)xxV_(1)=1xx8.314xx522.27`
`V_(1)=4.342xx10^(-4)m^(3)`
(b) Assuming pressure to remain constant, for 2 moles of the ideal gas.
`T=(522.27)/(2)=261.13 K`.
20.

One mole of an ideal gas undergoes a process (10L, 4atm) to (4L, 10 atm ). Theprocess may be represented by straight line in P-V diagram The correct statementfor the process :

Answer»

The temperature of gas is CONSTANT throughout the PROCESS
Thetemperature of gas will first increase and then DECREASE in the process
The temperature of gas will first decrease and then increase in the process
The INITIAL and final temperature if gas are different

Answer :B
21.

One mole of an ideal gas undergoes change of state from (4 L, 3 atm) to (6 L, 5 atm). If the change in internal energy is 45 L-atm then the change of enthalpy for the process is

Answer»

27 L-atm
45 L-atm
55 L-atm
63 L-atm

Answer :D
22.

One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown In the graph below. If the work done along the solid line path is W_(s) and that along the the dotted line path is w_(d), then the integer closest to the ratio w_(d)//w_(s) is

Answer»


Solution :Horizontal dotted LINES indicate the work of constant pressure, each GIVEN by `PDeltaV`. Vertical dotted lines indicate that work done is zero because volume remains constant
`W_(d)=(4.0)xx1.5)+(1.0xx1.0)+(0.75xx2.5)L"atm"=6+1+1.815=8.875 L "atm"`
Solid line indicates isothermal reversible expansion. It is isothermal because `PV` is constant
`:. W_(s)=2.303nRT log V_(2)//V_(1)`
`=2.303PVlog V_(2)//V_(1) (PV=nRT)`
`=2.303xx2log"(5.5)/(0.5)(PV=2 L "atm")`
`=4.61 L"atm"`
Hence `=(W_(d))/(W_(s))=(8.875)/(4.61)=2`
23.

One mole of an ideal gas is expanded isothermally at 300 K from 10 atm to 1 atm. Find the values of DeltaS_("sys"),DeltaS_("surr")&DeltaS_("total") under the following conditions. (i) Expansion is carried out reversibly. (ii) Expansion is carried out irreversibly (iii) Expansion is free.

Answer»

Solution :(i) `DeltaS_("sys")=nC_(p)"ln"T_(2)/T_(1)+nR"ln"P_(1)/P_(2)=0+Rln10=Rln10`
`DeltaS_("surr")=-DeltaS_("sys")=-Rln10`
`DeltaS_("total")=0`
(ii) `DeltaS_("sys")=Rln10`
`DeltaS_("surr")=(-q_("IRR"))/T`
`DeltaU=0=Q+w`
`q_("irr")=p_("EXT")(v_(2)-v_(1))`
`q_("irr")=p_("ext")[(RT)/p_(2)-(RT)/p_(1)]=RT[1/1-1/10]`
= `RTxx9/10=9/10xxRxx300=270R`
`DeltaS_("surr")=(-q_("irr"))/T=(-9R)/10`
`DeltaS_("total")=DeltaS_("sys")+DeltaS_("sur")=Rln10-(9R)/10`
(iii) Free expansion `DeltaT = 0`
w = 0
q = 0
`DeltaS_("sys")=Rln10`
`DeltaS_("surr")=(-q_("irr"))/T=0`
`DeltaS_("total")=Rln10`
24.

One mole of an ideal gas is taken from a to b along two parths dented by the solid and the dashed lines as shown in the graph below. If the work done along the solid line pathis W_(s)and that along the dotted line path is W_(d) then the interger closet to the ratio W_(d) // W_(s) is

Answer»


ANSWER :-2
25.

One mole of an ideal gas is expanded isothermally from 1 dm^(3) to 10 dm^(3) at 300 K. DeltaG will be equal to

Answer»

5.744 kJ
57.44 J
574.4 J
`-5744J`

ANSWER :D
26.

One mole of an idealgas iscompressedfrom 500cm^(3) againsta constantpressureof 1.216 xx 10^(5)Pa. Theworkinvolved in theprocessis 35.50 J. calculatethe finalvolume .

Answer»


Solution :Given :Number of themolesa GAS= n= 1mol
Constantexternalpressure `= P_(EX)`
`= 1.216 xx10^(5)Pa = 1.216 xx 10^(5)NM^(-2)`
Forcompressionworkobtained= W=+36 .5 J
Initialvolume= `V_(1)= 500cm^(3)= 500xx 10^( -6)m^(3)( :'1 m^(3)=10^(6) cm^(3))`
Finalvolume `=V_(2)= ?`
Sincethe COMPRESSION thetakesplaceagainsta constantpressureit isan irreversibleprocess andworkobtained is
`W= -P_(ex) (V_(2)-V_(1))`
`36.5 =-1.216xx 10^(5)xx (V_(2) - 500xx 10^(-6))`
`:.(36.5)/(1.216 xx 10^(5))= - (V_(2) -50 xx 10^(5))`
`30 xx 10^(-5)=-(V_(2) -50 xx 10^(-5))`
`:. 3010^(5) = - V_(2) + 50 xx 10^(-5)`
`:. V_(2)= (50 -300) xx 10^(-5)=20 xx 10^(-5)m^(3)`
`= 20 xx 10^(-5) xx 10^(6) cm^(3) = 200 cm^(3)`
27.

One mole of an ideal gas is expanded isothermally at 300 K from 10 atm to 1 atm. Calculate q, w, DeltaU&DeltaH under the following conditions. (i) Expansion is carried out reversibly. (ii) Expansion is carried out irreversibly

Answer»

SOLUTION :Isothermal process
(i) For ideal gas `DeltaU=0" "DeltaH=0`
Q = -W
`w_("rev")=-nRT"ln"P_(1)/P_(2)=-1xxRxx300"ln"10/1=-690.9R`
(ii) `W_("irrev")=-P_("ext")(V_(2)-V_(1))=-P_("ext")((nRT)/P_(2)-(nRT)/P_(1))`
= `-1xx1xxRxx300[1/1-1/10]=-270R`
28.

One mole of an ideal gas is allowed to expand freely and adiabatically into vacuum until its volume has doubled. Astatement which is not true concerning this expression is

Answer»

`DELTAH = 0`
`DELTAS = 0`
`DeltaE = 0`
W = 0

Solution :W=0 is not true.
29.

One mole of an ideal gas going from state-A to state-B through different processes. Column-I shows graph of the process and column-II shows change in parameter in these process. Match the following:

Answer»


ANSWER :a-p,R,s,t; b-p,t; c-p, r,s,t; d-p
30.

One mole of an ideal gas is allowed to expand reversibly and adibatically from a temperature of 27^(@)C. If the work done during the process is 3 kJ, then final temperature of the gas is (C_(v)=20 J//K)

Answer»

100 K
150 K
195 K
255 K

Solution :Given nubmer of moles =1
Initial temperature = `27^(@)C=300K`
Work DONE by the SYSTEM = 3kJ = 3000 J
It will be (-) because work is done by the system
HEAT capacity at constant volume (CV) = 20 J/K
We know that work done
`W=-nC_(V)(T_(2)-T_(1)), 3000 =-1xx20(T_(2)-300)`
`3000=-20T_(2)+6000`
`20T_(2)=3000, T_(2)=(3000)/(20)=150 K`
31.

One mole of an ideal gas (C_(v) = 3/2R) is heated at constant pressure of 1 atmosphere from 25^(@) C to 100^(@) C. Calculate DeltaE and DeltaH.

Answer»

Solution :As `(C_(P) - C_(v)) = R rArr C_(p)=C_(v) + R`
`rArr C_(p) = 3/2 R + R= 5/2 R`
Heat given at CONSTANT PRESSURE.
`(DeltaH) = nC_(p)DeltaT = 1 x 5/2 R xx (373 - 298)`
`rArr (DeltaH) = 1 xx 5/2 xx 1.987 xx 75 = 372.56` cal
Work done in the process = `-PDeltaV`
`=-P(V_(2)-V_(1))`
`=-P((nRT_(2))/P -(nRT_(1))/P)`(As PV = nRT)
`=-NR(T_(2)-T_(1))`
`=-1 xx 1.987 xx (373 - 298)`
`=-149.05` cal.
From FIRST law of thermodynamics.
`DeltaE = q+W`
`=372.56 - 149.05`
=223.51 cal.
32.

One mole of an ideal gas expands reversibly and isothermally at 300 K from 5dm^(3)" to "50dm^(3). The work done by the gas for the process is equal to

Answer»

`-1.382` KCAL
`+1.382` kcal
`-1381.8` kcal
`+1382` kcal

Answer :A
33.

One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 litre to 10 litres. The DeltaE for this process is (R = 2 cal mol^(-1)K^(-1))-

Answer»

163.7 cal
zero
138.1 cal
9 LIT atm.

Answer :B
34.

One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 L to 10 L. The DeltaE for this process is (R=2 cal . "mol"^(-1)K^(-1))

Answer»

163.7cal
zero
1381.1cal
9lit. Atm

Solution :For a ISOTHERMAL PROCESS `DeltaE=0`
35.

One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surrounding (DeltaS_("surr")) in JK^(-1) is (1 L atm = 101.3 J)

Answer»

5.763
1.013
`-1.013`
`-5.763`

SOLUTION :Isothermal PROCESS, `DeltaU=0`
`dq=-dW=P_(ext)(V_(2)-V_(1))=3L-atm=3xx101.3" JOULE"`
`DeltaS_("SURROUNDING")=-(3xx101.3)/(300)" Joule "K^(-1)=-1.013 " Joule "K^(-1)`
`therefore DeltaS_("SURR")=-1.013" Joule "K^(-1)`
36.

One mole of an ideal gas at 300 K expands isothermally and reversibly from 5 to 20 litres. Calculate the work done and heat absorbed by the gas.

Answer»


ANSWER :(-832 KJ, 832 kJ)
37.

One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (DeltaS_("max"))" in "JK^(-1) is

Answer»

5.763
1.013
`-1.013`
`-5.763`

ANSWER :C
38.

One mole of an ideal gas at 22.4 litres is expanded isothermally and reversibly at 300 K to a volume of 224 litres at a constant pressure. Calculate W,q,DeltaH, DeltaG and Delta S.

Answer»


ANSWER :`(-5.74 KJ, 5.74 kJ, 0,-5.74 kJ, +19.1 J/K)`
39.

One mole of an ideal at 300 K is expanded isothermaly from an initial volume of 1 liter to 10 liters. The triangleE for this process is [ R= 2 cal k^-1 mol^-1]

Answer»

163.7 cal
1381.1 cal
91 H-atm
zero

Answer :B
40.

One mole of an element contains 4.2xx10^(24) electrons. What is the atomic number of the element?

Answer»


ANSWER :7
41.

one mole of an anydrose salt AB dissolves in water with the evolution of 21.0mol^(-1)of heat. If the heat of hydration of AB is -29.4 J mol^(-1) , then the heat of dissociation of hydrated salt AB is

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` 50.4 Jmol ^(-1)`
`8.4 Jmol^(-1)`
`-50.4 Jmol ^(-1)`
`-8.4Jmol^(-1)`

Solution :`(i)AB(s)+(AQ)toAB(aq),DELTAH=-21 J mol^(-1)`
`(ii) AB(s)+xH_(2)O to AB.xH_(2)O(s),DeltaH=-29.4 J mol ^(-1)`
REQUIRED equation is `AB.xh_(2)O(s)+ (aq) to AB(aq), DeltaH=?`
EQ.(i) is equivalent to
` AB(s)+ xH_(2)O to AB.x H_(2)O(s), DeltaH=DeltaH_(1)`
`AB.xH_(2)O(s) + (aq)to AB(aq), DeltaH=DeltaH_(2)`
DeltaH_(1) + DeltaH_(2)=-21`
` -29.4 + DeltaH_(2)=-21 or DeltaH_(2) =8.4 mol^(-1)`
42.

One mole of an amine (A) consumes two moles of methyl bromide to give a quaternary ammonium salt. The amine (A) is

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`(CH_(3))_(3)C CH_(2)NH_(2)`
`(CH_(3))_(2)NCH_(2)CH_(3)`

Answer :C
43.

One mole of ammonia was completely absorbed in one litre solution each of (a) 1 M HCI, (b) 1 MCH_3COOH and (c) 1 M H_2SO_4 at 298 K. The decreasing order for the pH of the resulting solution is (Given K_(b) (NH_3) = 4.74)

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`b GT c gt a `
`a gt b gt c `
`b gt a gt c `
`c gt b gt a `

Solution :a) `NH_3 + HCL to NH_4Cl`
1M of `NH_3` completely reacts with 1M HCl. The SALT formed is ACIDIC
b) `CH_3COOH + NH_3 to CH_3COO NH_4`. The salt formed is neutral
c) `2NH_3 + H_2SO_4 to (NH_4)_2 SO_4 `. 1 M of `NH_3` need 0.5 M of `H_2SO_4 `, 0.5 M still REMAIN in solution.
The salt is acidic.
44.

One mole of alkene on ozonolysis gives 2 moles of butanone. The alkene is :

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3, 4-dimethylhex - 2- ENE
2, 3 - dimetylhex - 3- ene
3, 4 -dimethylhex - 3- ene
2, 3- dimethylhex - 2- ene

Answer :B
45.

One mole of acetal on complete hydrolysis gives

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MOLE of aldyhyde,1 mole of alcohol
1 mole of aldehyde ,2 MOLES of alcohol
2 moles of aldehyde ,1 mole of alcohol
2 moles of aldehyde and 2 moles of alcohol

Answer :B
46.

One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44u. The alkene is:

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Ethene
Propene
1-butene
2-butene

Solution :MOLECULAR MASS of ALDEHYDE is 44 U so it is `CH_3CHO`
47.

One mole of a prefectputthroughta cycleconsitingofthefollowingthreereversiblesteps :(CA) Isothermal comoressionfrom 2atm and10litres to20 atmandlitre.(AB ) Isobaric expansion to returnthe gasto the origanlvolume of 10litrewithT going fromT_(1) to T_(2)(BC)Coolingat constantvolume to bringthegastotheoriganlpressurelt brgtandtemperature .The stepsare shownschematicallyin the figureshown. CalculateT_(1) to T_(2).(b) CalculateDeltaU, qandwill be in calories , foreachstep andfor thecycle .

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Solution :(a)`""(PV)/(R)= (20 XX1)/(0.0821) = 243.6 K`
`T_(2) = (20 xx 10 )/(0.021) = 2436.0 K`
(b)CA(Isothermal )
`W_(1) = - 2.303 xx 2 xx 243.6 log .(2)/(20)= 1122`
BC(Isochoric )
`W_(2) = 0`
AB(Isochoric)
`W_(3) = - RDeltaT = - 4384.9 cal`
For complete CYCLE
`W = W_(1) + W_(2) + W_(3) = 3262. 888 cal`
48.

One mole of a symmetrical alkene on hydrolyses gives two moles of an aldehyde having a molecular mas of 44 u. The alkene is

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2-butene
ethene
propene
1-butene.

Solution :The aldehyde with molecular MASS will 44u is `CH_(3)CHO` (ACETALDEHYDE) `therefore` symmetrical ALKENE is 2-butane
`underset("2-butene")(CH_(3)CH=CHCH_(3))OVERSET((i)O_(3))underset((ii)Zn//H_(2)O)rarrundersetunderset("Mol. mass=44")("Acetaldehyde")(2CH_(3)CHO)`
49.

One mole of a perfect gas expands isothermally to ten times of its original volume. The change in entropy is

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0.1 R
2.303 R
10.0 R
100.0 R

Solution :We KNOW that, `DeltaS=2.303" NR log "(V_(2))/(V_(1))`
`therefore DeltaS=2.303R" log"(10)/(1)`
`DeltaS=2.303R""(because log10=1)`
50.

One "mole" of a nonideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) to (4 atm, 5L, 245 K) with a change in internal energy, Deltau=30.0L atm. The change in enthalpy, DeltaH, of the process in L. atm is

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<P>`40.0`
`42.3`
`44.0`
not DEFINED , because pressure is not constant

SOLUTION :`DeltaH=Deltau+(p_(2)V_(2)-p_(1)V_(1))`