Introduction:
In this blog, I give past years questions of chemical thermodynamics and energetic chapter.A well as definition of important terms for entrance exam. This is very helpful Mht Cet .for more questions and chapter mcq plase follow this blog.
Past years questions of previous chapter
(2) solid state
*Chemical Thermodynamics & Energetics
Shortcuts, Important Results & Formulae:
Thermodynamie Termns:
Branch of science which deals with the study of different forms of energy and the quantiative
relationships between them.
*System
Specified part of the universe which is under investigation.
*Surroundings
Remaining part of the universe which can interact with the system.
*Open system : Can exchange both matter and energy.
*Closed system: Can exchange energy but not matter.
*Isolated system: Neither matter nor energy can be exchanged.
Chemical apparatus.
(1) For a particular reaction, system absorbs 6 KJ, of heat and does 1.5 KJ or work on its
surrounding. What is enthalpy change
[Mht cet 2019]
(A) 1.5 K
(B)+4.5KJ
(C) +6.0 KJ
(D)+7.5 KJ
Ans:
Use first law of thermodynamics
q = +6 kJ
W = -1.5 kJ (work is done on its surrounding so negative)
Now,
∆U = q + W = 6 - 1.5
∆U = +4.5 kJ
that's why option (b) is right.
(2) Which of the following properties is extensive ? [Mht cet 2019]
(A) Volume
(B) Density
(D) Boiling Point
(C) Melting point
Ans:
A property of a system, whose magnitude depends upon the amount of matter, is known as extensive property. Properties like mass, volume, internal energy, heat content, free energy, enthalpy, entropy, heat capacity, surface area, energy, etc. are dependent upon the mass of the system and are called extensive properties.
that's why option (a) is right.
(3) A gas performs 0.320 kJ work on surrounding and absorbs 120 J of heat from the surrounding
Hence change in internal energy is
[MHT-CET 2019]
(A)440 J
(B) 120.32 J
(C) 200J
(D)-200J
Ans:
option (d) is right.
(4) Consider the following reaction Hag+Cho2HCl+44 Kcal. Calculate heat of formation
for 36.5 g of HCI
[MHT-CET 2019]
(A)-44 K Cal
(C)-22 K Cal
(B)-88 K Cal
(D)11 K Cal
Ans:
H2(g)+Cl2(g)→2HCl(g)+44kcalmol−1
Heat of formation =ΔHf=−44kcalmol−1
that's why option (a) is right.
(5) For a reaction to be non-spontaneous at al temperatures, values of ∆H and ∆S respectively are
(A)negative, negative
(B) positive, positive
(C)negative, positive
(D)positive, negative
[MHT-CET 2019]
Ans:
ΔH and ΔG should be negative.
For a process that occurs at constant temperature and pressure, spontaneity can be determined using the change in Gibbs free energy, which is given by:
ΔG=ΔH−TΔS
where the sign of ΔG depends on the signs of the changes in enthalpy (ΔH) and entropy (ΔS), as well as on the absolute temperature (T). The sign of ΔG will change from positive to negative (or vice versa) where T=ΔSΔH.
In cases where ΔG is:
negative, the process is spontaneous and may proceed in the forward direction as written.positive, the process is non-spontaneous as written, but it may proceed spontaneously in the reverse direction.zero, the process is at equilibrium, with no net change taking place over time.
This set of rules can be used to determine four distinct cases by examining the signs of the ΔS and ΔH.
When ΔS > 0 and ΔH < 0, the process is always spontaneous as written.
When ΔS < 0 and ΔH > 0, the process is never spontaneous, but the reverse process is always spontaneous.
When ΔS > 0 and ΔH > 0, the process will be spontaneous at high temperatures and non-spontaneous at low temperatures.
When ΔS < 0 and ΔH < 0, the process will be spontaneous at low temperatures and non-spontaneous at high temperatures.
For the latter two cases, the temperature at which the spontaneity changes will be determined by the relative magnitudes of ΔS and ΔH.
that's why option (a) is right.
(6) Which among the following equations represents the first law of thermodynamics under isobarnie
conditions [MHT-CET 2018]
(A) AU - Pa.AV
(B) q, AU
(C)AU W
(D) W-q
Ans:
The internal energy change ΔU=qp+W
ΔU=qp+(−Pex⋅ΔV) (∵W=−Pex⋅ΔV)
ΔU=qp−Pex⋅ΔV
that's why option (a) is right.
(7) Two moles of an ideal gas are allowed to expand from a volume of 10 dm' to 2m at 300
against a pressure of 101.325 KPa. Calculate the work done.
[MHT-CET 2018]
(A)-201.6 k
(B) 13.22 kJ
(C)-810.6 J
(D)-18.96 kJ
Ans:
ΔV=V2−V1=2 m3−(10 dm3×10−3 m3/dm3)=1.99 m3
P=101.325 kPa=101.325×103 Pa
W=−PΔV
W=−101.325×103 Pa×1.99 m3
W=−201.6×103 J
W=−201.6 kJ
that's why option (a) is right.
(8) Calculate the work done during combustion of 0.138 kg of ethanol, CH,OH) at 300 K.
Given: R 8.314 Jk mol, molar mass of ethanol 46 g mol". [MHT-CET 2018]
(A)-7482 JJ
(B) 7482 J
(C)-2494 J
(D)2494 J
Ans:
C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)
0.138 kg of ethanol =46 g/mol0.138 kg×1000 g/kg=3 mol
3C2H5OH+9O2→6CO2+9H2O
Δn=6−9=−3
Work W=−ΔnRT
W=−(−3)×8.314×300
W=7482 J
that's why option (b) is right.
(9) The work done during combustion of 9 x 10 Kg of ethane, CaHs (g) at 300 Kis
(Given R-8.314 J deg mof, atomic mass C 12, H 1). [MHT-CET 2017]
(A)6.236 k
(B)18.71 kJ
(C)-6.236 kJ
(D)-18.71 k
Ans:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
C2H6(g)+3.5O2(g)→2CO2(g)+3H2O(l)
Δn=2−(1+3.5)=−2.5
Molar mass of ethane =2(12)+6(1)=30 g/mol
9×10−2kg ethane =9×10−2kg×1000g/kg=90g
90 g ethane =30g/mol90g=3mol ethane.
For 1 mole of ethane, Δn=−2.5
For 3 mole of ethane, Δn=−2.5×3=−7.5
Work done ,W=−ΔnRT
W=(−7.5)×8.314J/mol/K×300K
W=18707J
W=18.71kJ
that's why option (d) is right.
(10) Calculate the work done during compression of 2 mol of an ideal gas from a volume of Im to
10 dm at 300 K against a pressure of 100 KPa. [MHT-CET 2017]
(A)-99 k
(B)+99 kJ
(C)+22.98 k
(D)-22.98 kJ
Ans:
The expression for work done is
W=−2.303nRTlogV1V2
W=−2.303×2mol×8.314J/mol/K×300×log1m3×1000dm3/m310dm3
W=+22977J
W=+22.98kJ (∵ 1 kJ = 1000
thats why option C is right
(11) The first law of thermodynamics for isothermal process is
[MHT-CET 2017]
(A)q-W
(B)AU W
(C)AU q
(D)AU-q
Ans:
According to the first law of thermodynamics
ΔU=q+W
For an isothermal process, ΔU=0
Hence,
q+W=0 or
q=−W
that's why option (d) is right.
(12) Mathematical equation of first law of thermodynamics for isochoric process is. [Mht cet 2016]
(A) AU
(B)q-W
(C)∆U -qv
(D)-AU-
Ans:
The first law of thermodynamics says,
ΔU=q+w
In an isochoric process, the work done is 0 since there is no change in volume so no work is done.
So, ΔU=qv
that's why option (c) is right.
(13) The criterion for a spontaneous process is. [MHT-CET 2016]
(A) AG0
(B) AG<0
(D) ASa0
(C)AG-0
Ans:
For a spontaneous process to occur, the free energy should always decrease from its initial value.
So, ΔG<0.
that's why option (b) is right.
(14) Identify an extensive property amongst the following [Mht cet 2016]
(A) Viscosity
(B) Heat capacity
(C) Density
(D) Surface tension
Ans:
Intensive properties do not depend on the amount of matter that is present.
Extensive properties do depend on the amount of matter that is present.
Viscosity, density and surface tension do not depend on mass or amount of matter, hence they are intensive properties.
Heat capacity (the amount of heat usually expressed in calories, kilocalories, or joules needed to raise the system's temperature by one degree ) is an extensive property.
that's why option (b) is right.
(15) What is the amount of work done when 0.5 mole of methane, CH4(g), is subjected to combustion
at 300 K ? (given, R- 8.314 J/Kmol)
[MHT-CET 2016]
(A)-2494J
(B)-4988 J
(C)+4988 J
(D)+2494J
Ans:
The balanced chemical equation for combustion reaction is
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
When 1 mole of methane undergoes combustion
Δn=1−(1+2)=−2
When 0.5 mole of methane undergoes combustion
Δn=−2×0.5=−1
At STP (1 atm and 273 K) , 1 mole of a gas occupies a volume of 22.4 L.
The volume change ΔV=22.4L×Δn×273K300K
ΔV=−24.6L
The work done W=−PΔV
W=−1atm×(−24.6L)
W=+24.6Latm
1 L atm =−101.33 J
W=−24.6Latm×−101.33J/Latm=+2494J
Hence, the amount of work done is +2494 J.
that's why option (d) is right
(16) Given R-8.314 JK mol the work done during combustion of 0.090 kg of ethane. [MHT-CET 2015](molar mass 30) at 300 K is
(A)6.234 k
(B)-6.234 k
(C)- 18.7kJ
(D) 18.7 k
Ans:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l) or
C2H6+7/2O2→2CO2+3H2O
Change in number of moles of gas is
Δn=(2−(1+7/2))=−2.5
Now,
Work done, W=−PΔV=ΔnRT
W=−ΔnRT=−2.5×8.314×300
W=−(−2.5×8.314×300)=6235.5 J
For one mole combustion, we get work done =6235.5 J
So, for 0.090 kg =0.090×1000/30 moles =3 moles
For 3 moles the work done =3×6235.5 J =18706.5=18.7 kJ
that's why option (d) is right
(17) Which among the following is a feature of adiabatic expansion ?
[Mht cet 2015]
(A)AT 0
(B)AU>0
(C)AU<0
(D)AV<0
Ans:
For an adiabatic expansion ΔT=0.
For an adiabatic free expansion of an ideal gas, the gas is contained in an insulated container and then allowed to expand in a vacuum. Because there is no external pressure for the gas to expand against, the work done by or on the system is zero. Since this process does not involve any heat transfer or work, the First Law of Thermodynamics then implies that the net internal energy change of the system is zero. For an ideal gas, the temperature remains constant because the internal energy only depends on the temperature in that case.
that's why option (a) is right
(18) What is the amount of work done when two moles of ideal gas is compressed from a volume of
Im to 10 dm' at 300 K against a pressure of 100 kPa? [MHT-CET 2015]
(A)-114.9 kJ
(B)99 k
(C)114.9 kJ
(D)-99 k
Ans:
V1=1m3,
Pressure =100 kPa =100000 Pa
V2=10dm3=(1/100)m3.
So, W=PΔV=100000×(V2−V1)
=105×(1/100−1)
=99000J=99 kJ.
that's why option (b) is right.
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(2)solid state
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 years questions for entrance exam){kind=link}
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