Top Questions with detailed solutions| Thermodynamics-II|CSIR NET, GATE
Thermodynamics-II
Top Questions with detailed solutions| Thermodynamics-II
Hello everyone today we will discuss Thermodynamics questions with answer series-2. Today we will discuss and solve a few difficult and most common problems in the thermodynamics chapter with detailed solutions.
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Q.9. For a system subjected to only P-V work, entropy is given by
(i) – (dG/dT)P (ii) (dG/dP)T (iii) – (dA/dV)T (iv) – (dA/dT)V
(A) (i) and (ii)
(B) (i) and (iv)
(C) (i) only
(D) (ii) only
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| Thermodynamics Q.9 |
Answer: (B) (i) and (iv)
Hints: we know that a system subjected to only P-V work... dG = - TdS at constant Pressure P,
dS = - (dG/dT)p
and dA = - TdS at constant Volume V,
dS = - (dA/dT)v
follow those condition in (i) and (iv)
Q.10. On reacting 1.55 g of a diol with an excess of methyl magnesium iodide, 1.12 L (corrected to STP) of methane gas is liberated. The molecular mass (g mol-1) of the diol is ____________.
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| Thermodynamics Q.10 |
Answer: 62 g.mol-1
Hints:
Q.11. Of the following inequalities, the criterion/criteria for spontaneity of a chemical reaction is/are (i) (ΔG)T,P < 0 (ii) (ΔU)S,V > 0 (iii) (ΔS)U,V > 0
(A) (i) only
(B)(ii) only
(C) (i) and (ii)
(D) (i) and (iii)
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| Thermodynamics Q.11 |
Answer: (D) (i) and (iii)
Hints: At a spontaneity of a chemical reaction must follow those conditions (i) (ΔG)T, P < 0
(ii) (ΔU)S,V < 0
(iii) (ΔS)U,V > 0
Q.12. For an elementary bimolecular gas phase reaction, activation energy is 5.5 kJ mol-1. Enthalpy of activation, in kJ mol-1, at 300 K is _____________. (R = 8.314 J K-1 mol-1)
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| Thermodynamics Q.12 |
Answer: 0.512 KJ
Hints:
For an elementary bimolecular gas-phase reaction, Activation energy (Ea) and Enthalpy activation (ΔH#) relation is
E_a=〖ΔH〗^#+2RT
5.5×1000=〖ΔH〗^#+2×8.314×300
〖ΔH〗^#=5500-4988=512 J.〖mol〗^(-1)=0.512 KJ.〖mol〗^(-1)
Q.13. One mole of an ideal gas is compressed from 5 L to 2 L at a constant temperature. The change in entropy, in J K-1, of the gas is ____________. (R = 8.314 J K-1 mol-1)
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| Thermodynamics Q.13 |
Answer: -7.618 J.K-1
Hints: At constant temperature, ΔU = 0,
According to thermodynamics first low we get, ΔU = Q + W or 0 = Q + W or W = - Q.
ΔS=Q/T
ΔS=-W/T
ΔS= nRln V_2/V_1
ΔS= 1 mol ×8.314×ln 2/5
ΔS=-7.61 J/K
Q.14. The dependence of rate constant k on temperature T (in K) of a reaction is given by the expression
lnk=[(-5000K)/T]+10
The activation energy of the reaction (in KJ mol-1) is ____________.
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| Thermodynamics Q.14 |
Answer: 41.57 KJ
Hints:
lnk=[(-5000K)/T]+10………… (1)
Both sides differenced equation (1) with respect to T, we get
dlnk/dT=((-5000K))/T^2 …………(2)
According to Arrhenius equation …
k=〖Ae〗^(- E_a/RT)…………… (3)
Where A= Arrhenius constant and k =rate constant, Ea =Activation energy at T (in K) temperature.
Both sides take to log in this equation (3), we get
lnk=lnA-E_a/RT ………….. (4)
Both sides differenced equation (4) with respect to T, we get
dlnk/dT=0-E_a/(RT^2 )
dlnk/dT=-E_a/(RT^2 )……………. (5)
From eqution (2) and (5) we get,
-E_a/(RT^2 )=((-5000K))/T^2
E_a=5000×R
Activation energy E_a=5000×8.314=41570 J=41.57 KJ
Q.15. Consider an ideal gas of volume V at temperature T and pressure P. If the entropy of the gas is S, the partial derivative (dP/dS)V is equal to
(A) (dT/dP)S (B) (dT/dV)P (C) – (dT/dV)S (D) (dT/dS)P
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| Thermodynamics Q.15 |
Answer: (C) – (dT/dV)S
Hints: we know that (dP/dS)V = – (dT/dV)S
This equation easy to prove it
According to Maxwell Relation, we know that dU = TdS – PdV ………. (1)
If S is constant then dS = 0, equation (1) differenced concerning V, at S constant. We get
〖(dU/dV)〗_S=-P…………. (2)
Again, differenced equation (3) concerning S, we get
d/dS dU/dV=-〖(dP/dS)〗_V ………….. (3)
If S is constant then dV = 0, equation (1) differenced concerning S, at V constant. We get
〖(dU/dS)〗_V=T …………… (4)
Again, differenced equation (4) with respect to V, we get
d/dV dU/dS=〖(dT/dV)〗_S ………….. (5)
Now, from equation (3) and (5) we get,
d/dS dU/dV=d/dV dU/dS
-(dP/dS)_V=(dT/dV)_S
Prove it.
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