Thermodynamics Question with detailed solutions| CSIR NET, GATE
Thermodynamics-I
Thermodynamics Question and Answer
Hello everyone today we will discuss the Thermodynamics Question series with detailed solutions for CSIR NET, GATE, IIT JAM aspirants. We have to briefly discuss and solve the most common and most difficult questions. That is helpful for an entrance examination aspirant.
How to solve Thermodynamics Problem?
Thermodynamics is the most popular and most versatile subject for any sciences student such as engineering or general B.sc and M.sc Students. Most topic of thermodynamics has been cleared in class 11 and 12 standards. And also, be cleared in B.sc and M.sc stream. So that is sufficient for CSIR NET and GATE relative examination.
You must solve the previous years' Question paper of the CSIR NET and GATE. You follow the question pattern of this examination, which is helpful for you.
How to solve the thermodynamics Question without wasting any time?
First of all, clear your concept of the thermodynamics chapter one by one. And more practice question answers using previous years' question series.
In examination Holl first, you follow and solve the most common and identical questions of the given question paper. Then solve the other difficult questions carefully.
What are the best books for Thermodynamics?
The best books for the Thermodynamics chapter are Principles of Physical Chemistry (Puri, Sharma, Pathiana), Physical chemistry (Atkins), etc. you can follow those books for entrance examinations like CSIR NET, GATE, IIT JAM, BARC, TIFR, NTPC, DRDO, etc. Use as a textbook.
Solve question with details solution:
Q.1. ΔH of a reaction is equal to the slope of the plot of
1. ΔG versus (1/T)
2. ΔG versus T
3. (ΔG/T) versus T
4. (ΔG/T) versus (1/T)
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| Thermodynamics-Q.1 |
Answer: 4. (ΔG/T) versus (1/T)
Hints: We know that ΔG = ΔH - TΔS ……. (i)
We can follow this equation y = MX + c, where M = slope of the grape and c = intersect point when plotting the y vs x grape.
Similarly, from equation (i) divided both sites by T, we get
(ΔG/T) = (ΔH/T) - ΔS
Then plot the (ΔG/T) vs (1/T) grape we get the slope = ΔH, and intersecting point = - ΔS.
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| plot the (ΔG/T) vs (1/T) grape |
Q.2. From the below Carnot cycle undergone by an ideal gas, identify the processes in which the change in internal energy is NON-ZERO.
(A) I and II
(B) II and IV
(C) II and III
(D) I and IV
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| Thermodynamics-Q.2 |
Answer: (B) II and IV
Hints: In the isothermal process Internal energy change is zero, ΔU=0, due to Temperature being constant. And in the adiabatic process of internal energy change, ΔU is non-zero.
We know that in the adiabatic process heat is zero (q=0), but ΔU is not equal to zero.
That is why the path of II and IV are adiabatic processes and ΔU is non-zero.
Q.3. Which one of the following defines the absolute temperature of a system?
A.
B.
C.
D.
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| Thermodynamics-Q.3 |
Answer: A. (dU/dS)V
Hints: we know that at constant pressure (P = constant) dS = dH/T and we get the absolute temperature of a system at constant P, is T = (dH/dS)P .
Similarly, at constant volume (V = constant) dS = dU/T and we get the absolute temperature of a system at constant V, is T = (dU/dS)V .
So, we connect the second one is Tabsolute = (dU/dS)V .
Q.4. Which of the following properties are characteristic of an ideal solution?
(i) (ΔmixG)T,P is negative
(ii) (ΔmixS)T,P is positive
(iii) (ΔmixV)T,P is positive
(iv) (ΔmixH)T,P is negative
(A) (i) and (iv) (B) (i) and (ii) (C) (i) and (iii) (D) (iii) and (iv)
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| Thermodynamics-Q.4 |
Answer: (B) (i) and (ii)
Hints: For an ideal solution the condition is following ΔG = -ve. So, we know that Gibbs Free energy ΔG = -TΔS. when ΔS =+ve then we get ΔG =-ve. That is why follow those conditions at Constant Temperature and Pressure….
(i) (ΔmixG)T,P is negative
(ii) (ΔmixS)T,P is positive
Q.5. The concentration of a reactant R varies with time for two different reactions as shown in the following plots:
The orders of these two reactions I and II, respectively, are
1. zero and one 2. one and zero
3. zero and two 4. two and zero
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| Thermodynamics-Q.5 |
Answer: 3. zero and two
Hints:
Q.6. The value of ΔU - ΔH for the reaction is
1.-3RT 2. +3RT
3. +RT 4. -RT
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| Thermodynamics-Q.6 |
Answer: 1. - 3RT
Hints: Give reaction we get Δn = n (gaseous product) – n (gaseous reactant).
∆n = (3 + 0) – (0 + 0) = +3
We know that ΔH = ΔU + ΔnRT
ΔU - ΔH = - ΔnRT = -3RT = - 3 x 8.314 J.K-1.mol-1 x 298K = - 7.432 KJ. At temperature 25oC.
Q.7. If the pressure p (system) is greater than the p (surroundings), then
1. work is done on the system by the surroundings
2. work is done on the surroundings by the system
3. work done on the system by the surroundings is equal to the work done on the surroundings by the system
4. internal energy of the system increases
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| Thermodynamics-Q.7 |
Answer: 2. work is done on the surroundings by the system.
Hints: If the pressure p (system) is greater than the p (surroundings), then work is done on the surroundings by the system (W = -ve).
If the pressure p (system) is less than the p (surroundings), then work is done on the system by the surroundings (W = +ve).
If the pressure p (system) is equal to the p (surroundings), then work done on the system by the surroundings is equal to the work done on the surroundings by the system (W = 0).
Q.8. Which one of the following plots represents an acceptable wavefunction?
A.
B.
C.
D.
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| Thermodynamics-Q.8 |
Answer: D.
Hints: For the condition of an acceptable wavefunction (ψ) are
- Single value
- Finite value
- Continuous
- Quadratically integrable over the integral (-∞,∞).
- dψ/dx is again continuous.
In case we can see that grape (A) and (C) are not continuous, this is not an acceptable wavefunction.
In case we can see that grape (B) has not any finite value, this is not an acceptable wavefunction.
In case we can see that grape (D) has a finite value, continuous, single value, Quadratically integrable over the integral, dψ/dx is again continuous. So, this is an acceptable wave function.
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