Arrange the following in order of magnitude of work done by the system / on the system at constant temperature : (a) |mathrmw_mathrmreversible| for expansion in infinite stage. (b) |mathrmw_mathrmirreversible| for expansion in single stage. (c) |mathrmw_mathrmreversible| for compression in infinite stage. (d) |mathrmw_mathrmirreversible| for compression in single stage. Choose the correct answer from the options given below:

Solution & Explanation

### Related Formula w_textrev = -nRT lnleft(fracV_mathrmfV_mathrmiright) w_textirrev = -P_textext left(V_mathrmf - V_mathrmiright) ### Core Logic For isothermal reversible and irreversible steps: 1. **Reversible Path**: Since a reversible compression path retraces the exact coordinates of the reversible expansion path, the magnitudes of work are equal: |w_textrev, expansion| = |w_textrev, compression| implies a = c
P-V indicator diagrams comparing reversible and irreversible expansion/compression
P-V indicator diagrams comparing reversible and irreversible expansion/compression
2. **Isothermal Expansion**: Reversible work magnitude is the maximum possible work. Hence, for expansion: |w_textrev, expansion| > |w_textirrev, expansion| implies a > b
P-V indicator diagrams comparing reversible and irreversible expansion/compression
P-V indicator diagrams comparing reversible and irreversible expansion/compression
3. **Isothermal Compression**: Irreversible compression requires more work than reversible compression because of sudden pressure adjustments against the surroundings: |w_textirrev, compression| > |w_textrev, compression| implies d > c
P-V indicator diagrams comparing reversible and irreversible expansion/compression
P-V indicator diagrams comparing reversible and irreversible expansion/compression
P-V indicator diagrams comparing reversible and irreversible expansion/compression
P-V indicator diagrams comparing reversible and irreversible expansion/compression
### Step 1: Combine the inequalities Combining the results: - We have a = c - We have d > c - We have a > b This leads to the strict inequality sequence: d > c = a > b ### Pattern Recognition Thermodynamics Principle: Reversible expansion is the most efficient (gives maximum work magnitude), whereas reversible compression is the most efficient (requires minimum work magnitude). Single-stage irreversible compression is always the least efficient, demanding the absolute highest work input. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

Reference Study Guides

More Chemical Thermodynamics Previous-Year Questions — Page 7

Q78 jee_main_2024_29_jan_morning Spontaneity and Gibbs Free Energy
  • A. Delta G text is negative for a spontaneous reaction
  • B. Delta G text is positive for a spontaneous reaction
  • C. Delta G text is zero for a reversible reaction
  • D. Delta G text is positive for a non-spontaneous reaction

Solution

### Core Logic According to the second law of thermodynamics, at constant temperature and pressure, the change in Gibbs free energy (Delta G) dictates the spontaneity of a process. - If Delta G lt 0 (negative), the process is spontaneous. - If Delta G gt 0 (positive), the process is non-spontaneous. - If Delta G = 0, the system is in equilibrium (reversible process). Therefore, the statement "Delta G is positive for a spontaneous reaction" is factually incorrect. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q83 jee_main_2024_30_january_evening Hess's Law of Constant Heat Summation
Two reactions are given below: 2mathrmFe_(s) + frac32mathrmO_2(g) rightarrow mathrmFe_2mathrmO_3(s), Delta mathrmH^circ = -822 mathrmkJ/mol mathrmC_(s) + frac12mathrmO_2(g) rightarrow mathrmCO_(g), Delta mathrmH^circ = -110 mathrmkJ/mol Then enthalpy change for following reaction 3mathrmC_(s) + mathrmFe_2mathrmO_3(s) rightarrow 2mathrmFe_(s) + 3mathrmCO_(g)
Numerical Answer. Answer: 492 to 492

Solution

### Related Formula According to Hess's Law, the net enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which it can be divided. ### Core Logic Let the given reactions be: (1) 2mathrmFe_(s) + frac32mathrmO_2(g) rightarrow mathrmFe_2mathrmO_3(s), quad Delta H_1 = -822 \, mathrmkJ/mol (2) mathrmC_(s) + frac12mathrmO_2(g) rightarrow mathrmCO_(g), quad Delta H_2 = -110 \, mathrmkJ/mol Target Reaction (3): 3mathrmC_(s) + mathrmFe_2mathrmO_3(s) rightarrow 2mathrmFe_(s) + 3mathrmCO_(g), quad Delta H_3 = ? To construct the target reaction: - We need 3 mathrmCO_(g) on the product side, so we multiply reaction (2) by 3. - We need mathrmFe_2mathrmO_3(s) on the reactant side and 2 mathrmFe_(s) on the product side, so we reverse reaction (1). ### Step 1: Calculate Net Enthalpy Target Reaction (3) = 3 times (2) - (1) Delta H_3 = 3 times Delta H_2 - Delta H_1 Delta H_3 = 3(-110) - (-822) Delta H_3 = -330 + 822 = 492 \, mathrmkJ/mol ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q83 jee_main_2024_30_jan_morning Work Done in Cyclic Process
An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path Arightarrow Brightarrow Crightarrow A as shown in the diagram. The total work done in the process is ________ J.
Work Done in Cyclic Process diagram for Q83 - JEE Main 2024 Morning
The image is a graph of Volume (dm3) vs Pressure (kPa) showing a triangular cyclic process starting from A(10,10) to B(10,30) to C(30,10) and back to A.
Numerical Answer. Answer: 200 to 200

Solution

### Related Formula W_textcyclic = textArea enclosed in P-V graph ### Core Logic The work done in a cyclic process is equal to the magnitude of the area enclosed by the cycle on a Pressure-Volume graph. Note that the provided graph is Volume (V) on the y-axis versus Pressure (P) on the x-axis. The path A rightarrow B rightarrow C rightarrow A is traced in a clockwise direction on the V-P graph. Clockwise on a V-P graph corresponds to anti-clockwise on a standard P-V graph, meaning net expansion work is done by the gas, making it positive conventionally (or negative depending on chemistry sign convention, but magnitude is asked for). ### Step 1: Calculating Area The enclosed region is a right-angled triangle. Base of triangle on P-axis = 30 - 10 = 20 text kPa Height of triangle on V-axis = 30 - 10 = 20 text dm^3 textArea = frac12 times textbase times textheight textArea = frac12 times 20 times 20 = 200 text kPacdottextdm^3 ### Step 2: Unit conversion 1 text kPa = 10^3 text Pa 1 text dm^3 = 1 text Litre = 10^-3 text m^3 W = 200 times 10^3 text Pa times 10^-3 text m^3 W = 200 text J ### Pattern Recognition 1 text kPa cdot 1 text L = 1 text Joule. This direct conversion saves time without converting explicitly to standard SI units (Pa and m^3). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q90 jee_main_2024_31_jan_evening Work Done in Isothermal Reversible Expansion
If 5text moles of an ideal gas expands from 10text L to a volume of 100text L at 300text K under isothermal and reversible condition then work w, is -xtext J. The value of x is ________ (Given R = 8.314text J K^-1textmol^-1)
Numerical Answer. Answer: 28720 to 28721

Solution

### Related Formula W = -2.303 \, nRT log left( fracV_2V_1 right) ### Core Logic For an isothermal and reversible expansion of an ideal gas, work is done by the system on the surroundings, hence it is negative by IUPAC convention. Given: n = 5text moles R = 8.314text J K^-1textmol^-1 T = 300text K V_1 = 10text L V_2 = 100text L ### Step 1: Calculating Work Done W = -2.303 times 5 times 8.314 times 300 times logleft( frac10010 right) W = -2.303 times 5 times 8.314 times 300 times log(10) W = -2.303 times 12471 times 1 W = -28720.713text J ### Step 2: Final Formatting The question asks for work w = -xtext J. So x = 28720.713, which rounds to 28721. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q88 jee_main_2024_31_jan_morning Gibbs Free Energy and Equilibrium
Consider the following reaction at 298 K. frac32O_2(g) rightleftharpoons O_3(g). quad K_p = 2.47 times 10^-29 Delta_rG^ominus for the reaction is ________ kJ. (Given R = 8.314 text J K^-1 mol^-1)
Numerical Answer. Answer: 163 to 164

Solution

### Related Formula Delta_rG^ominus = -RT ln K_p ### Step 1: Calculation Delta_rG^ominus = -8.314 times 10^-3 text kJ K^-1 mol^-1 times 298 text K times ln(2.47 times 10^-29) = -8.314 times 10^-3 times 298 times (-65.87) = 163.19 text kJ ### Step 2: Nearest Integer Rounding 163.19 to the nearest integer gives 163. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

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