10 mole of oxygen is heated at constant volume from 30^circtextC to 40^circtextC. The change in the internal energy of the gas is ____ cal. (The molecular specific heat of oxygen at constant pressure, C_p = 7text cal./mol ^circtextC and R = 2text cal./mol ^circtextC.)

Numerical Answer Type:
Enter a numerical value Answer: 500 to 500 +4 marks

Solution & Explanation

### Related Formula Delta U = n C_v Delta T C_v = C_p - R ### Core Logic Given values: n = 10text moles Delta T = 40^circtextC - 30^circtextC = 10^circtextC C_p = 7text cal/molcdot^circtextC R = 2text cal/molcdot^circtextC First, find C_v using Mayer's relation: C_v = C_p - R = 7 - 2 = 5text cal/molcdot^circtextC ### Step 1: Calculate Internal Energy Change Delta U = n C_v Delta T Delta U = 10 times (7 - 2) times (40 - 30) Delta U = 10 times 5 times 10 = 500text cal ### Pattern Recognition Change in internal energy of an ideal gas is ALWAYS Delta U = n C_v Delta T, regardless of the process (constant volume or not). Use C_v = C_p - R when C_p is given. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics Class 11 Physics: Kinetic Theory of Gases

Reference Study Guides

Thermodynamics

More Thermodynamics Previous-Year Questions — Page 3

Q1 jee_main_2025_03_april_morning Phase Change and Melting
During the melting of a slab of ice at 273mathrm~K at atmospheric pressure:
  • A. Internal energy of ice-water system remains unchanged.
  • B. Positive work is done by the ice-water system on the atmosphere.
  • C. Internal energy of the ice-water system decreases.
  • D. Positive work is done on the ice-water system by the atmosphere.

Solution

### Related Formula Delta U = Delta Q + Delta W_texton system where, Delta U = change in internal energy, Delta Q = heat exchange, Delta W_texton system = work done on the system. ### Core Logic During the melting of ice at 273mathrm~K, the density of water is higher than the density of ice. This means the volume of the ice-water system decreases during melting: V_f < V_i implies Delta V < 0 Since the system contracts, the atmosphere performs positive work on it: W_texton system = -P Delta V > 0 Additionally, heat is absorbed by the system to melt the ice, so Delta Q > 0. By the first law of thermodynamics, since both Delta Q and Delta W_texton system are positive, the internal energy of the system increases: Delta U = Delta Q + W_texton system > 0 ### Step 1: Evaluation of Options Let's check the given options: 1. Internal energy remains unchanged rightarrow False (it increases). 2. Positive work is done by the system rightarrow False (work done by the system is negative since it contracts). 3. Internal energy decreases rightarrow False. 4. Positive work is done on the ice-water system by the atmosphere rightarrow True (since volume decreases under atmospheric pressure). ### Pattern Recognition Remember: Ice contracts upon melting (unlike most solids). Shrinking volume (V downarrow) under positive pressure (P > 0) means the surroundings (atmosphere) compress it, performing positive work on the system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter Class 11 Physics: Thermodynamics
Q11 jee_main_2025_03_april_morning Isothermal Expansion with Non-Linear Spring
A piston of mass M is hung from a massless spring whose restoring force law goes as F = -kx^3, where k is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with 'n' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height L_0 to L_1, the total energy delivered by the filament is (Assume spring to be in its natural length before heating)
Piston connected to spring with gas underneath for Q11
A schematic of a piston of mass M connected to a spring inside a vertical chamber, separating gas at the bottom from vacuum/atmosphere at the top.
  • A. 3nRTln left(fracL_1L_0right) + 2Mg(L_1 - L_0) + frack3 (L_1^3 - L_0^3)
  • B. nRTln left(fracL_1^2L_0^2right) + fracMg2 (L_1 - L_0) + frack4 (L_1^4 - L_0^4)
  • C. nRTln left(fracL_1L_0right) + Mg(L_1 - L_0) + frack4 (L_1^4 - L_0^4)
  • D. nRTln left(fracL_1L_0right) + Mg(L_1 - L_0) + frac3k4 (L_1^4 - L_0^4)

Solution

### Related Formula First Law of Thermodynamics: Delta Q = Delta U + W_textby gas Work done by an ideal gas during isothermal expansion: W_textgas = nRTlnleft(fracV_1V_0right) = nRTlnleft(fracL_1L_0 ight) Conservation of Energy (Work-Energy Theorem): Total energy delivered by the heating filament (W_textfilament) must equal the total work needed to lift the piston against gravity and compress the non-linear spring. ### Core Logic Since the process is isothermal, the change in internal energy of the ideal gas is zero (Delta U = 0). Hence: Q = W_textgas By the Work-Energy Theorem for the piston: W_textgas + W_textfilament = Delta U_textgravity + Delta U_textspring Let's evaluate each term: - Increase in gravitational potential energy: Delta U_textgravity = Mg(L_1 - L_0) - Increase in spring potential energy: U_textspring = -int_L_0^L_1 F_textrestoring dx = int_L_0^L_1 kx^3 dx = frack4(L_1^4 - L_0^4) ### Step 1: Finding Total Energy Delivered Isolating W_textfilament (the net external energy delivered to the gas system): W_textfilament = W_textgas + Mg(L_1 - L_0) + frack4(L_1^4 - L_0^4) Since W_textgas = nRTlnleft(fracL_1L_0 ight): W_textfilament = nRTlnleft(fracL_1L_0 ight) + Mg(L_1 - L_0) + frack4(L_1^4 - L_0^4) ### Pattern Recognition Notice how energy conservation instantly frames this complex thermodynamics question. The heating filament's energy simply goes into three distinct stores: the isothermal work of gas expansion, raising the mass against gravity (Mgh), and the potential energy of the spring (integrated from kx^3). Keeping this total energy ledger in mind prevents tedious mathematical tangents. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics: First Law Class 11 Physics: Work, Energy and Power: Variable Force Integration
Q12 jee_main_2025_03_april_morning Adiabatic Compression
A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800~mathrmcm^3 and temperature 27^circmathrmC . The change in temperature when the gas is adiabatically compressed to 200~mathrmcm^3 is: (Take gamma = 1.5)
  • A. 327mathrm~K
  • B. 600mathrm~K
  • C. 522mathrm~K
  • D. 300mathrm~K

Solution

### Related Formula For an adiabatic process: T V^gamma - 1 = textconstant where, T = absolute temperature in Kelvin, V = volume of the gas, gamma = adiabatic exponent. ### Core Logic Given values: - Initial volume, V_1 = 800mathrm~cm^3 - Final volume, V_2 = 200mathrm~cm^3 - Initial temperature, T_1 = 27^circmathrmC = 27 + 273 = 300mathrm~K - Adiabatic exponent, gamma = 1.5 implies gamma - 1 = 0.5 ### Step 1: Calculating Final Temperature Apply the adiabatic relation: T_1 V_1^gamma - 1 = T_2 V_2^gamma - 1 T_2 = T_1 left(fracV_1V_2right)^gamma - 1 Substitute the values: T_2 = 300 left(frac800200right)^0.5 = 300 times (4)^0.5 T_2 = 300 times 2 = 600mathrm~K ### Step 2: Calculating Change in Temperature Now compute the change in temperature (Delta T): Delta T = T_2 - T_1 = 600mathrm~K - 300mathrm~K = 300mathrm~K ### Pattern Recognition Always read carefully to see if the question asks for the **final temperature** or the **change in temperature**. Many students lose marks by choosing 600mathrm~K (the final temperature) instead of the difference 300mathrm~K! Stay sharp. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics
Q17 jee_main_2025_04_april_evening Thermodynamic Processes
Match List-I with List-II:
List-IList-II
(A) Isobaric(I) Delta Q=Delta W
(B) Isochoric(II) Delta Q=Delta U
(C) Adiabatic(III) Delta Q=textzero
(D) Isothermal(IV) Delta Q=Delta U+PDelta V
Choose the correct answer from the options given below:
  • A. \text{(A)-(IV), (B)-(III), (C)-(II), (D)-(I)}
  • B. \text{(A)-(IV), (B)-(I), (C)-(III), (D)-(II)}
  • C. \text{(A)-(IV), (B)-(II), (C)-(III), (D)-(I)}
  • D. \text{(A)-(II), (B)-(IV), (C)-(III), (D)-(I)}

Solution

### Related Formula First Law of Thermodynamics: Delta Q = Delta U + Delta W ### Core Logic - **Isobaric**: Pressure is constant, work done Delta W = PDelta V. Thus, Delta Q = Delta U + PDelta V (Matches IV). - **Isochoric**: Volume is constant, Delta V = 0 implies Delta W = 0. Thus, Delta Q = Delta U (Matches II). - **Adiabatic**: No heat transfer, Delta Q = 0 (Matches III). - **Isothermal**: Temperature is constant, internal energy change Delta U = 0 for an ideal gas. Thus, Delta Q = Delta W (Matches I). ### Step 1: Match Compilation Combining everything gives: (A)-(IV), (B)-(II), (C)-(III), (D)-(I). ### Pattern Recognition Adiabatic definition is always heat-isolated (Q=0), isochoric implies rigid boundaries (W=0). Identifying these two immediately isolates the correct match permutation in seconds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics
Q17 jee_main_2025_07_april_evening Thermodynamic Processes
Match List-I with List-II.
List-IList-II
(A) Isothermal(I) Delta W (work done) =0
(B) Adiabatic(II) Delta Q (supplied heat) =0
(C) Isobaric(III) Delta U (change in internal energy) ne0
(D) Isochoric(IV) Delta U=0
Choose the correct answer from the options given below: [cite: 151, 152]
  • A. (A)-(III), (B)-(II), (C)-(I), (D)-(IV) [cite: 153]
  • B. (A)-(IV), (B)-(I), (C)-(III), (D)-(II) [cite: 154]
  • C. (A)-(IV), (B)-(II), (C)-(III), (D)-(I) [cite: 155]
  • D. (A)-(II), (B)-(IV), (C)-(I), (D)-(III) [cite: 156]

Solution

### Core Logic Let's evaluate each process condition based on the first law of thermodynamics: * **(A) Isothermal:** Continuous constant temperature (Delta T = 0) implies that the internal energy change of an ideal gas is zero, so Delta U = 0 implies text(IV) [cite: 730]. * **(B) Adiabatic:** No thermal energy transfer occurs between the system and surroundings, meaning Delta Q = 0 implies text(II) [cite: 731]. * **(C) Isobaric:** Constant pressure process where both volume and temperature typically vary, so internal energy changes continuously, Delta U neq 0 implies text(III) [cite: 732]. * **(D) Isochoric:** Rigid boundary condition at constant volume (Delta V = 0) ensures work done Delta W = PDelta V = 0 implies text(I) [cite: 733]. Putting these together yields: (A)-(IV), (B)-(II), (C)-(III), (D)-(I)[cite: 155, 729]. ### Pattern Recognition Matching 'Isochoric' with zero work done (Delta W=0) or 'Adiabatic' with zero heat exchange (Delta Q=0) are fundamental definitions that let you rapidly break down multi-choice grids[cite: 731, 733]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

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