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Thermodynamics

JEE Main 2025 (Online) 22nd January Morning ShiftMCQEasy📈 Topic in 13% papers+4−1
Ques. 1JEE Main 2025, 22 January (Morning Shift)

A liquid when kept inside a thermally insulated closed vessel at 25 was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ?

  1. U = 0, q < 0, w > 0
  2. U > 0, q = 0, w > 0
  3. U = 0, q = 0, w = 0
  4. U < 0, q = 0, w > 0

Answer

(2) U > 0, q = 0, w > 0

Solution

Insulated vessel q = 0; stirring does work on the liquid w > 0

U = q + w = w > 0

Answer: (2)

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Ques. 2JEE Main 2025, 22 January (Evening Shift)

Match List - I with List - II. Choose the correct answer from the options given below :

  1. A-I, B-II, C-IV, D-III
  2. A-II, B-III, C-I, D-IV
  3. A-II, B-I, C-IV, D-III
  4. A-II, B-I, C-III, D-IV
Ques. 3JEE Main 2025, 22 January (Evening Shift)

Consider the following cases of standard enthalpy of reaction ( Hr in kJ mol-1) C2H6(g)+ 72O2(g) 2CO2(g)+3H2O(l) H1 =-1550 C(graphite)+O2(g) CO2(g) H2 =-393.5 H2(g)+ 12O2(g) H2O(l) H3 =-286 The magnitude of H f C2H6(g) is kJ mol-1 (Nearest integer).

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