NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
1 Mole of C O 2 gas at 300 K expanded under the reversible adiabatic condition such that its volume becomes 27 times. The magnitude of work done i n k J / m o l is: (Given γ = 1.33 and C v = 25.10 J m o l – 1 K – 1 f o r C O 2 ) report your answer by rounding it up to nearest whole number
Correct answer
5
Step-by-step solution
Number of moles = 1 T 1 = 300 K V 2 = 27 V 1 T V 1 γ - 1 = T 2 V 2 γ - 1 T 1 T 2 = V 2 V 1 r – 1 T 2 = 300 1 27 1 3 T 2 = 100 K Adiabatic condition, q = 0 ; Δ E = w = n C v T 2 – T 1 w = 1 × 25.10 100 – 300 = - 5020 J / m o l e w = - 5.02 k J / m o l e