JEE Main20198 Apr 2019Evening ShiftChemistryThermodynamics (C)Actual
5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K . If C V = 28 J K - 1 , calculate ∆ U and ∆ p V for the process. ( R = 8.0 J K - 1 m o l - 1 )
Options
- A∆ U = 14 J ; ∆ p V = 0.8 J
- B∆ U = 14   kJ ;   ∆ p V = 18   kJ
- C∆ U = 14   kJ ;   ∆ p V = 4   kJ
- D∆ U = 2.8   kJ ;   ∆ p V = 0.8   kJ
Correct answer
C. ∆ U = 14   kJ ;   ∆ p V = 4   kJ
Step-by-step solution
Δ U = n C V Δ T = 5 × 28 × 100 = 14000   J = 14   kJ Δ ( P V ) = n R Δ T = 5 × 8 × 100 = 4000   J = 4   k J