In adiabatic conditions, 2 mole of CO ₂ gas at 300 K is expanded such that its volume becomes 27 times. Then, the work done is ( C _ V =6 cal mol ⁻¹ . and . =1.33 )

Step-by-step solution

Given, T ₁=300 ~K , ~V ₂=27 ~V ₁, n =1 ~mol C_V=6 cal mol ⁻¹, =1.33 . In adiabatic conditions T ₂ ~T ₁ = ( V ₁ ~V ₂ )^ -1 or, T ₂ ~T ₁ = ( 1 27 )^ 1.33-1 = ( 1 27 )^ 0.33 = ( 1 27 )^ 1 / 3 = 1 3 or, T ₂=300 1 3 =100 ~K Thus, T ₂ T ₁ hence, cooling takes place due to expansion under adiabatic condition. E = q + W E W ( q =0 for adiabatic change ) E =- ve because gas expands. Then, W =- E =- C _ V ( T ₂- T ₁ )=-6(100-300)=1200 cal