1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27 ° C . The work done is 3 kJ mol - 1 . The final temperature of the gas is _____ K (Nearest integer). Given C v = 20 J mol - 1 K - 1

Step-by-step solution

From first law of thermodynamics: ∆ U   =   q   +   W We know for adiabatic process, ∴ q = 0 ∴ ΔU = w ⇒ m   ×   C v   ×   ∆ T   =   w ⇒ 1 × 20 × T 2 - 300 = - 3000 ⇒ T 2 - 300 = - 150 ∴ T 2 = 150   K