The work done (in Cal) in adiabatic compression of 2 moles of an ideal monatomic gas by the constant external pressure of 2 atm starting from an initial pressure of 1 atm and an initial temperature of 300 K is: [ R = 2 cal / mol − K ]

Step-by-step solution

To calculate the work done, we neet to know the final temperature. T 1 = 300 K , P 1 = 1 atm , n = 2 mole , P 2 = 2 atm , V 1 = nRT 1 P 1 w = Δ Y = nC v T 2 − T 1 = − P 2 V 2 − V 1 For a monatomic gas, C v = 3 2 R ⇒ 3 2 nR T 2 − T 1 = − P 2 V 2 − V 1 3 2 nR T 2 − T 1 = − P 2 nR T 2 P 2 − nR T 1 P 1 3 2 T 2 − T 1 = − T 2 + P 2 P 1 ⇒ T 2 = 2 3 P 2 P 1 + 3 2 T 1 T 2 = 0 .4 2 + 1 .5 300 = 420 K Work done, W = nC v ΔΤ = 2 × 3 2 R × 420 − 300 ⇒ W = 2 × 3 2 × 2 × 120 = 720 cal