One mole of a monatomic ideal gas at $300 K$ and $4.0 atm$ expands adiabatically against a constant external pressure of $1.0 atm$ until it reaches mechanical equilibrium (pressure $1.0 atm$ ). The final temperature is

Answer: B

Adiabatic:

Δ U = w

n C_{V} (T_{2} - T_{1}) = - P_{e x t} (V_{2} - V_{1}) = - R (T_{2} - T_{1} P_{2} / P_{1})

, so

1.5 (T_{2} - 300) = - (T_{2} - 75)

giving

T_{2} = (450 + 75) /2.5 = 210 K

. The reversible adiabatic result 119 K is lower because less work is done by the gas in the irreversible path.