One mole of a rigid diatomic ideal gas ( $γ = 1.4$ ) at $T_{1}$ and $P_{1}$ expands reversibly and adiabatically to twice its initial volume. Select all correct statements.

A $T_{2} \approx 0.76 T_{1}$
B $P_{2} \approx 0.379 P_{1}$
C $P_{2} = P_{1} /2$
D $P_{2}$ is lower than the final pressure after a reversible isothermal expansion to $2 V_{1}$

Answer: A, B, D

For a reversible adiabatic path

T V^{γ - 1}

and

P V^{γ}

are constant. So

T_{2} / T_{1} = 2^{- 0.4} = 0.758

and

P_{2} / P_{1} = 2^{- 1.4} = 0.379

. Isothermal expansion would give

P_{2} = 0.5 P_{1}

, which is higher than

0.379 P_{1}

, because the adiabatic gas also cools. So statements 1, 2 and 4 are correct and statement 3 is false.