For an adiabatic process, we have T1V1γ−1 = T2V2γ−1 For monatomic gas γ = 5/3, and it is given that initial volume V1 = 5.6 L and temperature T1. Therefore, T2 = T1(
V1
V2
)γ−1 = T1(
5.6
0.7
)(
5
3
)−1 = 8
2
3
×T1 For an adiabatic process, the work done is W =
nR
γ−1
(T2−T1) However, at STP, 1 mol of substance occupies 22.4 L. Therefore, n =