Question

Calculate q, w, ∆U and ∆H if 1.65 mol of an ideal gas with Cv = R 3/2
undergoes a reversible
adiabatic expansion from an initial volume Vi = 7.75 m3
to a final volume Vf = 20.5 m3
. The initial
temperature is 300 K

Answer 1

W = P$_{external}$ ∆V

= 1 × (20.5-7.75) = 12.75J

∆U = n𝐶𝑣,𝑚∆T = 1.65 × 3/2 × 8.314 × 355k

= 7.305 × 10³J

q = ∆U - W = 7305 - 12.75

= 7292.13

= 7.29213 × 10³J