Question

In an adiabatic expansion of 2 moles of a gas, the initial pressure was 1.013 × 105 Pa, the initial volume was 22.4 L, the final pressure was 3.191 × 104 Pa and the final volume was 44.8 L. Find the work done by the gas on its surroundings. Taken `γ = 5/3`.

Answer 1

n = 2, P_i = 1.013 × 10⁵ Pa, P_f = 3.191 × 10⁴ Pa V_i = 22.4 L = 22.4 × 10⁻³ m³, V_f = 44.8 L = 44.8 × 10⁻³ m^3, `γ = 5/3`

The work done by the gas on its surroundings,

W = `(P_iV_i - P_fV_f)/(γ - 1)`

= `((1.013 × 10^5) (22.4 × 10^-3) - (3.191 × 10^4) (44.8 × 10^-3))/(5/3 - 1)`

= `(2269 - 1430)/(2/3)`

= `((839) (3))/2`

= 1258.5 J