Question

I compress an ideal gas under a constant pressure of 3.2 × 105 Pa from 6.31 mL to 5.12 mL.

i. What is the work done in this process?

ii. If this is an adiabatic process, what is q and ∆U for this process?

Answer 1

Given pressure , P = 3.2x10⁵ Pa

                            = 3.2x10⁵ N/m²

Initial volume , V₁ = 6.31 mL

Final volume , V₂ = 5.12 mL

Change in volume , V = final - initial

                                    = 5.12-6.31 mL

                                    = -1.19 mL

                                    = -1.19x10^-3 L    Since 1 mL = 10 ^-3 L

                                    = -1.19x10^-3 x 10^-3 m³         Since 1 L = 10^-3 m³

                                    = -1.19x10^-6 m³

(1)Wordk done , W = PV

                        = (3.2x10⁵ N/m²)x (-1.19x10^-6 m³)

                        = -0.381 Nm

                        =- 0.381 J                          Since 1 J = 1 Nm

(2) In an adiabatic process heta energy remains constant so q = 0

From first law of thermodynamics , q = U + W

                                                   0 = U +(-0.381)

                                                 U = +0.381 J