Question

300 mmol of perfect gas occupies 13 dm³ at 320 K. Calculate the work done in joules when the gas expands :

(a) Isothermally against a constant external pressure of 0.20 bar,

(b) Isothermal and reversible process,

(c) Into vacuum until the volume of gas is increased by 3 dm³. (R = 8.314 J mol^-1 K^-1)

Answer 1

Given :

Number of moles of a gas = n 

= 300 mmol 

= 0.3 mol

Initial volume = V₁ = 13 dm³

Increase in volume = ΔV = 3 dm³

Pressure = P_ex = 0.2 atm

Temperature = 320 K

(a) Expansion against constant pressure is an irreversible process.

∴ W = -P_ex × ΔV

= - 0.2 × 3

= - 0.6 dm³ bar

= - 0.6 × 100 J

= - 60 J

(b) For isothermal reversible process,

W_max = 2.303 nRT log₁₀\(\frac{V_2}{V_1}\)

Now, 

V₂ = V₁ + ΔV = 13 + 3 = 16 dm³

W_max = – 2.303 × 0.3 × 8.314 × 320 log₁₀\(\frac{16}{13}\)

= -165.4 J

(c) In vacuum,

P_ex = 0

∴ W = -P_ex × ΔV

= - 0 × 3

= 0