Question

A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically from 500 cm3 to 2.10 dm3. What is the work done by the gas in the units of J? Enter your answer with the sign ( + or -) to 3 significant figures. DO NOT include the units.

Answer 1

182 to 3 s.f

Step-by-step explanation:

Workdone for an adiabatic process is given as

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats. For carbon dioxide, γ = 1.28

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

K = P₁V₁ʸ

We need to calculate the P₁ using ideal gas equation

P₁V₁ = mRT₁

P₁ = (mRT₁/V₁)

m = 2.80 g = 0.0028 kg

R = 188.92 J/kg.K

T₁ = 27°C = 300 K

V₁ = 500 cm³ = 0.0005 m³

P₁ = (0.0028)(188.92)(300)/0.0005

P₁ = 317385.6 Pa

K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

V₁ = 0.0005 m³

V₂ = 2.10 dm³ = 0.002 m³

1 - γ = 1 - 1.28 = - 0.28

W =

18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)

W = -67.47 (5.698 - 8.4)

W = 182.3 = 182 to 3 s.f