Question

Suppose 2.40 mol of an ideal gas of volume V1 = 4.00 m3 at T1 = 290 K is allowed to expand isothermally to V2 = 24.0 m3 at T2 = 290 K .

Part A

Determine the work done by the gas.

W=

Answer 1

(a)
Work done ON the gas is given by the integral:
W = - ∫ p dV from V₁ to V₂

We have a constant temperature process on an ideal gas, that means
p∙V = n∙R∙T = constant throughout the whole process.

Hence:
p∙V = p₁∙V₁
<=>
p = p₁∙V₁/V

Substitute this expression to the work integral:
W = - ∫ p₁∙V₁/V dV from V₁ to V₂
= - p₁∙V₁∙ ∫ 1/V dV from V₁ to V₂
= - p₁∙V₁ ∙ ln( V₂/V₁ )

since p₁∙V₁ = n∙R∙T₁
W = - n∙R∙T₁ ∙ ln( V₂/V₁ )
= - 2.40mol ∙ 8.3145J/molK ∙ 290K ∙ ln(24.0m³ / 4m³ )
= -10368.71J

So 10368.71 Joules of work are done BY the gas.