Final volume is: V₂ = n∙R∙T / P₂ = 4.20 mol ∙ 8.3145 J∙mol⁻¹∙K⁻¹ ∙ 295 K / 145×10³ Pa = 7.1×10⁻² m³ = 71 L

(b) Assuming reversible process the work done by the gas is given by the integral W = ∫ P dV from V₁ to V₂ (i.e. initial to final volume) Using ideal gas law you can express P in terms of V: P = n∙R∙T / V Hence, W = ∫ n∙R∙T / V dV from V₁ to V₂ Since n, R and T are constant for our process: W = n∙R∙T ∙ ∫ 1/V dV from V₁ to V₂ = n∙R∙T∙ln(V₂/V₁)

Using ideal gas law you can substitute volume ratio by a pressure ratio: P∙V = n∙R∙T = constant => P₁∙V₁ = P₂∙V₂ => V₂/V₁ = P₁/P₂

So the isothermal work is given by: W = n∙R∙T∙ln(P₁/P₂) = 4.20 mol ∙ 8.3145 J∙mol⁻¹∙K⁻¹ ∙ 295 K ∙ (101 kPa / 145 kPa) = - 7175.6 J (minus sign indicates that work is done ON the gas)

(c) Change in internal energy of the gas equals heat added to it mus work done by it: ΔU = Q - W

Assuming ideal gas behavior, th internal energy of the gas is a function of temperature alone: U = n∙Cv∙T => ΔU = n∙Cv∙ΔT That means there is no change in internal energy for constant temperature process on an ideal gas: ΔU = Q - W

Hence, Q = W = - 7175.6 J (minus sign indicates that heats is removed from the gas)

Apr 20th, 2015

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