find the volume, work done, and heat?

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Apr 20th, 2015

Use ideal gas law. 
P∙V = n∙R∙T 
=> 
V = n∙R∙T / P 

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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