find the volume, work done, and heat?
Physics

Tutor: None Selected  Time limit: 1 Day 
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)
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