Question

Expand 1.00 mol of a monatomic gas, initially at 3.60 kPa and 313 K, from initial volume V_i = 0.723 m³ to final volume V_f = 2.70 m³. At any instant during the expansion, the pressure p and volume V of the gas are related by p = 3.60 exp[(V_i - V)/a], with p in kilopascals, V_i and V are in cubic meters, and a = 2.10 m³. What are the final (a) pressure and (b)temperature of the gas? (c) How much work is done by the gas during the expansion? (d) What is the change in entropy of the gas for the expansion? (Hint: Use two simple reversible processes to find the entropy change.)

Answer 1

(a)

Since pressure and and volume at any instant of the process are related as.

P = 3.60 kPa ∙ exp( (Vi - V)/a )

or

p = pi ∙ exp( (Vi - V)/a )

we get the final pressure by applying the final volume: :

pf = pi ∙ exp( (Vi - Vf)/a )

= 3.6 kPa ∙ exp( (0.723 m³ - 2.70m³) / 2.10 m³ )

= 1.40 kPa

(b)

Assuming ideal gas behavior:

pf∙Vf = n∙R∙Tf

=>

Tf = pf∙Vf / (n∙R)

= 1.40 ×10³ Pa ∙ 2.70 m³ / (1.00 mol ∙ 8.3144 J∙K⁻¹∙mol⁻¹)

= 456 K

c)

Generally work done by the gas in a reversible process is given by the integral

W = ∫ p dV from initial to final volume

With the pressure volume for the process the work integral for this process becomes::

...... Vf

W = ∫ pi ∙ exp( (Vi - V)/a ) dV

...... Vi

With the anti-derivative

∫ pi ∙ exp( (Vi - V)/a ) dV = - a pi exp( (Vi - V)/a )

you get the solution:

W = - a∙pi∙exp( (Vi - Vf)/a ) + a∙pi∙exp( (Vi - Vi)/a )

Since

exp( (Vi - Vi)/a ) = exp(0) = 1

and

pi∙exp( (Vi - Vf)/a ) = pf (as shown in part (a) )

you may rewrite:

W = a ∙(pi - pf)

= 2.10 m³ ∙ (3.6 kPa – 1.4 kPa)

= 4.62 kPa∙m³

= 4.62 kJ

(d)

The entropy of the gas is state function, that means the change in entropy is independent from the process path. So we may consider a different path between state (pi,Vi) and state (pf;Vf). Let's break up the process into two steps:

step 1: change from Vi to Vf at constant pressure pi

step 2: change from pi to pf at constant volume Vf

The change in entropy can be found from heat transferred:

dS = dQ/T

Step1 :For an ideal gas undergoing constant volume process:

dS = dU/T = n∙Cv∙dT/T

=>

∆S₁ = ∫ dS = ∫ n∙Cv/T dT from initial temperature Ti₁ to final temperature Tf₁

with the antiderivative

∫ n∙Cv/T dT = n∙Cv∙ln(T)

you get:

∆S₁ = n∙Cv∙(ln(Tf₁) - ln(Ti₁) ) = n∙Cv∙ln(Tf₁/Ti₁)

Using ideal gas law you can express the temperature ratio in terms of initial and final pressure Since volume in is constant it follows from ideal gas law:

p/T = n∙R/V = constant

Hence,

pi/Ti₁ = pf/Tf₁∙

=>

TF₁/Ti₁ = pf/pi

=>

∆S₁ = n∙Cv∙ln(pf/pi)

Step2 :For an ideal gas undergoing constant pressure process:

dS = dU/T = n∙Cp∙dT/T

=>

∆S₂ = ∫ dS = ∫ n∙Cp/T dT from initial temperature Ti₂ to final temperature Tf₂

= n∙Cp∙(ln(Tf₂) - ln(Ti₂) ) = n∙Cp∙ln(Tf₂/Ti₂)

Using ideal gas law you can express the temperature ratio in terms of initial and final volume Since pressure in is constant it follows from ideal gas law:

V/T = n∙R/p = constant

Hence,

Vi/Ti₂ = Vf/Tf₂∙

=>

Tf₂/Ti₂ = Vf/Vi

=>

∆S₂ = n∙Cp∙ln(Vf/Vi)

The change in entropy for whole two step process is sum of the changes for each step:

∆S = ∆S₁ + ∆S₂ = n∙Cv∙ln(pf/pi) + n∙Cp∙ln(Vf/Vi)

= n∙( Cv∙ln(pf/pi) + Cp∙ln(Vf/Vi) )

The molar heat capacities for a monatomic iideal gas are:

Cv = (3/2)∙R

Cp = (5/2)∙R

So the change in entropy for this process is:

∆S = n∙R∙( (3/2)∙ln(pf/pi) + (5/2)∙ln(Vf/Vi) )

= 1.00 mol ∙ 8.3144 J∙K⁻¹∙mol⁻¹ ∙ ( (3/2)∙ln(1.4kPa/3.6kPa) + (5/2)∙ln(2.70m³/0.723m3) )

= 15.6 J∙K⁻¹

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