Question

ofthepistonis1.0m3 ofairat400K,3bar.Ontheothersideis1.0m3 of air at 400 K, 1.5 bar. The piston is released and equilibrium is attained, with the piston experiencing no change of state. Employing the ideal gas model for the air, determine a. the final temperature of the air, in K. b. the final pressure of the air, in bar. c. the amount of entropy produced, in kJ/K.

Answer 1

The entropy indicates the amount of thermodynamic energy that is not

available for work.

a. The final temperature is 400 K

b. The final pressure is 2.25 bar

c. The amount entropy produced, Δs is approximately 0.063712 kJ/K

Reason:

The given parameters are;

Type of box = Insulated

Volumes into which the box is divided = Equal volumes

Type of piston = Frictionless, thermally conducting

Volume on each side of the piston = 1.0 m³

Temperature on each side of the piston, T₁ = T₂ = 400 K

Pressure on one side of the piston = 3.0 bar

Pressure on the other side of the piston = 1.5 bar

a. The final temperature of the air;

Given that the piston is frictionless and allowed move freely, and  air is

contained on both sides of the piston, we have;

The system is equivalent to the piston being removed such that the gas

moves freely in the box.

Therefore, the high pressured side drops in temperature while the low

pressure side gains temperature, such that the change in temperature is

zero.

Which gives; T₁ = T₂ = $T_f$

Therefore;

The final temperature, $\mathbf{T_f}$ = T₁ = T₂ = 400 K

b. The final pressure is given as follows;

We have;

$\dfrac{P_{1}\times V_{1}}{T_{1}} = \dfrac{P_{f}\times (V_{1} -V) }{T_{f}}$

$\dfrac{3\times 1}{400} = \dfrac{P_{f}\times (1+V) }{T_{f}}$

$\dfrac{P_{2}\times V_{2}}{T_{2}} = \dfrac{P_{f}\times (V_{2} +V) }{T_{f}}$

$\dfrac{1.5\times 1}{400} = \dfrac{P_{f}\times (1 -V) }{T_{f}}$

$\dfrac{1 + V}{1 - V} = 2$

Which gives;

$V = \dfrac{1}{3}$

$T_f$ = 400 K, gives;

$\dfrac{1.5\times 1}{400} = \dfrac{P_{f}\times \left (1 -\dfrac{1}{3} \right) }{400} = \dfrac{\dfrac{2}{3} \cdot P_f}{400}$

$1.5\times 1} =\dfrac{2}{3} \cdot P_f}$

$P_f = \dfrac{3 \times 1.5\times 1 }{2} = 2.25$

The final pressure, $\mathbf{P_f}$ = 2.25 bar

c. The amount of entropy produced, in kJ/K

The change in entropy is given by the formula;

$\Delta s = \Delta s_1 + \Delta s_1 = -R \cdot \dfrac{P_1 \cdot V_1}{R \cdot T_1} \cdot ln\dfrac{P_f}{P_1} + \left(-R \cdot \dfrac{P_2 \cdot V_2}{R \cdot T_2} \cdot ln\dfrac{P_f}{P_2} \right)$

$\Delta s = -\dfrac{P_1 \cdot \dfrac{V}{2} }{ T_f} \cdot ln\dfrac{P_f}{P_1} + \left(-\dfrac{P_2 \cdot \dfrac{V}{2} }{ T_f} \cdot ln\dfrac{P_f}{P_2} \right)$

Which gives;

$\Delta s = -\dfrac{3 \times \dfrac{2}{2} }{ 400} \cdot ln\dfrac{2.25}{3} + \left(-\dfrac{1.5 \cdot \dfrac{2}{2} }{ 400} \cdot ln\dfrac{2.25}{1.5} \right) \approx 6.371 \times 10^{-4}$

Δs = 6.371 × 10⁻⁴ bar·m³/K = 63.712 J/K = 0.063712 J/K

The entropy produced, Δs = 0.063712 kJ/K.

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