Question

Compressed gases aren't ideal. Let's consider a gas that's non-ideal only because the volume available to each of the N molecules is reduced because each other molecule occupies volume v. Instead of pV=NkT, we get: p(V-Nb)=NkT. Let b=7 × 10^-29 m3. Let's look at 3 moles of this gas at T=300K starting in 0.001 m3 volume.

1) What's the initial value of the pressure? p_initial = ?

2) The gas expands isothermally to 0.002 m3. What's the final pressure? p_final = ?

3) How much work did the gas do in this isothermal expansion? W = ?

Answer 1

Answer:

8563732.58906 Pa

3992793.23326 Pa

5708.00923 J

Explanation:

V = Volume

N = Number of molecules = $3\times 6.023\times 10^{23}$

T = Temperature = 300 K

b = $7\times 10^{-29}\ m^3$

$k_$ = Boltzmann constant = $1.38\times 10^{-23}\ J/K$

P = Pressure

We have the equation

$P(V-Nb)=NkT\\\Rightarrow P=\dfrac{NkT}{V-Nb}\\\Rightarrow P=\dfrac{3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P=8563732.58906\ Pa$

The pressure is 8563732.58906 Pa

For isothermal expansion

$P_1(V_1-Nb)=P_2(V_2-Nb)\\\Rightarrow P_2=\dfrac{P_1(V_1-Nb)}{V_2-Nb}\\\Rightarrow P_2=\dfrac{8563732.58906(0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29})}{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P_2=3992793.23326\ Pa$

The pressure is 3992793.23326 Pa

Work done is given by

$dw=Pdv\\\Rightarrow W=\int_{v_1}^{v_2}\dfrac{NkT}{V-Nb}dv\\\Rightarrow W=NkTln\dfrac{V_2-Nb}{V_1-Nb}\\\Rightarrow W=3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300ln\dfrac{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow W=5708.00923\ J$

The work done is 5708.00923 J

Answer 2

The initial pressure, final pressure and work done in the isothermal expansion are respectively; 8563.733 kPa, 3992.793 kPa and 5708 J

What is the pressure in isothermal expansion?

1) The formula to find the initial pressure is;

P = NkT/(V - Nb)

where;

N is number of molecules = 3 * 6.023 * 10²³

T is temperature = 300 K

b = 7 × 10⁻²⁹ m³

V is volume = 0.001 m³

k is  Boltzmann constant  = 1.38 * 10⁻²³ J/K

Thus;

P = (3 * 6.023 * 10²³ * 1.38 * 10⁻²³ * 300)/(0.001 - (3 * 6.023 * 10²³ * 1.38 * 10⁻²³))

P_initial = 8563.733 kPa

2) The formula to get the final pressure in Isothermal Expansion is;

P_final = (P_initial * (V₁ - Nb))/(V₂ - Nb)

P_final = (8563.733 * (0.001 - (3 * 6.023 * 10²³ * 1.38 * 10⁻²³))/(0.002 - (3 * 6.023 * 10²³ * 1.38 * 10⁻²³))

P_final = 3992.793 kPa

3) The workdone is gotten from the formula;

W = NkT In [(V₂ - Nb)/(V₁ - Nb)]

Plugging in the relevant values and solving gives us;

W = 5708 J

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