container of 3 m³ capacity contains 10 kg of $CO _2$ at 27°C. Estimate the pressure exerted by $CO _2$ by using :

(i) Perfect gas equation

(ii) Van der Waals’ equation

(iii) Beattie Bridgeman equation.

Question

container of 3 m³ capacity contains 10 kg of [latex]CO _2[/latex] at 27°C. Estimate the pressure exerted by [latex]CO _2[/latex] by using :

(i) Perfect gas equation

(ii) Van der Waals’ equation

(iii) Beattie Bridgeman equation.

Accepted Answer

Capacity of the container, V = 3 m³

Mass of [latex]CO _2[/latex], m = 10 kg

Temperature of [latex]CO _2[/latex], T = 27 + 273 = 300 K

Pressure exerted by [latex]CO _2[/latex], p :

(i) Using perfect gas equation :

Characteristic gas constant, [latex]R=\frac{R_0}{M}=\frac{8314}{44}=188.95[/latex] Nm/kg K (for [latex]CO _2[/latex])

Using perfect gas equation

[latex]p V=m R T[/latex]

∴ [latex]p =\frac{m R T}{V}=\frac{10 imes 188.95 imes 300}{3}[/latex]

= 188950 N/m² or 1.889 bar.

(ii) Using Van der Waals’ equation :

[latex]\left(p+\frac{a}{\bar{v}^2} ight)(\bar{v}-b)=R_0 T[/latex]

[latex]p=\frac{R_0 T}{\bar{v}-b}-\frac{a}{\bar{v}^2}[/latex]

From Table 8.1

For [latex]CO _2[/latex] : a = 362850 [latex]Nm ^4 /( kg - mol )^2[/latex]

b = 0.0423 m³/(kg-mol)

[latex]\bar{v}=[/latex] Molar specific volume [latex]=\frac{3 imes 44}{10}=13.2[/latex] m³/kg-mol

Now substituting the values in the above equation, we get

[latex]p=\frac{8314 imes 300}{13.2-0.0423}-\frac{362850}{(13.2)^2}[/latex]

= 189562 – 2082.5 = 187479.5 N/m² or 1.875 bar.

(iii) Using Beattie Bridgeman equation :

[latex]p=\frac{R_0 T(1-e)}{(\bar{v})^2}(\bar{v}+B)-\frac{A}{\bar{v}^2}[/latex]

where p = pressure, [latex]A=A_0\left(1-\frac{a}{\bar{v}} ight), B=B_0\left(1-\frac{b}{\bar{v}} ight) ext { and } e=\frac{c}{\bar{v} T^3}[/latex]

From Table 8.2 [latex]A_0=507.2836, a=0.07132[/latex]

[latex]B_0=0.10476, \quad b=0.07235[/latex]

[latex]C=66 imes 10^4[/latex]

∴ [latex]A=507.2836\left(1-\frac{0.07132}{13.2} ight)=504.5[/latex]

[latex]B=0.10476\left(1-\frac{0.07235}{13.2} ight)=0.1042[/latex]

[latex]C=\frac{66 imes 10^4}{13.2 imes(300)^3}=0.001852[/latex]

Now substituting the various values in the above equation, we get

[latex]p =\frac{8314 imes 300(1-0.001852)}{(13.2)^2}(13.2+0.1042)-\frac{504.5}{(13.2)^2}[/latex]

= [latex]190093-2.89 \simeq 1.9 imes 10^5[/latex] N/m² = 1.9 bar.

Table 8.1. Constants of Van der Waals’ Equation
S.No.	Substance	a [latex]Nm ^4 /( kg - mol )^2[/latex]	b [latex]m^3 / kg -m o l[/latex]
1	Hydrogen [latex]\left( H _2 ight)[/latex]	25105	0.0262
2	Oxygen [latex]\left( O _2 ight)[/latex]	139250	0.0314
3	Carbon dioxide [latex]\left( CO _2 ight)[/latex]	362850	0.0423
4	Helium (He)	3417620	0.0228
5	Air	135522	0.0362
6	Water [latex]\left( H _2 O ight)[/latex] vapour	551130	0.03
7	Mercury (Hg) vapour	2031940	0.0657

Table 8.2. Constants of the Beattie-Bridgeman Equation of State
Gas	[latex]A_0[/latex]	a	[latex]B_0[/latex]	b	[latex]c imes 10^{-4}[/latex]
Hydrogen [latex]\left( H _2 ight)[/latex]	20.0117	-0.00506	0.02096	-0.04359	0.0504
Oxygen [latex]\left( O _2 ight)[/latex]	151.0857	0.02562	0.04624	0.004208	4.80
Carbon dioxide [latex]\left( CO _2 ight)[/latex]	507.2836	0.07132	0.10476	0.07235	66.0
Helium (He)	2.1886	0.05984	0.014	0.0	0.004
Air	131.8441	0.01931	0.04611	-0.001101	4.34
Nitrogen	136.2315	0.02617	0.05046	-0.00691	4.20
Argon	130.7802	0.02328	0.03931	0.0	5.99

Question 8.10: container of 3 m³ capacity contains 10 kg of CO2 at 27°C. Es...

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