One kg of CO2 has a volume of 1 m³ at 100°C. Compute the pressure by (i) Van der Waals’ equation (ii) Perfect gas equation.

Question

One kg of [latex]CO _2[/latex] has a volume of 1 m³ at 100°C. Compute the pressure by

(i) Van der Waals’ equation

(ii) Perfect gas equation.

Accepted Answer

(i) Using Van der Waals’ equation :

Molar specific volume, [latex]\bar{v}=1 imes 44=44[/latex] m³/kg-mol [latex]\left(\because M ext { for } CO _2=44 ight)[/latex]

Temperature, T = 100 + 273 = 373 K

The values of a and b for [latex]CO _2[/latex] (from Table 8.1)

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

and b = 0.0423 m³/kg-mol

[latex]R_0=8314[/latex] Nm/kg-mol K

Van der Waals’ equation is written as

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

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

Substituting the values in the above equation, we get

∴ [latex]p =\frac{8314 imes 373}{44-0.0423}-\frac{362850}{44^2}[/latex]

= 70548 – 187 = 70361 N/m² or 0.7036 bar. (Ans.)

(ii) Using perfect gas equation :

[latex]p \bar{v}=R_0 T[/latex]

∴ [latex]p =\frac{R_0 T}{\bar{v}}=\frac{8314 imes 373}{44}= 7 0 4 8 0[/latex] N/m² or 0.7048 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

Question 8.9: One kg of CO2 has a volume of 1 m³ at 100°C. Compute the pre...