Question

A vessel of 3 m^3 capacity contains a mixture of nitrogen and carbon dioxide , the analysis by volume showing equal quantities of each. The temperature ia 15'C and the total pressure is 3.5 bar. Determine the mass of each constituent.

Answer 1

Answer on Question #56577, Engineering / Mechanical Engineering

A vessel of $3\mathrm{m}^3$ capacity contains a mixture of nitrogen and carbon dioxide, the analysis by volume showing equal quantities of each. The temperature is $15^{\circ}\mathrm{C}$ and the total pressure is 3.5 bar. Determine the mass of each constituent.

Solution:

An ideal gas can be characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them may be deduced from kinetic theory and is called the ideal gas law:

$$PV = nRT$$

where $n$ = number of moles,

$$n = \frac{m}{M}$$

$R$ = universal gas constant = 8.3145 J/mol K

In our case,

$$P_1 = \frac{nRT}{V}$$

and

$$P_2 = \frac{nRT}{V}$$

Dalton's law of partial pressures states that the total pressure exerted by a mixture of gases is the sum of partial pressure of each individual gas present. Each gas is assumed to be an ideal gas.

$$P_{total} = P_1 + P_2$$

Hence,

$$P_{total} = P_1 + P_2 = 2P_1$$ $$P_1 = \frac{P_{total}}{2} = \frac{3.5}{2} = 1.75\ \mathrm{bar} = 1.75 \cdot 10^5\ \mathrm{Pa}$$

Thus,

$$\frac{m_{N_2}}{M_{N_2}} = \frac{P_1 V}{RT}$$ $$m_{N_2} = \frac{P_1 V}{RT} M_{N_2}$$

$M_{N_2}$ = molar mass of nitrogen is 28.0134 g/mol.

$$m_{N_2} = \frac{1.75 \cdot 10^5 \cdot 3 \cdot 28.0134 \cdot 10^{-3}}{8.315 \cdot 288} = 6.14\ \mathrm{kg}$$

For carbon dioxide:

Molar mass of $\mathrm{CO_2} = 44.0095\ \mathrm{g/mol}$

$$m_{\mathrm{CO_2}} = \frac{P_1 V}{RT} M_{\mathrm{CO_2}}$$ $$m_{\mathrm{CO_2}} = \frac{1.75 \cdot 10^5 \cdot 3 \cdot 44.0095 \cdot 10^{-3}}{8.315 \cdot 288} = 9.65\ \mathrm{kg}$$

Answer: $m_{N_2} = 6.14\ \mathrm{kg}$; $m_{\mathrm{CO_2}} = 9.65\ \mathrm{kg}$.

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