Question

18m^3 of air having a pressure of 4 bar absolute and a temperature of 90 degrees Celsius are to be compressed together with 28m^3 of air having a pressure of 3 bar absolute and a temperature of 15 degrees Celsius, in a vessel whose volume is 22m^3. What will be the pemperature of the mixture if it's pressure is 7 bar absolute?

Answer 1

Answer on Question #64196, Physics / Molecular Physics | Thermodynamics

Question:

$18\mathrm{m}^3$ of air having a pressure of 4 bar absolute and a temperature of 90 degrees Celsius are to be compressed together with $28\mathrm{m}^3$ of air having a pressure of 3 bar absolute and a temperature of 15 degrees Celsius, in a vessel whose volume is $22\mathrm{m}^3$. What will be the temperature of the mixture if its pressure is 7 bar absolute?

Solution:

Let's treat air as an ideal gas. Then we may apply the ideal gas law:

$$PV = \frac{m}{M} RT,$$

where $P$ is the pressure of air, $V$ — volume, $m$ — mass, $M$ — molar mass, $R$ — gas constant (8.314 J·K$^{-1}$·mol$^{-1}$), and $T$ — absolute temperature.

We may write three equations:

$$P_1 V_1 = \frac{m_1}{M} RT_1 \quad (1)$$ $$P_2 V_2 = \frac{m_2}{M} RT_2 \quad (2)$$ $$P_{tot} V_{tot} = \frac{m_1 + m_2}{M} RT_{tot} \quad (3)$$

From equations (1) and (2) we may evaluate masses and substitute them into equation (3).

$$m_1 = \frac{P_1 V_1}{RT_1} M, \quad m_2 = \frac{P_2 V_2}{RT_2} M, \quad P_{tot} V_{tot} = \frac{\frac{P_1 V_1}{RT_1} M + \frac{P_2 V_2}{RT_2} M}{M} RT_{tot} = \left(\frac{P_1 V_1}{T_1} + \frac{P_2 V_2}{T_2}\right) T_{tot}$$

Therefore $T_{tot} = \frac{P_{tot} V_{tot}}{\frac{P_1 V_1}{T_1} + \frac{P_2 V_2}{T_2}}$.

$$\begin{array}{l} P_1 = 4 \text{ bar abs.} = 4 \cdot 10^5 \text{ Pa}, \quad P_2 = 3 \text{ bar abs.} = 3 \cdot 10^5 \text{ Pa}, \quad P_{tot} = 7 \text{ bar abs.} = 7 \cdot 10^5 \text{ Pa} \\ T_1 = 90{}^\circ \text{C} = 363 \text{ K}, \quad T_2 = 15{}^\circ \text{C} = 288 \text{ K} \\ V_1 = 18 \text{ m}^3, \quad V_2 = 28 \text{ m}^3, \quad V_{tot} = 22 \text{ m}^3 \\ T_{tot} = \frac{7 \cdot 10^5 \cdot 22}{\frac{4 \cdot 10^5 \cdot 18}{363} + \frac{3 \cdot 10^5 \cdot 28}{288}} = 314 \text{ K} = 41{}^\circ \text{C} \\ \end{array}$$

Answer:

$$41{}^\circ \text{C}$$

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