# Find gas pressures of injection knowing mixture pressure without knowing volume

Let's say we have a gas 1 moving inside a pipe (we don't know the geometry) at mass flow rate $$Q_1$$ and temperature $$T_1.$$ We inject a gas 2 at some point inside the pipe with gas 1 at mass flow rate $$Q_2$$ and at temperature $$T_2.$$

We know that the resulting mixture is moving at the mass flow rate $$Q_\mathrm{mix} = Q_1 + Q_2$$ at the temperature $$T_\mathrm{mix}$$ and is at pressure $$p_\mathrm{mix}.$$ All gases are ideal gases.

Is it possible to find the gas 1 pressure $$p_1$$ and the gas 2 pressure $$p_2?$$ And if so, what is the relation?

So far I haven't found any clues to solve this problem. I thought about partial pressure, but I feel wrong to say that $$p_1$$ and $$p_2$$ are respectively equal the partial pressures of gas 1 and gas 2 in the gas mixture because there is no mention of volume in this problem.

• Can parameter $\ce{M_w}$ molecular weight of the gas used to answer this question? – Avon97 Mar 3 at 11:58
• @Avon97 Yes it can be used. But I don't see where I can use molecular weight of each gases to answer the question – Jonses Mar 3 at 12:10

Of course it is a pipe and it has a infinite volume hypotactically,

To use ideal gas equations is a bit tricky with mass flow rate need to convert to molar flow rate of each gas thus need to divide by $$\ce{M_w}$$ molecular weight respective to each gas.

Consider $$\ce{1}$$ time unit where inside of pipe gets

$$\frac{Q_1}{M_{w_1}}$$ $$\frac{Q_2}{M_{w_2}}$$

moles of gas to one volume section of pipe

At one time unit according to the ideal gas equation

$$\ce{PV=nRT}$$

Gas 1 will consume $$\ce{V1}$$ amount of volume $$V_1 = \frac{Q_1RT_1}{M_{w_1}p_1}$$ Gas 2 will consume $$\ce{V2}$$ amount of volume $$V_2 = \frac{Q_2RT_2}{M_{w_2}p_2}$$

Total of $$\ce{V}$$ volume per each time unit $$V = V_1 + V_2$$

But $$\ce{V}$$ is just a section of the pipe it just shifts to what ever direction the gas is flowing and keeps going for ever in the infinite pipe.

But as further the gases go away just like you said turbulence will stabilize and final temperature and pressure inside pipe will be almost equal to $$\ce{T_{mix}}$$ and $$\ce{p_{mix}}$$

When it is stabilized,

Gas 1 will consume $$\ce{V'_1}$$ amount of volume $$V'_1 = \frac{Q_1RT_{mix}}{M_{w_1}p'_1}$$ Gas 2 will consume $$\ce{V'_2}$$ amount of volume $$V'_2 = \frac{Q_2RT_{mix}}{M_{w_2}p'_2}$$

Total of $$\ce{V'}$$ volume per each time unit $$V' = V'_1 + V'_2$$

The important factor here is that number moles of gas will shift to next stage will be the same thus molar fraction for each gas is a constant. Think as $$\ce{V'}$$ stage in pipe

Molar fraction can be easily calculated

$$\ce{X_1 = \frac{(\frac{Q1}{M_{w_1}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})}}$$ $$\ce{X_2 = \frac{(\frac{Q2}{M_{w_2}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})}}$$

Using the partial pressure equation

$$\ce{P_A = X_A P_{tot}}$$

what is the relation?

$$\ce{p'_1 = \frac{(\frac{Q1}{M_{w_1}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})} p_{mix}}$$ $$\ce{p'_2 = \frac{(\frac{Q2}{M_{w_2}})}{(\frac{Q1}{M_{w_1}})+(\frac{Q2}{M_{w_2}})} p_{mix}}$$

There you have it final answer but consider that its one time unit thus take only the magnitude with respective unit of mass flow rate * unit time $$\ce{|Q1| |Q2|}$$ symbol ` is there to show gases are stabilized.

• Thank you. So does it mean that $p_1$ and $p_2$ I am looking for are equal to $p'_1$ and to $p'_2$ (or at least very close) ? – Jonses Mar 3 at 18:41
• Well that depends on $\ce{T_{1} and T_{2}}$ difference the closer the values are faster the partial pressure of gas gets to $\ce{p_{1} p_{2}}$ you mean stable value yes – Avon97 Mar 4 at 11:03
• So for instance, if $T_1 =T_2$, then we can say $p_1=p'_1$ and $p_2 = p'_2$ . But when $T_1 >> T_2$ it is impossible to find $p_1$ and $p_2$. But what if we consider both $X_2 << X_1$ and $T_1>> T_2$ ? Can we say in that specific case that $p_1=p'_1$ and $p_2 = p'_2$ ? – Jonses Mar 4 at 12:04
• Yes if you say $\ce{ X_{2} \ll X_{1}}$ there is hardly any gas 2 in the pipe so yes $\ce{p1 = p1^' and p2 = p2^'}$ also $\ce{T_1 \approx T_{mix} p_1 \approx p_{mix}}$ – Avon97 Mar 4 at 13:20