What is the final pressure of helium and neon gas mixture in this case?

• MHB
• WMDhamnekar
In summary, the given answer for the final pressure is incorrect and the correct calculation takes into account the volumes of each gas using the formula $pV=nRT$. The final pressure should be $6.67\,\text{bar}$.
WMDhamnekar
MHB

The answer given to the above question is Final pressure$=\frac{(10.0mol\times 5.00bar) + (5.00 mol \times 20.0 bar)}{(10.0 mol +5.00 mol)}= 10.0 bar$ Is this answer correct? if yes, How and why?

My question is while computing this answer, the volumes of each gas is not considered.

Note:- While computing the volumes of each gas, temperature is not considered as it is a constant.The computed volume of helium gas $=\frac{(10.0mol \times 8.314472 J/mol\cdot K)}{5.00 bar}=0.16628944 Litres$

The computed volume of neon gas $= \frac{(5.00 mols \times 8.314472 J/mol\cdot K)}{20.0 bar}= 0.02078618 Litres$

How to use this additional information to compute the final pressure of the gas mixture?

We can treat noble gasses as ideal gasses, which means we can use the formula $p V = n R T$, which is independent of the actual type of gas.
Let the first container have $n_1$ moles, pressure $p_1$, and volume $V_1$.
Let the second container have $n_2$, $p_2$, and $V_2$.
Let the temperature be $T$, and let the final pressure be $p$.

Then we have:
\begin{cases}p_1V_1 = n_1 R T \\ p_2 V_2 = n_2 R T \\ p(V_1+V_2) = (n_1+n_2) R T \end{cases}
Since we do not know the volumes, we will eliminate them from the equations:
$$\begin{cases}V_1 = \frac{n_1 R T}{p_1} \\ V_2 = \frac{n_2 R T}{p_2} \\ p(V_1+V_2) = (n_1+n_2) R T \end{cases} \implies p = \frac{(n_1+n_2) R T}{V_1+V_2} = \frac{(n_1+n_2) R T}{\frac{n_1 R T}{p_1}+\frac{n_2 R T}{p_2}} =\frac{n_1+n_2}{\frac{n_1}{p_1}+\frac{n_2}{p_2}}= \frac{10.0+5.00}{\frac{10.0}{5.00}+\frac{5.00}{20.0}}=6.67\,\text{bar}$$

So it looks as if the given answer is incorrect.
The final pressure should be $p=6.67\,\text{bar}$.

1. What is the final pressure of the helium and neon gas mixture?

The final pressure of a mixture of gases is determined by the combined pressures of each gas, according to Dalton's Law of Partial Pressures. In this case, the final pressure would depend on the relative amounts of helium and neon present in the mixture.

2. How does the temperature affect the final pressure of the gas mixture?

The temperature of a gas directly affects its pressure, according to the Ideal Gas Law. As the temperature increases, the particles in the gas have more kinetic energy and collide with the container walls more frequently, resulting in a higher pressure. Therefore, the final pressure of the gas mixture would also be affected by the temperature at which it is measured.

3. Does the size of the container affect the final pressure of the gas mixture?

According to Boyle's Law, the volume of a gas is inversely proportional to its pressure at a constant temperature. Therefore, if the size of the container is changed, the final pressure of the gas mixture would also change. However, the relative amounts of helium and neon in the mixture would remain the same.

4. Can the final pressure of the gas mixture be calculated using the individual pressures of helium and neon?

No, the final pressure of a gas mixture cannot be calculated simply by adding the individual pressures of each gas. This is because the gases will interact with each other, leading to a different final pressure. The ideal gas law and Dalton's Law of Partial Pressures must be used to accurately calculate the final pressure of a gas mixture.

5. How can the final pressure of the gas mixture be measured experimentally?

The final pressure of a gas mixture can be measured using a variety of methods, such as a manometer, barometer, or pressure gauge. These instruments measure the pressure of the gas in the container and can be used to calculate the final pressure of the mixture. Additionally, the final pressure can also be determined by using the ideal gas law and measuring the volume, temperature, and number of moles of each gas in the mixture.

Replies
3
Views
3K
Replies
3
Views
1K
Replies
9
Views
353
Replies
2
Views
1K
Replies
12
Views
1K
Replies
3
Views
1K
Replies
5
Views
366
Replies
12
Views
891
Replies
4
Views
1K
Replies
2
Views
3K