# How to find the final pressure excerted by a mixture of gasses which are then released from a vessel after succesive fillings?

The problem is as follows:

$$5\ \mathrm l$$ of oxygen at $$2\ \mathrm{atm}$$ of pressure and $$10\ \mathrm l$$ of nitrogen at $$4\ \mathrm{atm}$$ are collected in a vessel of $$2\ \mathrm l$$ in capacity. Then $$25\ \mathrm l$$ of the mixture is released from the vessel at the pressure of $$760\ \mathrm{mmHg}$$. Find the final pressure of the mixture if the temperature doesn't change.

What I've attempted to do to solve this problem was to use this equation from gasses. However, it seems tricky as there's no specific information of the temperature only it mentions that it doesn't change.

$$pV=nRT$$

The way how I thought to approach this problem was that if I could obtain the moles of each gas and add them both I could find the final pressure in the container.

Then as the temperature is held constant I could use this equation using Boyle's law to get the new pressure.

$$\frac{p_{1}V_{1}}{T_{1}}=\frac{p_{2}V_{2}}{T_{2}}$$

(Assuming $$T_1=T_2$$). However, I'm confused at exactly why was I given the pressure of $$760\ \mathrm{mmHg}$$. Is it to stress the fact that gas can escape from the vessel if the pressure outside is lower?

Well here's what I did:

For oxygen:

$$(2\ \mathrm{atm})(5\ \mathrm l)=\left(n_{\ce{O2}}\right)(0.082\ \mathrm{atm\ L \ mol^{-1}\ K^{-1}})(T_1)$$

$$n_{\ce{O2}}= \frac{121.9512}{T_1}$$

For nitrogen:

$$(4\ \mathrm{atm})(10\ \mathrm l)=\left(n_{\ce{N2}}\right)(0.082\ \mathrm{atm\ L\ mol^{-1}\ K^{-1}})(T_1)$$

$$n_{\ce{N2}}= \frac{487.8048}{T_1}$$

Then adding those moles together results into:

$$n_\text{total}= \frac{121.9512}{T_1} + \frac{487.8048}{T_1}$$

$$n_\text{total}= \frac{609.7560}{T_1}$$

Using this amount of moles, I am given the volume of the vessel hence I can know the pressure.

$$(p)(2\ \mathrm l)=\frac{609.7560}{T_1} (0.082\ \mathrm{atm\ L\ mol^{-1}\ K^{-1}})(T_1)$$

$$p= 25\ \mathrm{atm}$$

Finally this pressure can be used to get the new one using the equation shown above:

$$\frac{25\ \mathrm{atm} \cdot 2\ \mathrm l}{T_{1}}=\frac{p_2\cdot 25\ \mathrm l}{T_1}$$

Therefore I'm obtaining $$2\ \mathrm{atm}$$.

But this option doesn't appear in any of the alternatives for this given problem. What could I be overlooking, can someone help me?

• Comments are not for extended discussion; this conversation has been moved to chat. – user7951 Dec 24 '19 at 19:29