Question

A container (volume V) has an ideal gas at 4 atm and 300 K. This is connected by a tube with another container of volume 2 V, which has a gas at 8 atm and 600 K. What will be final pressure of each container if the temperature is maintained - the container with volume V at 300 K and the 2V one at 600 K?

Answer 1

When the tubes are connected, pressures become the same.
Also, the total number of moles before and after the connection are the same.
$n_1=\frac{4V}{300R},n_2=\frac{16V}{600R}$, total moles = $n_1+n_2=\frac{V}{25R}$
When the tube is connected:
$P_1=P_2;\frac{n_1\times 300R}{V}=\frac{n_2\times 600R}{2V}$ i.e. $n_1=n_2=\frac{V}{50R}$
So, P = $\frac{\frac{V}{50R}\times 300R}{V}=6atm$