Question

A container has a volume of `3 L` and holds `2 mol` of gas. If the container is expanded such that its new volume is `8 L`, how many moles of gas must be injected into the container to maintain a constant temperature and pressure?

Answer 1

The number of moles to be injected is `=3.33mol`

Explanation:

Apply Avogadro's Law

`"Volume"/"Number of moles"= " Constant"`

At constant temperature and pressure

`V_1/n_1=V_2/n_2`

The initial volume is `=3L`

The initial number of moles is `n_1=2mol`

The final volume is `V_2=8L`

The final number of moles is

`n_2=V_2/V_1*n_1=8/3*2=5.33mol`

The number of moles to be injected is

`=n_2-n_1=5.33-2=3.33mol`

Answer 2

`10/3 \ \text{moles`

Explanation:

Assume the gas is ideal

Let `P_1` & `T_1` be the initial pressure & temperature of given gas (assumed ideal) having initial volume `V_1=3\ L` & moles `n_1=2 \ \text{moles`

Now, from ideal gas equation,

`PV=nRT`

`\implies P_1V_1=n_1RT_1`

`P_1(3)=2RT_1\ .........(1)`

Let `x\ \text{moles` of same gas be injected to the container such that final volume is `V_2=8\ L` & final moles in container become `n_2=2+x` such that the final temperature & pressure remain same t.e. `T_2=T_1` & `P_2=P_1`

Now, from ideal gas equation,

`P_2V_2=n_2RT_2`

`P_1(8)=(x+2)RT_1\ .........(2)`

Dividing (2) by (1) as follows

`\frac{P_1(8)}{P_1(3)}=\frac{(x+2)RT_1}{2RT_1}`

`8/3=\frac{x+2}{2}`

`x+2=16/3`

`x=16/3-2`

`x=10/3`

Thus, `10/3 \ \text{moles` can be added to the container to maintain same temperature & pressure