# 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?

Jul 25, 2018

The number of moles to be injected is $= 3.33 m o l$

#### Explanation:

$\text{Volume"/"Number of moles"= " Constant}$

At constant temperature and pressure

${V}_{1} / {n}_{1} = {V}_{2} / {n}_{2}$

The initial volume is $= 3 L$

The initial number of moles is ${n}_{1} = 2 m o l$

The final volume is ${V}_{2} = 8 L$

The final number of moles is

${n}_{2} = {V}_{2} / {V}_{1} \cdot {n}_{1} = \frac{8}{3} \cdot 2 = 5.33 m o l$

The number of moles to be injected is

$= {n}_{2} - {n}_{1} = 5.33 - 2 = 3.33 m o l$

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 ${V}_{1} = 3 \setminus L$ & moles n_1=2 \ \text{moles

Now, from ideal gas equation,

$P V = n R T$

$\setminus \implies {P}_{1} {V}_{1} = {n}_{1} R {T}_{1}$

${P}_{1} \left(3\right) = 2 R {T}_{1} \setminus \ldots \ldots \ldots \left(1\right)$

Let x\ \text{moles of same gas be injected to the container such that final volume is ${V}_{2} = 8 \setminus L$ & 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,

${P}_{2} {V}_{2} = {n}_{2} R {T}_{2}$

${P}_{1} \left(8\right) = \left(x + 2\right) R {T}_{1} \setminus \ldots \ldots \ldots \left(2\right)$

Dividing (2) by (1) as follows

$\setminus \frac{{P}_{1} \left(8\right)}{{P}_{1} \left(3\right)} = \setminus \frac{\left(x + 2\right) R {T}_{1}}{2 R {T}_{1}}$

$\frac{8}{3} = \setminus \frac{x + 2}{2}$

$x + 2 = \frac{16}{3}$

$x = \frac{16}{3} - 2$

$x = \frac{10}{3}$

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