# 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 13 L, how many moles of gas must be injected into the container to maintain a constant temperature and pressure?

Apr 22, 2016

$= 6.67 m o l$

#### Explanation:

By The equation of state for real gas we know

$P V = n R T$

Where
P = Pressure, V = Volume, n = number of moles of gas , T = Absolute temperature of the gas and R = universal gas constant.
In our case P and T are constant in both the case

So $V \propto n$
Here ${V}_{1} = 3 L \mathmr{and} {V}_{2} = 13 L$
n_1=2mol and n_2 =?
${n}_{2} / {n}_{1} = {V}_{2} / {V}_{1} \implies {n}_{2} = {n}_{1} \times {V}_{2} / {V}_{1} = 2 \times \frac{13}{3} \approx 8.67 m o l$

So No of moles of gas injected $= {n}_{2} - {n}_{1} = 8.67 - 2 = 6.67 m o l$