Question

A $3\text{L}$ container has $4\text{atm}$ pressure at $27{}^{\circ }\text{C}$. If $2\text{L}$ of gas at $2\text{atm}$ is added to this container at same temperature, then what will be the new pressure in the container?

• A. $4\text{atm}$
• B. $4.6\text{atm}$
• C. $5.3\text{atm}$
• D. $6.06\text{atm}$

Answer 1

The correct option is C $5.3\text{atm}$

Initially moles of gas at $4\text{atm}$
$n_1=\frac{PV}{RT}=\frac{4\times 3}{RT}...\left(1\right)$
Moles of gas in $2\text{L}$ container at $2\text{atm}$
$n_2=\frac{2\times 2}{RT}...\left(2\right)$
On adding total moles $=n_1+n_2$
$\frac{4\times 3}{RT}+\frac{2\times 2}{RT}$
New pressure in the container of volume $3\text{L}$
$P=\frac{nRT}{V}=\left[\frac{4\times 3+2\times 2}{RT\times 3}\right]RT=5.33\text{atm}$