Question

A $8.3\text{L}$ cylinder contains $128.0\text{g}$ of $O_2$ gas at ${27}^{\circ }\text{C}$. What mass of $O_2$ gas must be released to reduce the pressure to 6.0 bar?

• A. $32.0\text{g}$
• B. $64.0\text{g}$
• C. $16.0\text{g}$
• D. $8.0\text{g}$

Answer 1

The correct option is B $64.0\text{g}$

We know,
$p_1{V}_1=n_{1}R{T}_1$

$\Rightarrow P_1=\frac{n_{1}R{T}_1}{V_1}=\frac{128\times 0.083\times 300}{32\times 8.3}=12\text{bar}$

Since the volume is constant, at constant temperature,
$\frac{P_2}{P_1}=\frac{n_2}{n_1}$
$\Rightarrow \frac{6}{12}=\frac{n_2}{4}$
Hence, $n_2=2\text{moles}$

Thus, mass of $O_2$ released$=2\times 32\text{g}=64g$