Question

In a deep sea expedition at depth of about $350\text{ft},$ a driver is subject to a pressure of approximately $10\text{atm}.$ A gas cylinder carried by him contains $51.2\text{g}$ of ${\text{O}}_2$ and $326.4\text{g}$ of $\text{He}$ and has a volume of $10\text{L}.$ What is the total pressure in the cylinder at $20{}^{\circ }\text{C}$ temperature?

• A. $100\text{atm}$
• B. $300\text{atm}$
• C. $200\text{atm}$
• D. $150\text{atm}$

Answer 1

The correct option is C $200\text{atm}$

Given,
$m_{He}=326.4\text{g};m_{O_2}=51.2\text{g};V_{cyl}=10\text{L}$

First we need to calculate number of moles of $\text{He}$ and${\text{O}}_2$

So,
$n_{He}=\frac{\text{mass}}{\text{molecular weight}}=\frac{326.4}{4.003}=81.5\text{mol}$

$n_{O_2}=\frac{51.2}{32}=1.6\text{mol}$

Using ideal gas law, the partial pressure can be calculated as,

$P_{He}=\frac{n_{He}RT}{V}$

$R=0.08205\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$

$\therefore P_{He}=\frac{81.5\times 0.08206\times 293}{10}=196\text{atm}$

$P_{O_2}=\frac{1.6\times 0.08205\times 293}{10}=3.85\text{atm}$

So, total pressure of the mixture,

$P_{tot}=P_{Hc}+P_{O_2}$

$=196+3.85=200\text{atm}$

Hence, option $\left(\text{C}\right)$ is correct.