Question

What will be the partial pressure of $\mathrm{He}\mathrm{and}{\mathrm{O}}_2$ respectively, if $200\mathrm{ml}$ of $\mathrm{He}$ at $0.66\mathrm{atm}$ and $400\mathrm{ml}$ of ${\mathrm{O}}_2$ at $0.52\mathrm{atm}$ pressure are mixed in a 400ml vessel at $20°\mathrm{C}$ (temperature).

Answer 1

Step 1: Boyle's law:

• “At constant temperature, the pressure of a fixed amount (i.e., number of moles n) of gas varies inversely with its volume. This is known as Boyle’s law.”
• $\mathrm{p}\propto \frac{1}{\mathrm{V}}$
• It is expressed as:

${\mathrm{p}}_1{\mathrm{V}}_1={\mathrm{p}}_2{\mathrm{V}}_2=\mathrm{constant}$

Step 2: Given Information:

• The partial pressure of $\mathrm{He}$ is $0.66\mathrm{atm}$.
• The partial pressure of ${\mathrm{O}}_2$ is $0.52\mathrm{atm}$.
• The initial volume of $\mathrm{He}$ is $200\mathrm{ml}$.
• The initial volume of ${\mathrm{O}}_2$ is $400\mathrm{ml}$.

Step 3: Calculation for partial pressure of $\mathrm{He}$:

$$\begin{gathered} {\mathrm{p}}_1{\mathrm{V}}_1={\mathrm{p}}_2{\mathrm{V}}_2 \\ 0.66\mathrm{atm}\times 200\mathrm{ml}={\mathrm{p}}_2\times 400\mathrm{ml} \\ {\mathrm{p}}_2=\frac{0.66\times 200}{400} \\ {\mathrm{p}}_2=0.33\mathrm{atm} \end{gathered}$$

Step 4: Calculation for partial pressure of ${\mathrm{O}}_2$:

$$\begin{gathered} {\mathrm{p}}_1{\mathrm{V}}_1={\mathrm{p}}_2{\mathrm{V}}_2 \\ 0.52\mathrm{atm}\times 400\mathrm{ml}={\mathrm{p}}_2\times 400\mathrm{ml} \\ {\mathrm{p}}_2=\frac{0.52\times 400}{400} \\ {\mathrm{p}}_2=0.52\mathrm{atm} \end{gathered}$$

Thus, the partial pressures of $\mathrm{He}$ and ${\mathrm{O}}_2$ are $0.33\mathrm{atm}\mathrm{and}0.52\mathrm{atm}$ respectively.