Question

The following arrangement of flasks is set up. Assuming no temperature change, determine the final pressure inside the system after all stopcocks are opened. The connecting tube has zero volume.


$\begin{array}{cccc}V\left(L\right)& 4.0,& 3.0,& 5.0\\ p\left(\text{atm}\right)& 0.792,& 1.23,& 2.51\end{array}$

• A. $1.51\text{atm}$
• B. $4.53\text{atm}$
• C. $1.62\text{atm}$
• D. $3.24\text{atm}$

Answer 1

The correct option is C $1.62\text{atm}$

Before opening the stopcocks,
$n_{N_2}=\frac{p_1{V}_1}{RT};n_{Ar}=\frac{p_2{V}_2}{RT};n_{O_2}=\frac{p_3{V}_3}{RT}$
After opening the stopcocks,
$n_{total}=\sum \frac{pV}{RT}=\frac{p_1{V}_1+p_2{V}_2+p_3{V}_3}{RT}$
Total pressure $=\frac{n_{total}RT}{V_{total}}=\frac{p_1{V}_1+p_2{V}_2+p_3{V}_3}{V_{total}}$
$=\frac{0.792\times 4+1.23\times 3+2.51\times 5}{12}=1.62\text{atm}$
$\therefore \text{Final pressure inside the system}=1.62\text{atm}$
Hence, the correct answer is option (c).