Question

2 L of $SO_2$ gas at 760 mm Hg are transferred to 10 L flask containing oxygen at a particular temperature, the partial pressure of $SO_2$ in the flask is

• A. 63.33 mm Hg
• B. 152 mm Hg
• C. 760 mm Hg
• D. 1330 mm Hg

Answer 1

The Correct Option is B

$pSO_2 = \frac{P \times V}{V_{Total}} = \frac{760 \times 2000}{10000} $ = 152 mm Hg

Answer 2

Given:
- Volume of \(SO_2\) gas initially = 2 L
- Pressure of \(SO_2\) gas initially = 760 mm Hg
- Total volume of the flask after transfer = 10 L

Convert the initial pressure of \(SO_2\) gas to atm:
\(P_{SO_2} = 760 \text{ mm Hg} \times \frac{1 \text{ atm}}{760 \text{ mm Hg}} = 1 \text{ atm}\)

Apply Dalton's Law of Partial Pressures:
According to Dalton's Law, the total pressure in the flask \(P_{\text{total}}\) is the sum of the partial pressures of \(SO_2\) and \(O_2\):
\(P_{\text{total}} = P_{SO_2} + P_{\text{O}_2}\)

Calculate the partial pressure of \(SO_2\) in the flask:
Initially, \(SO_2\) gas was at 1atm and was transferred to a total volume of 10 L.
Since \(SO_2\) gas and \(O_2\) gas are ideal gases and assuming no chemical reaction or significant temperature change, the partial pressure of \(SO_2\) after transfer can be approximated by the ratio of its initial volume to the total volume:

\(P_{SO_2, \text{final}} = P_{SO_2} \times \frac{V_{SO_2}}{V_{\text{total}}}\)

\(P_{SO_2, \text{final}} = 1 \text{ atm} \times \frac{2 \text{ L}}{10 \text{ L}}\)

\(P_{SO_2, \text{final}} = 1 \text{ atm} \times 0.2\)
\(P_{SO_2, \text{final}} = 0.2 \text{ atm}\)

Convert the final pressure of \(SO_2\) gas to mm Hg (since the options are in mm Hg):
\(P_{SO_2, \text{final}} = 0.2 \text{ atm} \times 760 \text{ mm Hg/atm}\)

\(P_{SO_2, \text{final}} = 152 \text{ mm Hg}\)

So, the correct option is (B): \(152 \text{ mm Hg}\)