Question:
A vessel containing \(10\) liters of an ideal gas at a pressure of \(760\text{ mm of Hg}\) is connected to an evacuated \(9\) liter vessel. The resultant pressure is
A vessel containing \(10\) liters of an ideal gas at a pressure of \(760\text{ mm of Hg}\) is connected to an evacuated \(9\) liter vessel. The resultant pressure is
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When gas expands into an evacuated vessel at constant temperature, use Boyle's law \(P_1V_1=P_2V_2\).
Updated On: May 7, 2026
- \(400\text{ mm of Hg}\)
- \(1440\text{ mm of Hg}\)
- \(40\text{ mm of Hg}\)
- \(760\text{ mm of Hg}\)
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The Correct Option is A
Solution and Explanation
Initial volume of gas:
\[
V_1=10\text{ L}.
\]
Initial pressure:
\[
P_1=760\text{ mm of Hg}.
\]
The gas is connected to an evacuated vessel of volume:
\[
9\text{ L}.
\]
So the final volume becomes:
\[
V_2=10+9=19\text{ L}.
\]
Since the temperature remains constant, Boyle's law applies:
\[
P_1V_1=P_2V_2.
\]
Substitute the values:
\[
760\times 10=P_2\times 19.
\]
\[
P_2=\frac{7600}{19}.
\]
\[
P_2=400\text{ mm of Hg}.
\]
Therefore, the resultant pressure is:
\[
400\text{ mm of Hg}.
\]
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