Question:

A gas occupies a volume of \(0.35\) dm\(^3\) at \(290\) K and \(2\) atm. Pressure. Calculate the volume of gas at STP.

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Use the combined gas law with STP of $273$ K and $1$ atm.
Updated On: Oct 1, 2026
  • \(3.11\) dm\(^3\)
  • \(0.66\) dm\(^3\)
  • \(0.11\) dm\(^3\)
  • \(3.5\) dm\(^3\)
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The Correct Option is B

Solution and Explanation

Step 1: Combined law
\(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\).

Step 2: Substitute
\(\frac{2\times0.35}{290}=\frac{1\times V_2}{273}\).
\(V_2=\frac{0.7\times273}{290}=0.659\approx0.66\) dm\(^3\). Option (B).

Final Answer:
The volume at STP is \(0.66\) dm\(^3\), option (B). \[ \boxed{\text{(B)}} \]
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