The volume of 0.354g of helium at 273°C and 114cmhg of mercury pressure is 2667cm$^3$. Calculate the volume at STP.

Analyze the question: This is a gas law problem. A fixed mass of helium changes from one set of conditions to STP, so we use the combined gas law:

P₁V₁/T₁ = P₂V₂/T₂

Given:

• P₁ = 114 cmHg
• V₁ = 2667 cm³
• T₁ = 273°C = 273 + 273 = 546 K
• STP: P₂ = 76 cmHg, T₂ = 273 K
• V₂ = ?

Solution:

V₂ = (P₁V₁T₂) / (T₁P₂)

V₂ = (114 × 2667 × 273) / (546 × 76)

Step 1: Simplify the temperature ratio: 273/546 = 1/2

Step 2: Simplify the pressure ratio: 114/76 = 1.5

Step 3: V₂ = 2667 × 1.5 × 0.5 = 2667 × 0.75 = 2000.25 cm³

Answer: V₂ = 2000.25 cm³, which is option A.

Tip: Always convert temperature to Kelvin first (add 273), and remember STP = 76 cmHg (760 mmHg) and 273 K. Simplifying the ratios before multiplying, as done above, saves time in the exam. The mass of helium (0.354 g) is a distractor since the question only asks about the volume change.