100 m3 of a gas at 3 atm pressure and 27 oc is transferred to a chamber of 300 m3 volume maintained at a temperature of 327 oc. What will be the pressure in chamber?
Correct answer: C. 2 atm
- A. 6 atm
- B. 4 atm
- C. 2 atm
- D. 1 atm
Explanation
Use the combined gas law.P₁V₁/T₁ = P₂V₂/T₂P₁ = initial pressure = 3 atmV₁ = initial volume = 100 m³T₁ = initial temperature= 27° C+273.15 =300.15KP₂ = final pressure = ? atmV₂ = final volume = 300 m³ T₂ final temperature= 327° C + 273.15 = 600.15 KCalculate the final pressure (P₂).P₂ = P₁V₁T₂/V₂T₁P₂ = (3 atm × 100 m³ × 600.15 K)/(300 m³ × 300.15 K) = 2 atm
This question appeared on the UHS MDCAT 2008 paper, which you can sit online with every answer explained.
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About Ideal Gas Equation
The ideal gas equation, PV = nRT, relates pressure, volume, amount and absolute temperature when gas particles are assumed to have negligible volume and no intermolecular forces. Applications include finding molar mass, density or an unknown gas variable, with careful use of Kelvin temperature and consistent units, and comparison with real-gas behaviour.
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