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A cylinder with a moveable piston contains 216 mL of nitrogen gas at a pressure of 1.42 atm and a temperature of 293 K.?

What must the final volume be for the pressure of the gas to be 1.52 atm at a temperature of 332 K?

3 Answers

Relevance
  • 3 weeks ago

    (216 mL) x (1.42 atm / 1.52 atm) x (332 K / 293 K) = 229 mL

  • Jim
    Lv 7
    3 weeks ago

    PV=nRT <<<memorize!!

    n is moles, R is gas constant (use correct one!), T is temp in K (C+273.15)

    Since you have a before and after with the same n &R simplifies to

    PV/T initial = nR = PV/T final

    Now sub in your numbers and solve

    SI Units R values:

    8.31446261815324 J⋅K−1⋅mol−1

    8.31446261815324 m3⋅Pa⋅K−1⋅mol−1

    8.31446261815324 kg⋅m2·K−1⋅mol−1s−2

    8.31446261815324×103 L⋅Pa⋅K−1⋅mol−1

    8.31446261815324×10−2 L⋅bar⋅K−1⋅mol−1

    0.082057 L atm mol-1K-1

    62.36L·torr/mol·K = L·mmHg/mol·K

  • 3 weeks ago

    P1V1/T1 = P2V2/T2

    1.42 atm (216 mL) / 293K = 1.52 atm (V2) / 332 K

    V2 = 229 mL

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