Question:easy

The translational kinetic energy of the molecules of \(22\) grams of \(CO_2\) at \(27^\circ C\) is \[ (R=8.314\ \text{J mol}^{-1}\text{K}^{-1}) \]

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For any ideal gas, \[ \text{Total Translational K.E.} = \frac32 nRT. \] It depends only on the number of moles and absolute temperature, not on the nature of the gas.
Updated On: Jul 29, 2026
  • \(1870.6\ \text{J}\)
  • \(164.7\ \text{J}\)
  • \(2000\ \text{J}\)
  • \(2200\ \text{J}\)
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The Correct Option is A

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