Question:

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

Solution and Explanation

Concept: The total translational kinetic energy of an ideal gas is \[ K=\frac{3}{2}nRT, \] where \[ n=\text{number of moles}, \quad R=\text{universal gas constant}, \quad T=\text{absolute temperature}. \]

Step 1: Calculate the number of moles of \(CO_2\). Molar mass of \(CO_2\): \[ M=44\ \text{g mol}^{-1}. \] Given mass, \[ m=22\ \text{g}. \] Hence, \[ n=\frac{m}{M} =\frac{22}{44} =\frac12. \]

Step 2: Convert temperature into Kelvin. \[ T=27^\circ C+273. \] \[ T=300\ K. \]

Step 3: Calculate the translational kinetic energy. \[ K = \frac32 nRT. \] Substituting, \[ K = \frac32 \left(\frac12\right) (8.314)(300). \] \[ K = \frac34(2494.2). \] \[ K = 1870.65\ \text{J}. \] \[ K \approx 1870.6\ \text{J}. \] Therefore, \[ \boxed{K=1870.6\ \text{J}} \] \[ \boxed{\text{Answer = (A)}} \]
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