Question

The average kinetic energy of a monoatomic gas molecule kept at temperature \(27^\circ\)C is \((\text{Boltzmann constant}=1.3\times 10^{-23}\ \text{J K}^{-1})\)

• A. \(5.85\times10^{-21}\) J
• B. \(4.12\times10^{-21}\) J
• C. \(3.75\times10^{-21}\) J
• D. \(2.85\times10^{-21}\) J
• E. \(7.55\times10^{-21}\) J

Hint:

For one molecule of a monoatomic gas: \[ E=\frac{3}{2}kT \] Use kelvin, not degree Celsius.

Answer 1

The Correct Option is A

Average kinetic energy of a monoatomic gas molecule: \[ E=\frac{3}{2}kT \] Given: \[ T=27^\circ\text{C}=300\text{ K} \] \[ E=\frac{3}{2}(1.3\times10^{-23})(300) \] \[ E=1.5\times 1.3\times 300\times10^{-23} \] \[ E=585\times10^{-23}=5.85\times10^{-21}\text{ J} \] Hence, \[ \boxed{(A)\ 5.85\times10^{-21}\text{ J}} \]