Question

The average kinetic energy of a gas molecule at ${27}^{\circ }$ is $6.21\times {10}^{-21}J.$ The average kinetic energy at ${227}^{\circ }$ will be

• A. $9.35\times {10}^{-21}J$
  
• B. $10.35\times {10}^{-21}J$
  
• C. $11.35\times {10}^{-21}J$
  
• D. $12.35\times {10}^{-21}J$

Answer 1

The correct option is B

$10.35\times {10}^{-21}J$

$E=\frac{1}{2}kT.Thus6.21\times {10}^{-21}=\frac{1}{2}k\times (273+27)150kgivingk=\frac{6.21\times {10}^{-21}}{150}.Thereforeat{227}^{\circ },thevalueofEisE=\frac{1}{2}\times \frac{6.21\times {10}^{-21}}{150}(273+227)=10.35\times {10}^{-21}J$
Hence, the correct choice is (b).