Question

Translational kinetic energy for 2 moles of gas at $ 27{}^\circ C $ is

• A. $ 7.48\times {{10}^{3}}J $
• B. $ 6.48\times {{10}^{3}}J $
• C. $ 5.48\times {{10}^{3}}J $
• D. $ 4.48\times {{10}^{3}}J $

Answer 1

Answer: (A)

: Translational kinetic energy $ =\frac{3nRT}{2} $ $ n= $ no. of moles $ R= $ gas constant $ T= $ Temperature $ KE=\frac{3\times 2\times 8.314\times 300}{2}=7.48\times {{10}^{3}}J $ .