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 $KE=\frac{3\times 2\times 8.314\times 300}{2}=7.48\times {10}^{3}J$ .