Question

The average K.E per mole an ideal gas at ${27}^{0}C$ is:

• A. 3.74 kJ
• B. 3.695 kJ
• C. 984.15 kJ
• D. $3.7\times {10}^{10}$

Answer 1

The correct option is A 3.74 kJ

Average $KE$ per mol $=\frac{3}{2}RT\times N_A$

$=\frac{3}{2}\times 1.38\times {10}^{-23}\times 300\times 6.022\times {10}^{23}$

$=3\times 1.38\times 150\times 6.022$

$\approx 3.74kJ$

Hence, the correct option is $\text{A}$