Question

The kinetic energy (in cal.) of 4 moles of nitrogen gas at ${127}^{o}C$ is:

[$R=$ 2 $calmol$${}^{-1}$$K$${}^{-1}$ ]

• A. $4400$
• B. $3200$
• C. $4800$
• D. $1524$

Answer 1

The correct option is C $4800$

$K.E=\frac{3}{2}nRT$

$K.E=\frac{3\times 4\times 2\times 400}{2}$

Thus, $KE=4800cal$

Hence, option C is correct.