Question

The kinetic energy of 4 moles of nitrogen gas at 127 is: (R=2 cal ${mol}^{-1}{K}^{-1}$)

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

Answer 1

The correct option is C 4800 cal

Answer:-

Kinetic theory of gases is defined as -

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

where,

n = no. of moles$=4moles$ (given)

T = temperature $=127℃$ (given) $=(273+127)K=400K$

$R=2cal{mol}^{-1}{K}^{-1}$

$\therefore K.E.=\frac{3}{2}\times 4\times 2\times 400=4800cal$