Question

The kinetic energy of 4 moles of nitrogen gas at $\text{127}{{\,}^{\text{o}}}\text{C}$ is ...... cals. $\text{(R= 2 cal mo}{{\text{l}}^{-1}}{{\text{K}}^{-1}}\text{)}$

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

Answer 1

Answer: (C)

4800

Answer 2

$E=\frac{3}{2}nRT$ Where, E = Kinetic energy n = number of moles = 4 R = Gas constant =2cal/mol/K T = Temperature in Kelvin = 127 + 273 = 400K $E=\frac{3}{2}\times 4\times 2\times 400$ = 4800 calorie