The kinetic energy of translation of the molecules in 50 g of $\mathrm{CO}_{2}$ gas at $17^{\circ} \

Question

The kinetic energy of translation of the molecules in 50 g of $\mathrm{CO}_{2}$ gas at $17^{\circ} \mathrm{C}$ is\

Professor PC Thomas Classes · Answer

(c) : The kinetic energy of translation of molecule,\ $(	ext { K.E. })_{T}=\frac{3}{2} 	imes n 	imes R 	imes T$\ or $(K . E .)_{T}=\frac{3}{2} 	imes \frac{m}{M} 	imes R 	imes T$\ $=\frac{3}{2} 	imes \frac{50}{44} 	imes 8.3 	imes(273+17)=4102.8 \mathrm{~J}$\