### Related Formula The total translational kinetic energy of a gas sample depends only on the number of moles and absolute temperature, regardless of whether the molecule is monoatomic or polyatomic: K.E._texttranslational = frac32 n R T$K.E._{\text{translational}} = \frac{3}{2} n R T$ ### Core Logic Given data: * Mass of mathrmCO_2$\mathrm{CO}_2$ gas, m = 50 text g$m = 50 \text{ g}$ * Molar mass of mathrmCO_2$\mathrm{CO}_2$, M = 44 text g/mol$M = 44 \text{ g/mol}$ * Absolute Temperature, T = 17 + 273.15 = 290.15 text K$T = 17 + 273.15 = 290.15 \text{ K}$ * Universal gas constant, R approx 8.314 text J/(molcdottextK)$R \approx 8.314 \text{ J/(mol}\cdot\text{K)}$ Calculate total moles n$n$: n = frac5044 approx 1.1364 text moles$n = \frac{50}{44} \approx 1.1364 \text{ moles}$ Substitute values into the expression : K.E._texttranslational = frac32 times frac5044 times 8.314 times 290.15$K.E._{\text{translational}} = \frac{3}{2} \times \frac{50}{44} \times 8.314 \times 290.15$ K.E._texttranslational = 1.5 times 1.1364 times 8.314 times 290.15 approx 4108.6 text J$K.E._{\text{translational}} = 1.5 \times 1.1364 \times 8.314 \times 290.15 \approx 4108.6 \text{ J}$ The closest matching value specified in the test alternatives is 4102.8 mathrmJ$4102.8 \mathrm{J}$. ### Pattern Recognition A common mistake is using frac52nRT$\frac{5}{2}nRT$ or frac72nRT$\frac{7}{2}nRT$ because mathrmCO_2$\mathrm{CO}_2$ is a triatomic linear molecule. Remember that **translational** kinetic energy is always frac32nRT$\frac{3}{2}nRT$ for any gas sample, as translation has exactly 3 degrees of freedom regardless of the molecular layout. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases