To calculate the average kinetic energy of a mole of [tex]$CO_2$[/tex] at a temperature of 272 K, we use the formula for the average kinetic energy of an ideal gas. The formula is given by:
[tex]\[ E_{\text{avg}} = \left(\frac{3}{2}\right) R T \][/tex]
where:
- [tex]\( R \)[/tex] is the universal gas constant, which is approximately [tex]\( 8.314 \)[/tex] J/(mol·K),
- [tex]\( T \)[/tex] is the temperature in Kelvin (K).
Given:
- [tex]\( T = 272 \)[/tex] K,
- [tex]\( R = 8.314 \)[/tex] J/(mol·K).
Now, using these values in the formula:
[tex]\[ E_{\text{avg}} = \left(\frac{3}{2}\right) \times 8.314 \times 272 \][/tex]
Following this computation, the value of the average kinetic energy is approximately:
[tex]\[ E_{\text{avg}} = 3392.112 \, \text{J/mol} \][/tex]
Expressing the result to three significant figures:
[tex]\[ E_{\text{avg}} \approx 3390 \, \text{J/mol} \][/tex]
Thus, the average kinetic energy of [tex]\( CO_2 \)[/tex] at 272 K is:
[tex]\[ E_{\text{avg}} = 3390 \, \text{J/mol} \][/tex]
