For one mole of an ideal gas,increasing the t | vedclass

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(D) Given,$T_{1} = 273 + 10 = 283 \ K$ and $T_{2} = 273 + 20 = 293 \ K$. Average kinetic energy $(KE)$ is directly proportional to temperature $(T)$:

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increases both the average kinetic energy and rms velocity,but not significantly

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(D) Given,$T_{1} = 273 + 10 = 283 \ K$ and $T_{2} = 273 + 20 = 293 \ K$. Average kinetic energy $(KE)$ is directly proportional to temperature $(T)$: $KE = \frac{3}{2} RT$. Ratio $\frac{KE_{2}}{KE_{1}} = \frac{293}{283} \approx 1.035$. Root mean square velocity $(v_{rms})$ is proportional to $\sqrt{T}$: $v_{rms} = \sqrt{\frac{3RT}{M}}$. Ratio $\frac{v_{rms,2}}{v_{rms,1}} = \sqrt{\frac{293}{283}} \approx 1.017$. Since the temperature change is small relative to the absolute temperature,both the average kinetic energy and the rms velocity increase,but not significantly.

For one mole of an ideal gas,increasing the temperature from $10^{\circ} C$ to $20^{\circ} C$:

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