Question 650: Thermodynamics & KTG

Question

The initial average kinetic energy of the molecules was E , when a gas sample is at $27^{\circ} \mathrm{C}$. When the gas is heated to $327^{\circ} \mathrm{C}$, then the final average kinetic energy will be

Accepted Answer

Initial temperature $\mathrm{T}_1=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
Final temperature $\mathrm{T}_2=327^{\circ} \mathrm{C}=327+273=600 \mathrm{~K}$
Average kinetic energy of an ideal gas molecule is directly propartional to the abosulate temperature T .
$$ \begin{array}{ll}  & \mathrm{KE} \propto \mathrm{~T} \ 	herefore & \frac{\mathrm{KE}_2}{\mathrm{KE}_1}=\frac{\mathrm{T}_2}{\mathrm{~T}_1} \Rightarrow \frac{\mathrm{KE}_2}{\mathrm{KE}_1}=\frac{600}{300} \ 	herefore & \frac{\mathrm{KE}_2}{\mathrm{KE}_1}=2 \ 	herefore & \frac{\mathrm{KE}_2}{\mathrm{E}}=2 \ 	herefore & \mathrm{KE}_2=2 \mathrm{E} \end{array} $$

Answer

$\quad \sqrt{2} \mathrm{E}$

Answer

2 E

Answer

300 E

Answer

327 E

Thermodynamics & KTG

The initial average kinetic energy of the molecules was E , when a gas sample is at 27^{\circ} \mathrm{C}. When the gas is heated to 327^{\circ} \mathrm{C}, then the final average kinetic energy will be

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