KCET · Physics · Kinetic Theory of Gases
At \(27^{\circ} \mathrm{C}\) temperature, the mean kinetic energy of the atoms of an ideal gas is \(\mathrm{E}_1\). If the temperature is increased to \(327^{\circ} \mathrm{C}\), then the mean kinetic energy of the atoms will be
- A \(\frac{E_1}{\sqrt{2}}\)
- B \(\sqrt{2} \mathrm{E}_1\)
- C \(2 \mathrm{E}_1\)
- D \(\frac{E_1}{2}\)
Answer & Solution
Correct Answer
(C) \(2 \mathrm{E}_1\)
Step-by-step Solution
Detailed explanation
KE mean \(\propto \mathrm{T}\). ( \(\mathrm{T} \rightarrow\) temperature in Klevin)
\(\frac{E_1}{E_2}=\frac{T_1}{T_2}\)
\(\begin{aligned} & \mathrm{E}_2=\mathrm{E}_1 \frac{\mathrm{~T}_2}{\mathrm{~T}_1} \\ & =\mathrm{E}_1\left(\frac{327+273}{27+273}\right) \\ & =\mathrm{E}_1\left(\frac{600}{300}\right)=2 \mathrm{E}_1\end{aligned}\)
\(\frac{E_1}{E_2}=\frac{T_1}{T_2}\)
\(\begin{aligned} & \mathrm{E}_2=\mathrm{E}_1 \frac{\mathrm{~T}_2}{\mathrm{~T}_1} \\ & =\mathrm{E}_1\left(\frac{327+273}{27+273}\right) \\ & =\mathrm{E}_1\left(\frac{600}{300}\right)=2 \mathrm{E}_1\end{aligned}\)
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