Question

If temperature is increased by four times how much kinetic energy will increase

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{temperature},\mathrm{T}\text{'}=4\mathrm{T} \\ {\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}}} \\ \mathrm{where},\mathrm{R}=\mathrm{gas}\mathrm{constant} \\ \mathrm{T}=\mathrm{temperature} \\ \mathrm{M}=\mathrm{mass} \\ \mathrm{v}=\mathrm{velocity} \\ {\mathrm{v}}_{\mathrm{rms}}\mathrm{\alpha }\sqrt{\mathrm{T}} \\ \frac{\mathrm{v}\text{'}}{\mathrm{v}}=\sqrt{\frac{\mathrm{T}\text{'}}{\mathrm{T}}} \\ \frac{\mathrm{v}\text{'}}{\mathrm{v}}=\sqrt{\frac{4\mathrm{T}}{\mathrm{T}}} \\ \frac{\mathrm{v}\text{'}}{\mathrm{v}}=\sqrt{4}=2 \\ \mathrm{v}\text{'}=2\mathrm{v} \\ \mathrm{We}\mathrm{know}\mathrm{that},\mathrm{kinetic}\mathrm{energy},\mathrm{K}=\frac{1}{2}{\mathrm{Mv}}^2 \\ \mathrm{K}\text{'}=\frac{1}{2}\mathrm{Mv}{\text{'}}^2 \\ \mathrm{K}\text{'}=\frac{1}{2}\mathrm{M}{\left(2\mathrm{v}\right)}^2 \\ \mathrm{K}\text{'}=\frac{1}{2}\mathrm{M}\left(4{\mathrm{v}}^2\right)=4\mathrm{K} \\ \mathrm{Hence},\mathrm{kinetic}\mathrm{energy}\mathrm{is}\mathrm{also}\mathrm{increased}\mathrm{by}\mathrm{four}\mathrm{times}. \end{gathered}$$
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