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  • The average translational kinetic energy and the rms speed of molecules in a sample of oxygen at <img alt="300 \mathrm{~K}" src="https://entrancecorner.oncodecogs.com/gif.latex?300%20%5Cmathrm%7B%7

The average translational kinetic energy and the rms speed of molecules in a sample of oxygen at 300 \mathrm{~K} are \mathrm{6.21 \times 10^{-21} \mathrm{~J}} and \mathrm{484 \mathrm{~m} / \mathrm{s}} respectively. The corresponding values at \mathrm{600 \mathrm{~K}}area nearly (assuming ideal gas behavior).

 

Option: 1

12.42 \times 10^{-21} \mathrm{~J}, 968 \mathrm{~m} / \mathrm{s}


Option: 2

8.78 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~m} / \mathrm{s}


Option: 3

6.21 \times 10^{-21} \mathrm{~J},968 \mathrm{~m} / \mathrm{s}


Option: 4

12.42 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~m} / \mathrm{s}


Answers (1)

best_answer

\mathrm{K E=\frac{3}{2} K T, }
\mathrm{v_{\text {rms }}=\sqrt{\frac{3 R T}{M}} }
\mathrm{\text { i.e. } \quad \frac{K E_{2}}{K E_{1}}=\frac{T_{2}}{T_{1}}=2 }
\mathrm{\therefore \quad \mathrm{KE}_{2}=2 \mathrm{KE}_{1}=2 \times 6.21 \times 10^{-21}=12.42 \times 10^{-21} \mathrm{~J} }

\mathrm{\qquad \frac{\mathrm{V}_{\mathrm{rms}, 2}}{\mathrm{~V}_{\mathrm{rms}, 1}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\sqrt{2}} }
\mathrm{ \therefore \mathrm{V}_{\mathrm{rms}, 2}=\sqrt{2} \times \mathrm{V}_{\mathrm{rms}, 1}}

\mathrm{ =684 \mathrm{~m} / \mathrm{s} }.

Posted by

Divya Prakash Singh

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