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  • of Nitrogen is enclosed in a vessel at a temperature of <

\mathrm{0.056 \mathrm{~kg}} of Nitrogen is enclosed in a vessel at a temperature of \mathrm{127 \, \mathrm{C}}. The amount of heat required to double the sped of its molecules is________ \mathrm{k\, cal}.
\mathrm{(Take \, \mathrm{R}=2 \, \mathrm{cal} \: mole ^{-1} \mathrm{~K}^{-1} )}

Option: 1

4


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{m=0.056 \mathrm{~kg}=56 \, \mathrm{gm}}
no of moles of \mathrm{N_{2}= n= 2\, mol}

\mathrm{T=127\, C= 400\, k}
\mathrm{V=\sqrt{\frac{3 \mathrm{~K}_{B} \mathrm{~T}}{m}}}
\mathrm{\frac{V_{i}}{V_{f}}=\sqrt{\frac{T_{i}}{T_{f}}}}

\mathrm{V_{f}=2 V_{i}(\text { Given })}
\mathrm{\frac{1}{2}=\frac{\sqrt{400}}{\sqrt{T_{f}}}}
\mathrm{T_{f}=800 \mathrm{~K}=527^{\circ} \mathrm{C}}

Vessel is closed
\mathrm{\therefore \: W= 0}

By First law of thermodynamics,
\mathrm{Q=\frac{\Delta U}{\gamma -1}=\frac{n R \Delta T}{\gamma -1}}
     \mathrm{=\frac{2\times 2\times \left ( 400 \right )}{\left ( \frac{7}{5}-1 \right )}}
\mathrm{Q =4 k \text { cal }}

The amount of heat required to double  the speed of its molecules is 4\, \mathrm{k} \, \mathrm{cal}

Posted by

Divya Prakash Singh

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