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A vessel contains $14 \mathrm{~g}$ of nitrogen gas at a temperature of $27^{\circ} \mathrm{C}$ . The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be :
Take $\mathrm{R}=8.32 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{k}^{-1}$ .

Option: 1
$2229 \mathrm{~J}$

Option: 2
$5616 \mathrm{~J}$

Option: 3
$9360 \mathrm{~J}$

Option: 4
$13,104 \mathrm{~J}$

Answers (1)

$\mathrm{m=149}$ $\mathrm{T=300\, k}$

$\mathrm{ V_{r m s}=\sqrt{\frac{3 R T}{M}} }$

To double the rms speed,

$\mathrm{ T_f=4 T_i=1200 \mathrm{~K} }$

From first law of thermodynamice,

$\mathrm{Q=\Delta U+W}$

$\mathrm{Q=\Delta U+0 \quad(V= constant\: of \: vessd )}$

$\mathrm{Q=\Delta U=\frac{n R \Delta T}{V-1}}$

$\mathrm{Q=\frac{(1 / 2)(8.32)}{\left(\frac{7}{5}-1\right)} \times(900)}$

$\mathrm{=\frac{5(8.32) \times 900}{4}}$

$\mathrm{Q=9360 \mathrm{~J}}$

Hence 3 is correct option.

Posted by

Ramraj Saini

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