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

5616 J

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9360 J

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2229 J

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13104 J

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Solution

The correct option is B $9360 J$
Moles of nitrogen $n = \frac{14}{28} = 0.5$
$T = 300$
For $v_{r m s}$ to be doubled

$T' = 4 T = 4 \times 300 = 1200$
$\Rightarrow$ Heat transferred

$Q = n C_{v} Δ T = (0.5) (\frac{5}{2}) (8.32) (1200 - 300)$

$Q = 9360 J$

Hence , option $(D)$ is correct.

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$0.014 k g$ of nitrogen is enclosed in a vessel at a temperature of $27^{\circ} C$ . The amount of heat energy to be supplied to the gas to double the rms speed of its molecules is approximately equal to

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. Amount of heat transfered to the gas, so that rms velocity of molecules is doubled is about
[Take

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A vessel contains 14 g of nitrogen gas at a temperature of 27∘C. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be: Take R=8.32 J mol–1 K–1.

The correct option is B 9360 J Moles of nitrogen n=1428=0.5 T=300 For vrms to be doubled T'=4T=4×300=1200 ⇒ Heat transferred Q=nCvΔT=(0.5)(52)(8.32)(1200−300) Q=9360 J Hence , option (D) is correct.

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

The correct option is B $9360 J$
Moles of nitrogen $n = \frac{14}{28} = 0.5$
$T = 300$
For $v_{r m s}$ to be doubled

$T' = 4 T = 4 \times 300 = 1200$
$\Rightarrow$ Heat transferred

$Q = n C_{v} Δ T = (0.5) (\frac{5}{2}) (8.32) (1200 - 300)$

$Q = 9360 J$

Hence , option $(D)$ is correct.