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Standard XII

Chemistry

Relating Translational KE and T

the temperatu...

Question

the temperature of an ideal gas is increased from $120 K$ to $480 K$ . If at $120 K$ .If at $120 K$ , the root mean square speed of the gas molecules is $^{'} v^{'}$ ,then

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Solution

We know that, the rms velocity is directly proportional to the square root of the absolute temperature,

$v \propto \sqrt{T} \Rightarrow \frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} \Rightarrow V_{2} = v \times \sqrt{\frac{480}{120}} = 2 V$

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the temperature of an ideal gas is increased from 120K to 480K. If at 120K .If at 120K , the root mean square speed of the gas molecules is ′v′,then

We know that, the rms velocity is directly proportional to the square root of the absolute temperature,v∝√T⇒v2v1=√T2T1⇒V2=v×√480120=2V

the temperature of an ideal gas is increased from $120 K$ to $480 K$ . If at $120 K$ .If at $120 K$ , the root mean square speed of the gas molecules is $^{'} v^{'}$ ,then

We know that, the rms velocity is directly proportional to the square root of the absolute temperature,

$v \propto \sqrt{T} \Rightarrow \frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} \Rightarrow V_{2} = v \times \sqrt{\frac{480}{120}} = 2 V$