0 . 056 kg of Nitrogen is enclosed in a vessel at a temperature of 127 ° C . The amount of heat required to double the speed of its molecules is_____ kcal . (Take R = 2 cal mole K - 1 - 1 )

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TestHub · Accepted Answer

The RMS speed of the molecule is given by $v_{R M S} = \sqrt{\frac{3 R T}{M}}$ . Therefore, to double the speed the temperature must become $4$ times.

Therefore, the change in the temperature is $∆ T = 3 T_{0}$ .

Now the process is isochoric (enclosed vessel) therefore, the heat required will be,

$H = n C_{v} ∆ T \Rightarrow H = \frac{56}{28} \times \frac{5}{2} R \times 3 \times (273 + 127) = 12000 cal = 12 kcal$

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