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 -

Question

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 )

TestHub · Accepted Answer

The RMS speed of a gas at a given temperature is given by, $v_{rms} = \sqrt{\frac{3 R T}{M}}$ .
Therefore, to double the RMS speed, the temperature should be increased to four times the initial temperature. Therefore,
$T_{f} = 1200 K, T_{i} = 300 K, and n = \frac{14}{28} = \frac{1}{2}$
The volume of the vessel will remain the same therefore, the process is an isochoric process. Therefore,
$Q = n C_{v} Δ T = \frac{1}{2} \times \frac{5 R}{2} \times 900$
$\Rightarrow Q = 9360 J$

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