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 )

Step-by-step solution

The RMS speed of a gas at a given temperature is given by, v rms = 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 = 14 28 = 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 = 1 2 × 5 R 2 × 900 ⇒ Q = 9360   J