JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature\[27{}^\circ C\]. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about: [Take\[\,R=8.3\] J/K mole] [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) 10 kJ

    B)                                       0.9 KJ

    C) 14 kJ                

    D)               6 kJ

    Correct Answer: A

    Solution :

    To double the rms speed, temp is increased to 4 times \[{{T}_{1}}=300\,K\] \[{{T}_{2}}=300\times 4=1200\text{ }K\] \[\Delta \,T=1200-300=900\text{ }K\] \[Q\,\,=\,\,n{{C}_{V}}\Delta T\,=\,\frac{15}{28}\,\times \frac{5}{2}\,\,\times \,\,R\,\,\times \,\,900\] \[Q=10\,KJ\]


You need to login to perform this action.
You will be redirected in 3 sec spinner