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JEE Main & Advanced Sample Paper JEE Main Sample Paper-30

  • question_answer
    \[0.04\,\,kg\] of nitrogen is enclosed in a vessel at a temperature of \[{{27}^{o}}C\]. How much heat has to be transferred to the gas to double the rms speed of its molecules \[[R=2\,cal/mol\,K]\]

    A)  \[2250\,cal\]               

    B)  \[2250\,J\]

    C)  \[4500\,J\]                              

    D)  \[6428\,cal\]  

    Correct Answer: D

    Solution :

    Here \[n=\frac{M'}{M}=\frac{0.04\times {{10}^{3}}}{28}\] mole  Nitrogen is a diatomic gas \[{{C}_{V}}=\frac{5}{2}R\] As to double the rms speed of molecules the temperature of gas should be raised to 4 times. \[\Delta T=1200-300=900k\] As the gas is enclosed in the vessel V = constant \[\therefore {{(\Delta Q)}_{v}}=n{{C}_{v}}\Delta T\]


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