JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer A cylinder with fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by \[20{}^\circ C\] is : [Given that \[R=8.31\text{ }J\text{ }mo{{l}^{1}}\text{ }{{K}^{1}}\]] [JEE Main 10-4-2019 Morning]

    A) 748 J  

    B) 374 J

    C) 350 J              

    D) 700 J

    Correct Answer: A

    Solution :

    \[\Delta Q=n{{C}_{V}}\Delta T=n\frac{3}{2}R\Delta T\]           \[=\left( \frac{67.2}{22.4} \right)\left( \frac{3}{2}\times 8.31 \right)(20)\]\[\approx 748J\]

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