JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Evening)

  • question_answer
    Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from\[20{}^\circ C\text{ }to\text{ }90{}^\circ C\]. Work done by gas is close to -(Gas constant \[R=8.31\text{ }J\] /mol. K) [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

    A) 581 J                           

    B) 73 J  

    C) 146 J   

    D)                  291 J

    Correct Answer: D

    Solution :

    \[\mu \] = 0.5 mole; P = 1 atm \[{{T}_{1}}=20{}^\circ C;\text{ }{{T}_{2}}=90{}^\circ C\] \[W=P\left[ {{V}_{2}}-{{V}_{1}} \right]\] \[=\mu R({{T}_{2}}-{{T}_{1}})=\frac{1}{2}\times 8.31\times 70=291\,J\]


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