JEE Main & Advanced Physics Thermodynamical Processes JEE PYQ-Thermodynamical Processes

  • 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 10-Jan-2019 Evening]

    A)  581 J              

    B)  73 J  

    C)  146 J  

    D)       291 J

    Correct Answer: D

    Solution :

    [d] \[\mu \] = 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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