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JEE Main & Advanced Sample Paper JEE Main Sample Paper-37

  • question_answer
    1 mole of gas X is present in a closed adiabatic vessel fitted with a movable frictionless piston. The initial temperature of gas X is 300 K. The vessel is maintained at constant pressure of 1 aim. Keeping the pressure constant at 1 atm the reaction \[(3X\,(g)\to 2Y(g);\,\Delta H=-30kJ/mol)\]is started with the help of negligible amount of electric energy If finally 75 mole % of X undergone reaction at constant pressure of 1 atm, find the final temperature (in K) of reaction  vessel. Given: \[{{C}_{p,m(X)}}=40J/K\,mole,\]\[{{C}_{p,m(Y)}}=30J/K\,mole,\]

    A)  \[600\,K\]

    B)  \[300\,K\]

    C)  \[1200\,K\]                 

    D)  \[1000\,K\]

    Correct Answer: A

    Solution :

                            Mole of X reacted \[=0.75\]mole Heat liberated at constant pressure  \[=0.75\times 10\,kJ=7.5kJ\] Mole of X remaining \[=\frac{1}{4}mole\] Mole of Y formed \[=\frac{3}{4}\times \frac{2}{3}=\frac{1}{2}\] \[\left( \frac{1}{4}{{C}_{p.m(X)}}+\frac{1}{2}{{C}_{p.m(Y)}} \right)\,\,(\Delta T)=q=7500\] \[\Delta T=\frac{7500}{\frac{40}{4}+\frac{30}{2}}=300\] Final temperature \[=300+300=600\text{ }K\]


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