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JEE Main & Advanced Sample Paper JEE Main Sample Paper-12

  • 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 amt. Keeping the pressure constant at 1 atm the reaction \[(3X(g)\to 2Y(g);\]\[\Delta H=-30\,\text{kJ}\,\text{/mol)}\] 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: \[{{C}_{p,m(X)}}=40\text{J/K}\,\text{mole,}\]\[{{\text{C}}_{\text{p,m(Y)}}}\text{=30J/K}\,\text{mole,}\]

    A)  600K                                    

    B)  300K

    C)  1200K                  

    D)  1000K

    Correct Answer: A

    Solution :

     Mole of X reacted = 0.75 mole Heat liberated at constant pressure \[=0.75\times 10\,\text{kJ}\] \[=7.5\,\text{kJ}\] Mole of X remaining \[=\frac{1}{4}\]mole Mole of 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 \[=300+300=600K\]


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