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5 moles of helium expand isothermally and reversibly from a pressure `40xx10^(-5)Nm^(-2)` to `4xx10^(-5)Nm^(-2)` at 300 K. Calculate the work done, change in internal energy and heat absorbed during the expansion . `(R=8.314Jk^(-1)"mol"^(-1))` |
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Answer» Given, Number of moles `=5` Initial pressure `P_(1)=40xx10^(-5) Nm^(-2)` Final pressure `P_(2)=4xx10^(-5)Nm^(-2)` Temperature `=300 K` Work done `=?` `W=-2.303 nRT log_(10).(P_(1))/(P_(2))` `=-2.303xx8.314xx5xx300 log_(10).(40xx10^(-5))/(4xx10^(-5))` `=-2.303xx12471xxlog_(10)10` `=-2.303xx12471xx1` `=-28720.713 J` `W=-28.720kJ` `rArr` For isothermal process change in internal energy `(DeltaU=0)=`zero `rArr DeltaU=q+W` `0=q+W` or `q=-W` `q=28.720 kJ` `:.28.720` kJ heat absorb during expansion. |
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