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Calculate the work done in an open vessel at `300K,` when `112g` ion reacts with dil. `HCl`. |
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Answer» `Fe+2HCl rarr FeCl_(2)+H_(2)` Work done `=-Pxx(V_(2)-V_(1))` Since, mole of `Fe` used `=112/56=2` `:.` Mole of `H_(2)` formed `=2` Work is done by giving out `2`moles of `H_(2)` `=-Pxx[V_(H_(2))-V_(i)] i.e., =-PxxV_(H_(2))` [`V_(i)=0` for initial condition] For `H_(2) PxxV_(H_(2))=nRT` `:. V_(H_(2))=(nRT)/P` `:.` Work done `=-Pxx(nRT)/P=-nRT` `=-2xx2xx300=-1200 cal` |
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