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Oxidation ofpropane isrepresented as C_(3)H_(8_((g))) + 50_(2_((g))) to 3CO_(2_((g)))+ 4H_(2)O_((g)) DeltaH^(@) =- 2043 kJ How muchpressurevolumework is doneand whatis thevaluneof DeltaU at constantpressureof 1 atmwhen thevolumechangeis +22.4 L |
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Answer» Change in volume`= Delta V = + 22.4 L` `Delta U = ?` `C_(3)H_(8_((g)) + 5O _(2_((g))) to3CO_(2_((g)))+ 4H_(2)O_((g))` `Delta n =(n_(2))_("gaseous products ") - (n_(1))_("gaseous reactants ")` `= (3 + 4)- (1+5)` `= 1"mol" ` Sincethere is theanincrease in NUMBEROF moles the workwill beofexpansion . `W=- P xx Delta V` `=- 1 xx 22.4 L atm` `=-1 xx 22.4 xx 101.3 J` `=- 2270 J` `=- 2.27 kJ` `Delta H = Delta U + PDelta V` `:.Delta U = Delta H- P Delta V` `=- 2043- (2.27 )` `=- 2045 .27 kJ` |
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