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For his 18th birthday in February Peter plants to turn a hut in the garden of his parents into a swimming pool with an artifical beach. In order to estimate the consts for heating the water and the house , peter obtains the data for the natural gas combustion and its price. The desity of natural gas is `0.740 g L^(-1) (1.013 xx 10^(5)Pa, 25^(@)C)` specified by PUC , the public utility company. (a) Calculate the amount of methane and ethane (in moles) in `1.00m^(3)` of natural gas (natural gas, methane and ethane are not ideal gases). (b) Calculate the combustion energy which is released as thermal energy during the buring of `1.00 m^(3)` of natural gas under standard conditions assuming that all products are gaseous .(if you do not have the amount from 1.2 a) assume that `1.00 m^(3)` natural gas corresponds to 40.00 mol natural gas) According to the PUC the combustion energy will be 9.981K Wh per `m^(3)` of natural gas if all products are gaseous . How large is the deviation (in percent) from the value you obtained in b? The swimming pool inside the house is 3.00 m wide ,5.00 m long and 1.50 m deep (below the floor) . The tap water temperature is `8.00^(@)C` and the air temperature in the house (dimensions given in the figure) is `10.0^(@)C.` Assuming a water density of `rho=1.00 Kg L^(-1)` and air behaving like an ideal gas. |
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Answer» Correct Answer - Amount of methane and ethane in `1 m^(3)` natural gas : `m= gt xx V=0.740 g dm^(-3) xx 1000 dm^(3) =740 g` `M_(av) = sum_(i) xx (i) M(ii)=(0.0024 xx44.01 g "mol"^(-1)) + (0.0134 xx28.02 g "mol"^(-1))+(0.9732 xx 16.05 g "mol"^(-1)) + (0.011 xx 30.08 g "mol"^(-1))=16.43 g "mol"^(-1)` `n_("tol") =m(Mav)^(-1) = 740 gxx (16.43 g//mol)^(-1)=45.83 "mol"` n(i) = x(i) . n_("tot") `n(Ch_(4)) = x(CH_(4)) xx n_("tot") =09732xx 45.04 "mol"=0.495 "mol"` `n(C_(2)H_(6))=x(C_(2)H_(6)xxn_("tot")=0.0110 xx 45.04 "mol" =0.495 "mol"` (b) Energy of combustion , deviation: Ecomb.`(H_(2)O(g)) =sum_(i)n(i) Delta_(c)H-(i)=` `=43.83 "mol" xx (-802.5 "kj mol"^(-1)) + 0.495 "mol" xx 0.5 xx(-2856.8 KJ "mol"^(-1))` `=-35881 KJ` `E_("comb")(H_(2)O(g))=-35881 KJ` Deviation from PUC EPUC `(H_(2)O(g)) =9.981 K Whm^(-3) xx 1 m^(3)xx 3600 KJ (kWh)^(-1) =35932 KJ` Deviation: `DeltaE=(E_("comb")(H_(2)O(g))-E_(PUC)(H_(2)(g)) xx 100%xx [Ecomb.(H2)(g))]^(-1)` `=(35881 KJ -35932 KJ ) xx 100% xx (35881 KJ )^(-1)=.^(-)C0.14%` |
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