Question #b4cdf

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Question #b4cdf

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Answer 1 · 2016-08-20T08:09:19

Given

$V \to Volume of C H_{4} (g) = 0.1 L$ $V->"Volume of "CH_4(g)=0.1L$
$P \to Pressure of C H_{4} (g) = 744 m m = \frac{744}{760} a t m$ $P->"Pressure of "CH_4(g)=744mm=744/760atm$
$T \to Temperature of C H_{4} (g) = 25^{\circ} C = 298 K$ $T->"Temperature of "CH_4(g)=25^@C=298K$
$R \to Universal gas constant = 0.082 L a t m K^{- 1} m o l^{- 1}$ $R->"Universal gas constant"=0.082LatmK^-1mol^-1$

Let

$n \to Number of moles C H_{4} (g) = ?$ $n->"Number of moles"CH_4(g) =?$

By equation of state of ideal gas

$n = \frac{P V}{R T} = \frac{744 \times 0.1}{760 \times 0.082 \times 293} = 0.004 mol$ $n=(PV)/(RT)=(744xx0.1)/(760xx0.082xx293)=0.004"mol"$

Now From combustion data the heat of combustion of methane is $890 k J m o l^{- 1}$ $890kJmol^-1$

$CH_4(g)+2O_2(g)->CO_2(g)+2H_2O(l),DeltaH^@=-890kJ"mol"^-1$

So amount of heat evolved by combustion of 0.004mol $CH_4(g)$ is $=0.004xx890kJ=3560J$

Now we know that melting of 1g ice to 1g water requires 336J heat without raising its temperature

So 9.53g of ice requires $9.53xx336J=3202.8J$ heat to melt into water of $0^@C$

As the previous calculation of complete combustion $CH_4(g)$ provides 3560J heat and which is greater than the value obtained from heat of melting of ice (3202.8J), It can be said that the combustion of $CH_4(g)$ is incomplete in this case.

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