# Question b4cdf

Aug 20, 2016

Given

• $V \to \text{Volume of } C {H}_{4} \left(g\right) = 0.1 L$
• $P \to \text{Pressure of } C {H}_{4} \left(g\right) = 744 m m = \frac{744}{760} a t m$
• $T \to \text{Temperature of } C {H}_{4} \left(g\right) = {25}^{\circ} C = 298 K$
• $R \to \text{Universal gas constant} = 0.082 L a t m {K}^{-} 1 m o {l}^{-} 1$

Let

• n->"Number of moles"CH_4(g) =?#

By equation of state of ideal gas

$n = \frac{P V}{R T} = \frac{744 \times 0.1}{760 \times 0.082 \times 293} = 0.004 \text{mol}$

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

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) , \Delta {H}^{\circ} = - 890 k J {\text{mol}}^{-} 1$

So amount of heat evolved by combustion of 0.004mol$C {H}_{4} \left(g\right)$ is $= 0.004 \times 890 k J = 3560 J$

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

So 9.53g of ice requires $9.53 \times 336 J = 3202.8 J$ heat to melt into water of ${0}^{\circ} C$

As the previous calculation of complete combustion $C {H}_{4} \left(g\right)$ 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 $C {H}_{4} \left(g\right)$ is incomplete in this case.