Assuming gasoline is pure \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) predict the signs of \(q\) and \(w\) for the process of combusting gasoline into \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\)

Short Answer

Expert verified
During the combustion of gasoline (C₈H₁₈) into CO₂(g) and H₂O(g), both q and w are negative. This means that heat is released to the surroundings (exothermic process) and the system does work against the surroundings due to the expansion of gases. Therefore, \(q < 0\) and \(w < 0\).

Step by step solution

01

1. Write the balanced chemical equation for the combustion of gasoline

For the combustion of C₈H₁₈, the balanced chemical equation is: \[C_{8}H_{18}(l) + 12.5O_{2}(g) \rightarrow 8CO_{2}(g) + 9H_{2}O(g)\]
02

2. Determine the signs of q

Combustion reactions are exothermic, meaning they release heat. As a result, the system loses heat to the surroundings, making the value of q negative. So, the sign of q is negative: \[q < 0\]
03

3. Determine the signs of w

To find the sign of the work (w) done during the combustion process, we need to consider the expansion of gases. In the balanced chemical equation, we see that 12.5 moles of O₂(g) are consumed and 17 moles of products (8 moles of CO₂ and 9 moles of H₂O) are formed, which means more gas molecules are produced. This results in an expansion of the gases, as more particles are created and occupy a larger volume. In an expansion, the system does work against the surroundings; thus, the sign of work is negative. So, the sign of w is negative: \[w < 0\]
04

4. Conclusion

During the combustion of gasoline (C₈H₁₈) into CO₂(g) and H₂O(g), both q and w are negative. This means that heat is released to the surroundings and the system does work against the surroundings in the expansion of gases. q is negative: \(q < 0\) w is negative: \(w < 0\)

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