Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) are gaseous fuels with enthalpies of combustion of \(-49.9 \mathrm{kJ} / \mathrm{g}\) and $-49.5 \mathrm{kJ} / \mathrm{g}$ , respectively. Compare the energy available from the combustion of a given volume of acetylene to the combustion energy from the same volume of butane at the same temperature and pressure.

We can use the ideal gas law, \(PV = nRT\), where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is the temperature. Since pressure, volume, and temperature are the same for both acetylene and butane, we can set up an equation to find the ratio of the number of moles of acetylene to butane: \(n_{C_2H_2} = \frac{P \times V}{RT}\) and \(n_{C_4H_{10}} = \frac{P \times V}{RT}\) Dividing the two equations, we get: \(\frac{n_{C_2H_2}}{n_{C_4H_{10}}} = \frac{M_{C_4H_{10}}}{M_{C_2H_2}}\) where \(M_{C_2H_2}\) and \(M_{C_4H_{10}}\) are the molar masses of acetylene and butane, respectively. To find the molar masses, we will use the fact that carbon has a molar mass of 12 g/mol and hydrogen has a molar mass of 1 g/mol.

For acetylene, \(C_2H_2\), there are 2 carbon atoms and 2 hydrogen atoms: \(M_{C_2H_2} = 2 \times 12 + 2 \times 1 = 24 + 2 = 26 \: g/mol\) For butane, \(C_4H_{10}\), there are 4 carbon atoms and 10 hydrogen atoms: \(M_{C_4H_{10}} = 4 \times 12 + 10 \times 1 = 48 + 10 = 58 \: g/mol\)

Now, we can use the molar masses to find the ratio of the number of moles of acetylene to butane: \(\frac{n_{C_2H_2}}{n_{C_4H_{10}}} = \frac{M_{C_4H_{10}}}{M_{C_2H_2}} = \frac{58}{26} \approx 2.23\)

Next, we will use the enthalpies of combustion and the number of moles to find the combustion energies for acetylene and butane: Energy from acetylene = -49.9 kJ/g * 26 g/mol * \(n_{C_2H_2}\) Energy from butane = -49.5 kJ/g * 58 g/mol * \(n_{C_4H_{10}}\)

Finally, we will compare the combustion energies from the same volume of butane and acetylene using the ratio of the number of moles: Energy ratio = \(\frac{\text{Energy from acetylene}}{\text{Energy from butane}} = \frac{-49.9 \times 26 \times n_{C_2H_2}}{-49.5 \times 58 \times n_{C_4H_{10}}} \times \frac{2.23}{1}\) Energy ratio = \(\frac{-49.9 \times 26}{-49.5 \times 58} \times 2.23 \approx 1.07\) This means that the combustion of a given volume of acetylene produces about 1.07 times more energy than the combustion of the same volume of butane at the same temperature and pressure.