91Ó°ÊÓ

Given that the normal boiling point of butane is \(-0.5^{\circ} \mathrm{C}\), when the cylinder reaches a temperature of \(35^{\circ} \mathrm{C}\), the liquid butane will vaporize. Since the gas has to be contained within the cylinder, the pressure inside the cylinder will build up. Thus, we can expect the pressure in the tank to be greater than atmospheric pressure because the butane is vaporizing, and the confined space causes the pressure to increase. The pressure within the tank depends on how much liquid butane is in it. The more liquid butane there is, the more butane will vaporize and the higher the pressure will become inside the tank.

When a few liters of butane are allowed to escape rapidly from the cylinder, the pressure inside the tank decreases. To maintain equilibrium, more liquid butane has to vaporize to balance the pressure drop. The vaporization of butane requires energy, which is taken from the remaining liquid butane. As a result, the temperature of the remaining liquid butane will decrease after opening the valve because energy is being used for vaporization.

We are given that 250 g of butane has to be vaporized and that its heat of vaporization is 21.3 kJ/mol. To find the heat required, we first need to calculate the number of moles of butane. The molar mass of butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is: \(4(12.01) + 10(1.01) = 58.12 \: \mathrm{g/mol}\). Now, we can find the moles of butane: \(\frac{250 \: \mathrm{g}}{58.12 \: \mathrm{g/mol}} = 4.3 \: \mathrm{mol}\). Using the heat of vaporization, we can calculate the heat required to vaporize the butane: Heat = moles × heat of vaporization = \(4.3 \: \mathrm{mol} \times 21.3 \: \mathrm{kJ/mol} = 91.59 \: \mathrm{kJ}\).

The volume that the vaporized butane occupies can be calculated using the ideal gas law equation, PV = nRT, where - P is the pressure - V is the volume - n is the amount of substance (moles) - R is the ideal gas constant - T is the temperature We're given that the pressure is 755 torr and the temperature is \(35^{\circ} \mathrm{C}\). We need to convert these values to their appropriate units: - 1 atm = 760 torr, so the pressure in atmospheres is \(\frac{755 \: \mathrm{torr}}{760 \: \mathrm{torr/atm}} = 0.9934 \: \mathrm{atm}\). - The temperature in Kelvin is \(35 + 273.15 = 308.15 \: \mathrm{K}\). The ideal gas constant, R, is 0.0821 Lâ‹…atm/molâ‹…K. Now we can plug the values into the ideal gas law equation and solve for the volume: \(V = \frac{nRT}{P} = \frac{(4.3 \: \mathrm{mol})(0.0821 \: \mathrm{L\cdot atm / mol \cdot K})(308.15 \: \mathrm{K})}{0.9934 \: \mathrm{atm}} = 108.85 \: \mathrm{L}\). Thus, the vaporized butane occupies a volume of 108.85 L at 755 torr and \(35^{\circ} \mathrm{C}\).