A cylinder with an interior cross-sectional area of \(70.0 \mathrm{cm}^{2}\) has a moveable piston of mass \(5.40 \mathrm{kg}\) at the top that can move up and down without friction. The cylinder contains $2.25 \times 10^{-3} \mathrm{mol}\( of an ideal gas at \)23.0^{\circ} \mathrm{C} .$ (a) What is the volume of the gas when the piston is in equilibrium? Assume the air pressure outside the cylinder is 1.00 atm. (b) By what factor does the volume change if the gas temperature is raised to \(223.0^{\circ} \mathrm{C}\) and the piston moves until it is again in equilibrium?

At equilibrium, the pressure of the gas inside the cylinder equals the pressure exerted by the atmosphere and the weight of the piston. We can write this equilibrium condition as: \(P_{gas} = P_{atm} + \frac{mg}{A}\), where \(A\) is the cross-sectional area of the cylinder, and \(mg\) is the weight of the piston. We are given the following values: \(P_{atm} = 1.00 \ \mathrm{atm} = 101325 \ \mathrm{Pa}\) Area, \(A = 70.0 \ \mathrm{cm^2} = 7.0 \times 10^{-3} \ \mathrm{m^2}\) Mass of piston, \(m = 5.40 \ \mathrm{kg}\) Gravitational acceleration, \(g = 9.81 \ \mathrm{m/s^2}\) We can now calculate the pressure of the gas inside the cylinder as follows. \(P_{gas} = 101325 \ \mathrm{Pa} + \frac{(5.40 \ \mathrm{kg})(9.81 \ \mathrm{m/s^2})}{7.0 \times 10^{-3} \ \mathrm{m^2}}\)

We can now use the ideal gas law to find the volume: \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature. We are given: Amount of gas, \(n = 2.25 \times 10^{-3} \ \mathrm{mol}\) Gas constant, \(R = 8.314 \ \mathrm{J/(mol \cdot K)}\) Initial temperature, \(T_1 = 23.0 + 273.15 = 296.15 \ \mathrm{K}\) (convert to Kelvin) Rearrange the ideal gas law to find the volume: \(V_1 = \frac{nRT_1}{P_{gas}}\)

First, find the final temperature: \(T_2 = 223.0 + 273.15 = 496.15 \ \mathrm{K}\) Using the ideal gas law again, we can find the final volume \(V_2\): \(V_2 = \frac{nRT_2}{P_{gas}}\) Now we can calculate the factor by which the volume changes: \(\textrm{Factor} = \frac{V_2}{V_1}\)