91Ó°ÊÓ

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: \(\textrm{Factor} = \frac{V_2}{V_1}\)