1 .00 mol of molecular nitrogen gas expands in volume very quickly, so no heat is exchanged with the environment during the process. If the volume increases from \(1.00 \mathrm{~L}\) to \(1.50 \mathrm{~L},\) determine the work done on the environment if the gas's temperature dropped from \(22.0^{\circ} \mathrm{C}\) to \(18.0^{\circ} \mathrm{C}\). Assume ideal gas behavior.

First, convert the initial and final temperatures from Celsius to Kelvin: Initial temperature: \(T_1 = 22.0^{\circ}C + 273.15 = 295.15 K\) Final temperature: \(T_2 = 18.0^{\circ}C + 273.15 = 291.15 K\)

From the data given, we know the initial and final volumes and temperatures. We can use the Ideal Gas Law to find the initial and final pressures of the gas. Recall the Ideal Gas Law: \(PV = nRT\) Initial pressure: \(P_1 = \frac{nRT_1}{V_1} = \frac{(1.00\mathrm{~mol})(8.314\mathrm{~J/mol\cdot K})(295.15\mathrm{~K})}{1.00\mathrm{~L}}\cdot\frac{1000\mathrm{~L}}{1\mathrm{~m^3}} = 24469.23\mathrm{~Pa}\) Final pressure: \(P_2 = \frac{nRT_2}{V_2} = \frac{(1.00\mathrm{~mol})(8.314\mathrm{~J/mol\cdot K})(291.15\mathrm{~K})}{1.50\mathrm{~L}}\cdot\frac{1000\mathrm{~L}}{1\mathrm{~m^3}} = 16140.75\mathrm{~Pa}\)

Now we will use the first law of thermodynamics, which states that the change in internal energy of the system is equal to the heat added to the system minus the work done by the system, or \(\Delta U = Q - W\). Since no heat is exchanged with the environment during the process (Q = 0), we can rewrite the equation as \(\Delta U = -W\). For an ideal gas, the change in internal energy is directly related to the change in temperature and can be represented by the equation \(\Delta U = nC_v\Delta T\). Here, \(C_v\) is the molar heat capacity at constant volume for molecular nitrogen, which is approximately 20.8 J/mol∙K. Calculate the change in internal energy: \(\Delta U = nC_v\Delta T = (1.00\mathrm{~mol})(20.8\mathrm{~J/mol\cdot K})(291.15\mathrm{~K} - 295.15\mathrm{~K}) = -83.20\mathrm{~J}\) Now, using the first law of thermodynamics equation, we can determine the work done on the environment: \(W = -\Delta U = -(-83.20\mathrm{~J}) = 83.20\mathrm{~J}\) So, the work done on the environment during this quick expansion process is 83.20 J.