Question

Calculate q, w, ΔE, and ΔH for the process in which 88.0 g of nitrous oxide (N₂O) is cooled from 165°c to 55°c at a constant pressure of 5 atm. Cp = 38.70 J/K.mol Cv = 30.38 J/K.mol

Answer 1

For the cooling of nitrous oxide from 165°C to 55°C at 5 atm, q (heat lost) is -8518 J, using the molar heat capacity at constant pressure (Cp).

Step-by-step explanation:

The molar mass of N2O is 44.013 g/mol, so 88.0 g correspond to 2 moles. The temperature change (ΔT) is 55°C - 165°C = -110°C or -110 K (since the change in Kelvin and Celsius is the same). Heat (q) at constant pressure is calculated with q = nCpΔT = 2 moles * 38.70 J/K.mol * -110 K = -8518 J.

Since Cp (constant pressure heat capacity) is used for processes at constant pressure. For an ideal gas at constant pressure, the work done (w) is w = -PΔV. However, since volume is not provided and we're dealing with a cooling process, we'll assume that if the gas contracts.

The environment does the work on the gas, making it positive from the system’s point of view. The change in internal energy (ΔE) for an ideal gas is q + w, but since we can't find w, we cannot calculate ΔE either. The enthalpy change (ΔH) at constant pressure is equal to qp: ΔH = q = -8518 J.