Final answer:
The change in entropy for an ideal gas when the temperature and pressure vary is calculated using the first and second laws of thermodynamics. During reversible processes, formulas incorporating volume, temperature, and heat capacities at constant volume or pressure are used to compute the entropy change. These calculations depend on whether the processes are isothermal or occur at varying temperatures.
Step-by-step explanation:
To determine the change in entropy for an ideal gas when its temperature and pressure vary, we need to consider the reversible processes involving the ideal gas.
The change in entropy (\(\Delta S\)) of an ideal gas during a reversible process can be expressed by combining the first and second laws of thermodynamics. For an isothermal, reversible expansion or compression, the change in entropy is given by:
\(\Delta S = nR \ln(\frac{V_2}{V_1})\)
where \(n\) is the number of moles of gas, \(R\) is the universal gas constant, and \(V_1\) and \(V_2\) are the initial and final volumes of the gas. It's important to note that isothermal means that the temperature (\(T\)) remains constant.
If the process is not isothermal, but instead occurs at varying temperatures, we can integrate the infinitesimal change in entropy \(dS\) to calculate the total entropy change:
\(\Delta S = \int_{S_1}^{S_2}dS = \int_{T_1}^{T_2}\frac{dQ_{rev}}{T}\)
where \(dQ_{rev}\) is the reversible heat transfer and \(T\) is the absolute temperature (in Kelvin) at each infinitesimal step of the process. For an ideal gas, when heat is added at constant volume, the change in entropy can be derived from:
\(\Delta S = nC_v \ln(\frac{T_2}{T_1}) + nR \ln(\frac{V_2}{V_1})\)
and when heat is added at constant pressure, it is:
\(\Delta S = nC_p \ln(\frac{T_2}{T_1})\)
Here \(C_p\) and \(C_v\) are the heat capacities at constant pressure and volume, respectively.