Internal Energy and Enthalpy Changes for an Ideal Gas

From phase rule, it can be shown that the state of single component single phase system, can be specified by two independent variables of the system; and any thermodynamic property can be written as a function of two independent variables. Therefore, let us consider \(U\) as a function of \(T\), and \(V\). \[U = U(T,V)\] Differentiating, \[dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV \tag*{(15)}\] For ideal gas, \(U\) is a function of \(T\) alone. i.e., \[\left(\frac{\partial U}{\partial V}\right)_T = 0\] Hence the above equation and Eqn.(13), Eqn.(15) can be written as, \[dU = \left(\frac{\partial U}{\partial T}\right)_V dT = C_V dT\] i.e., \[dU = C_V dT \qquad \tag{for ideal gas}\]

As like above derivations, by considering \(H\) as a function of \(T\) and \(P\), we can write, \[dH = \left(\frac{\partial H}{\partial T}\right)_P dT + \left(\frac{\partial H}{\partial P}\right)_T dP \tag*{(16)}\] For ideal gases, \(H\) is also a function of \(T\) alone (as like \(U\)). Hence the above equation can be written as, \[dH = C_P dT \qquad \tag{for ideal gas}\] Internal energy and enthalpy are state functions. Hence for any kind of process with ideal gas, we can calculate internal energy and enthalpy changes with the following equations: \[\begin{align*} dU &= C_VdT \tag*{(17)} \\ dH &= C_PdT \tag*{(18)}\end{align*}\] For an ideal gas, it can be proved that \[C_P - C_V = R \tag*{(19)}\]

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