Question

For ideal gases, the development of the constant pressure specific heat yields $$\left(\frac{\partial h}{\partial P}\right)_{T}=0$$ Prove this by using the definitions of pressure and temperature, \(T=(\partial u / \partial s)_{v}\) and \(P=-(\partial u / \partial v)_{s}\).

Short answer

**Prompt:** As a teacher, create a short answer question based on the information given in the solution above: Question: For an ideal gas, enthalpy (h) can be expressed as a function of internal energy (u), pressure (P), and specific volume (v). Provide the formula for enthalpy in terms of these variables.

Step-by-step solution

Identify the necessary properties and relations

For an ideal gas, we can express the enthalpy \((h)\) as: $$h = u+Pv$$ where \(u\) is the internal energy, \(P\) is the pressure, and \(v\) is the specific volume. We are given that the temperature, \(T\), is defined as: $$T = \left(\frac{\partial u}{\partial s}\right)_{v}$$ And the pressure, \(P\), is defined as: $$P = -\left(\frac{\partial u}{\partial v}\right)_{s}$$ We need to prove that: $$\left(\frac{\partial h}{\partial P}\right)_{T} = 0$$

Differentiate enthalpy with respect to pressure

Apply the chain rule to differentiate enthalpy \((h)\) with respect to pressure \((P)\) while keeping the temperature \((T)\) constant: $$\frac{\partial h}{\partial P} = \frac{\partial (u+Pv)}{\partial P}$$ Using chain rule: $$\frac{\partial h}{\partial P} = \frac{\partial u}{\partial P} + \frac{\partial (Pv)}{\partial P}$$

Use the given definitions of pressure and temperature

Use the given definitions of temperature \(T=(\partial u / \partial s)_{v}\) and pressure \(P=-(\partial u / \partial v)_{s}\) to substitute in the previous expression: $$\frac{\partial h}{\partial P} = \frac{\partial u}{-\partial (\frac{\partial u}{\partial v})} + \frac{\partial (Pv)}{\partial P}$$ Now, we can simplify the expression: $$\frac{\partial h}{\partial P} = -\frac{\partial u}{\partial (\frac{\partial u}{\partial v})} + \frac{\partial (Pv)}{\partial P}$$

Evaluate the derivatives and simplify

Now, we evaluate each of the partial derivatives in the expression: $$-\frac{\partial u}{\partial (\frac{\partial u}{\partial v})} = 0$$ $$\frac{\partial (Pv)}{\partial P} = v$$ Substitute these values into the expression: $$\frac{\partial h}{\partial P} = 0 + v$$

Conclusion

Since the derivative of enthalpy with respect to pressure is non-zero, the expression: $$\left(\frac{\partial h}{\partial P}\right)_{T} = 0$$ cannot be proved using the given definitions of pressure and temperature provided. There might be a mistake in the given exercise statement or a need for further assumptions to prove the statement true.