Question

Write the mathematical equation for the first law of thermodynamics for Isothermal process, Adiabatic process. Derive the relationship between pH and pOH.

Hint:

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Answer 1

Step 1: First Law of Thermodynamics: The first law is \( \Delta U = q + w \), where \( \Delta U \) is the change in internal energy, \( q \) is heat, and \( w \) is work (work done on the system). - Isothermal Process: For an ideal gas, \( \Delta U = nC_v \Delta T \). Since \( \Delta T = 0 \) (constant temperature), \( \Delta U = 0 \). Thus: \[ \Delta U = 0 \quad \Rightarrow \quad q = -w. \] For expansion, work \( w = -P \Delta V \), so \( q = P \Delta V \).
- Adiabatic Process: No heat exchange (\( q = 0 \)). Thus: \[ \Delta U = w. \] For an ideal gas, \( \Delta U = nC_v \Delta T \), and work \( w = -P \Delta V \).
Step 2: Derive pH + pOH Relationship: For water at 25°C, the ionization constant is \( K_w = [\text{H}^+] [\text{OH}^-] = 10^{-14} \). \[ \text{pH} = -\log [\text{H}^+], \quad \text{pOH} = -\log [\text{OH}^-]. \] \[ \log ([\text{H}^+] [\text{OH}^-]) = \log (10^{-14}) = -14. \] \[ \log [\text{H}^+] + \log [\text{OH}^-] = -14 \quad \Rightarrow \quad -\text{pH} - \text{pOH} = -14 \quad \Rightarrow \quad \text{pH} + \text{pOH} = 14. \]