Question

The energy equation for a reversible non-flow process can be expressed as \(\delta q = du + p\,dv\), where \(q\) is the heat transfer per unit mass, \(u\) is the internal energy per unit mass, \(p\) is the pressure, and \(v\) is the mass specific volume. This energy equation is not in exact differential form. It can be made exact differential by multiplying with the following integrating factor: (The temperature \(T\) is absolute temperature)

• A. ( \dfrac{1}{p} \)
• B. ( \dfrac{1}{v} \)
• C. ( \dfrac{1}{T} \)
• D. ( \dfrac{1}{uT} \)

Hint:

Remember: For reversible processes, dividing heat by temperature gives entropy. This makes the thermodynamic energy equation exact.

Answer 1

The Correct Option is C

For a reversible process, the Clausius expression relates heat transfer to entropy: \[ \delta q_{\text{rev}} = T\,ds \] Given the energy equation: \[ \delta q = du + p\,dv \] Divide the entire equation by temperature \(T\): \[ \frac{\delta q}{T} = \frac{du}{T} + \frac{p\,dv}{T} \] But for a reversible process: \[ \frac{\delta q}{T} = ds \] Thus the right-hand side becomes an exact differential: \[ ds = \frac{du}{T} + \frac{p\,dv}{T} \] Hence the integrating factor that converts \(\delta q = du + p\,dv\) into an exact differential is: \[ \boxed{\frac{1}{T}} \]