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For a constant pressure reversible process, the enthalpy change (ΔH) of the system is
Answer & Solution
Correct Answer:
Option
C
By the definition of $${C_P}$$ we can write $${C_P} = \mathop {\lim }\limits_{\delta T \to 0} {\left( {\frac{{\delta Q}}{{\delta T}}} \right)_P}.$$
Under constant pressure process, and by first law of thermodynamics $$Q = dH.$$
$$\eqalign{ & \Rightarrow {\left( {\frac{{\partial H}}{{\partial T}}} \right)_P} = {C_P} \cr & \Rightarrow \Delta H = \int {{C_P}.dT} \cr} $$
Under constant pressure process, and by first law of thermodynamics $$Q = dH.$$
$$\eqalign{ & \Rightarrow {\left( {\frac{{\partial H}}{{\partial T}}} \right)_P} = {C_P} \cr & \Rightarrow \Delta H = \int {{C_P}.dT} \cr} $$
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