Name: Spontaneity and Entropy
Uploaded: 2019-07-19
Duration: 34 min 9 s
Description: What is the relation between spontaneity and entropy of a process? Watch this video to find the answer. It is an important topic from thermodynamics.

Question

Show that the heat absorbed at constant volume is equal to the increase in internal energy of the system whereas that at constant pressure is equal to the increase in enthalpy of the system.

Accepted Answer

The process which is carried out at constant volume is known as isochoric process. In isochoric process $PdV = 0$ . Hence, from the equation of first law of thermodynamics,

$∆ U = q + P ∆ V ∆ U = q_{v}$

Where $q_{v}$ is the heat absorbed at constant volume. Thus, heat supplied to a system at constant volume is used in increasing internal energy.

The process which is carried out at constant pressure is known as isobaric process. In isobaric process $∆ P = 0$ .Consider a system showing an increase in volume from $V_{1} to V_{2}$ at constant pressure $P$ , during absorption of heat $q$ The work done by the system is $w = - P ∆ V$ .

Hence, from the first law of thermodynamics,

$∆ U = q + w q_{p} = ∆ U - (- P ∆ V) = ∆ U + P ∆ V q_{p} = U_{2} - U_{1} + P (V_{2} - V_{1}) = (U_{2} + {PV}_{2}) - (U_{1} + {PV}_{1}) q_{p} = H_{2} - H_{1} = ∆ H$

Where $H_{2}, H_{1}$ are enthalpies of the system in initial state and final state. Thus, heat supplied to a system at constant pressure is used in increasing enthalpy.

Show that the heat absorbed at constant volume is equal to the increase in internal energy of the system whereas that at constant pressure is equal to the increase in enthalpy of the system.

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