Thermodynamics | Analysis of Thermodynamic Processes#
1. Isochoric Process#
- In an isochoric process, volume remains constant.
\[V = constant\]
\[dV = 0\]
\[So,\ work\ done\ by\ gas:\]
\[W = 0\]
\[Using\ first\ law\ of\ thermodynamics:\]
\[ΔU = q + W\]
\[q = ΔU\]
2. Isobaric Process#
- In an isobaric process, pressure remains constant.
\[P = constant\]
\[W = - \int\limits_{V_1}^{V_2} P_{ext} dV\]
\[Since,\ P_{ext}\ is\ constant:\]
\[W = -P_{ext} (V_2 - V_1) = -P_{ext}ΔV\]
\[Using\ first\ law\ of\ thermodynamics:\]
\[ΔU = q + W\]
\[ΔU = q - P_{ext}ΔV\]
3. Isothermal Process#
- In an isothermal process, temperature remains constant.
\[T = constant\]
\[dT = 0\]
\[Change\ in\ internal\ energy,\ ΔU:\]
\[ΔU = \int nC_vdT = 0\]
\[Using\ first\ law\ of\ thermodynamics:\]
\[ΔU = q + W\]
\[0 = q + W\]
\[q = -W\]
It means that heat absorbed = work done by the system or work done on the system = heat released.
- Work done by gas in isothermal reversible process:
\[W_{rev} = - \int\limits_{V_1}^{V_2} P_{ext}dV\]
\[W_{rev} = - \int\limits_{V_1}^{V_2} {nRT \over V}dV\]
\[W_{rev} = - nRT \int\limits_{V_1}^{V_2} {1 \over V}dV\]
\[W_{rev} = -nRT\ ln{V_2 \over V_1}\]
We know that P is inversely proportional to V. So, we can also write:
\[W_{rev} = -nRT\ ln{P_1 \over P_2}\]
- Work done by gas in isothermal irreversible process:
\[W_{irr} = -P_{ext} (V_2 - V_1)\]
\[W_{irr} = -P_{ext} ({nRT \over P_2} - {nRT \over P_1})\]
\[W_{irr} = -P_{ext}nRT ({1 \over P_2} - {1 \over P_1})\]
4. Adiabatic Process#
- An adiabatic process is a process in which there is no exchange of heat between system and surrounding.
\[q = 0\]
\[Using\ first\ law\ of\ thermodynamics:\]
\[ΔU = q + W\]
\[ΔU = 0 + W\]
\[ΔU = W\]
- Work done for an adiabatic Process:
\[W = ΔU\]
\[W = \int\limits_{T_1}^{T_2} nC_vdT\]
\[W = nC_v(T_2 - T_1)\]
If the process is irreversible, we can also calculate work done as:
\[W = -P_{ext}(V_2 - V_1)\]
\[W = -P_{ext}({nRT_2 \over P_2} - {nRT_1 \over P_1})\]
- Equation of state for an adiabatic Process:
The equation of state of an adiabatic process can be expressed in three forms as given below -
\[PV^{γ} = constant\ (First\ Form)\]
\[Since,\ V = {nRT \over P}\]
\[V ∝ {T \over P}\]
\[P({T \over P})^γ = constant\]
\[P^{1-γ}T^γ = constant\ (Second\ form)\]
\[Since,\ P = {nRT \over V}\]
\[P ∝ {T \over V}\]
\[PV^γ = constant\]
\[ {T \over V} V^γ = constant\]
\[TV^{γ-1} = constant\ (Third\ form)\]
- Joule Thomson Effect: A gas on adiabatic expansion shows cooling. In other words, the temperature of a gas or a system decreases if the gas is expanded adiabatically.
Proof
In case of expansion, work done is negative:
\[W < 0\]
Since, it is an adiabatic process, q = 0. So, W = ΔU
\[ΔU < 0\]
\[nC_v(T_2 - T_1) < 0\]
\[T_2 < T_1\]
Thus, final temperature, T2 is less than initial temperature T1. Hence, a gas on adiabatic expansion shows cooling.
5. Free Expansion#
- Expansion against zero external pressure is known as free expansion.
- In free expansion, work done is zero.
\[W = - \int\limits_{V_1}^{V_2}P_{ext}dV\]
\[W = 0\ (Since,\ P_{ext} = 0)\]
- No heat is supplied or released during free expansion.
\[q = 0\]
- Applying first law of thermodynamics, we can calculate change in internal energy as:
\[ΔU = q + W\]
\[ΔU = 0\]
\[ \int\limits_{T_2}^{T_1} nC_vdT = 0\]
\[nC_v(T_2 - T_1) = 0\]
\[T_2 = T_1\]
Thus, final temperature = initial temperature
So, in free expansion, temperature does not change.