4.5 Adiabatic process

Adiabatic process is defined as one with no heat transfer into or out of a system : $Q = 0$. We can prevent heat flow either by surrounding the system with thermally insulating material or carrying out the process so quickly that there is not enough time for appreciable heat flow. From the first law we find thatfor every adiabatic process,

$$\begin{equation} \begin{aligned} W = - \Delta U\ \ (as\ Q = 0) \\ = - n{C_V}\Delta T \\ = - n{C_V}\left( {{T_f} - {T_i}} \right) \\ = n\left( {\frac{R}{{\gamma - 1}}} \right)\left( {{T_i} - {T_f}} \right)(as\ {C_V} = \frac{R}{{\gamma - 1}}) \\ = \frac{{{P_i}{V_i} - {P_f}{V_f}}}{{\gamma - 1}}(as\ nRT = PV) \\\end{aligned} \end{equation} $$

Thus in an adiabatic process,$$Q = 0\ \ \ and\ \ \ W = - \Delta U = \frac{{{P_i}{V_i} - {P_f}{V_f}}}{{\gamma - 1}}$$

In an adiabatic process $W = - \Delta U$. Therefore, if the work done by the gas is positive( volume of the gas is increasing ), then $\Delta U$ will be negative. That is $U$ and hence $T$ will decrease. The cooling of air can be experienced practically during bursting of a tyre. The process is so fast that it can be assumed as adiabatic. As the gas expands. Therefore, it cools. On the other the compression stroke in an internal combustion engine is approximately an adiabatic process.The temperature rises as the air fuel mixture in the cylinder is compressed.

Question 14. An ideal monoatomic gas at $33 K$ expands adiabatically to twice its volume. What is the final temperature?

Solution: For an ideal monoatomic gas,$$\gamma = \frac{5}{3}$$In an adiabatic process,$$\begin{equation} \begin{aligned} T{V^{\gamma - 1}} = cons\tan t \\ \therefore \;{T_f}{V_f}^{\gamma - 1} = {T_i}{V_i}^{\gamma - 1} \\ {T_f} = {T_i}{\left( {\frac{{{V_i}}}{{{V_f}}}} \right)^{\gamma - 1}} \\ = 300{\left( {\frac{1}{2}} \right)^{\frac{5}{3} - 1}} = 189\;K \\\end{aligned} \end{equation} $$

In an adiabatic process $dQ$=0$$\begin{equation} \begin{aligned} dW = - dU \\ PdV = - {C_V}dT(for\ n = 1) \\ dT = - \frac{{PdV}}{{{C_V}}}\ \ \ ....(1) \\\end{aligned} \end{equation} $$

Also, For 1 mole of an ideal gas,$$\begin{equation} \begin{aligned} d\left( {PV} \right) = d\left( {RT} \right) \\ PdV + VdP = RdT \\ dT = \frac{{PdV + VdP}}{R}\ \ \ ....(2) \\\end{aligned} \end{equation} $$

From equations (1) and (2)$$\begin{equation} \begin{aligned} {C_V}VdP + \left( {{C_V} + R} \right)PdV = 0 \\ {C_V}VdP + {C_P}PdV = 0 \\\end{aligned} \end{equation} $$or

Dividing this equation by $PV$, we are laft with$$\begin{equation} \begin{aligned} {C_V}\frac{{dP}}{P} + {C_P}\frac{{dV}}{V} = 0 \\ or\ \ \ \frac{{dP}}{P} + \gamma \frac{{dV}}{V} = 0 \\ or\ \ \ \int {\frac{{dP}}{P} + \gamma \int {\frac{{dV}}{V} = 0} } \\ or\ \ \ \ln \left( P \right) + \gamma \ln \left( V \right) = constant\end{aligned} \end{equation} $$

We can write this equation in the form$$P{V^\gamma } = constant$$

This equation is the condition that must be obeyed by an ideal gas in an adiabatic process. For example, if an ideal gas makes an adiabatic transition from a state with pressure and volume ${{P_i}}$ and ${{V_i}}$ to a state with ${{P_f}}$ and ${{V_f}}$, then$${P_i}{V_i}^\gamma = {P_f}V_f^\gamma $$ The equation $P{V^\gamma } = constant$ can be written in terms of other pairs of thermodynamic variables by combining it with the ideal gas law $\left( {PV = nRT} \right)$ . In doing so we wil find that,$$T{V^{\gamma - 1}} = constant\ \ \ and\ \ \ {T^\gamma }{P^{\gamma - 1}} = constant$$

In an adiabatic process ($P{V^\gamma } = constant$), the slope of $PV$ diagram at any point is $$\begin{equation} \begin{aligned} \frac{{dP}}{{dV}} = \frac{d}{{dV}}\left( {\frac{{cons\tan t}}{{{V^\gamma }}}} \right) = - \gamma \left( {\frac{P}{V}} \right) \\ {(slope)_{adiabatic}} = - \gamma \left( {\frac{P}{V}} \right) \\\end{aligned} \end{equation} $$ Similarly in an isothermal process$\left( {PV = constant} \right)$, the slope of $PV$ diagram at any point is,

$$\frac{{dP}}{{dV}} = \frac{d}{{dV}}\left( {\frac{{cons\tan t}}{V}} \right) = - \left( {\frac{P}{V}} \right)$$$$or\ \ \ {(slope)_{isothermal}} = - \left( {\frac{P}{V}} \right)$$

Because $\gamma > 1$ ,the isothermal curve is not as steep as that for the adiabatic expansion.