## Adiabatic Expansion This is a process in which there is no heat transfer. Processes may be adiabatic when they are carried out in insulated systems. Some processes are so rapid that there is no enough time for any significant heat transfer to occur. Such processes may be assumed to be adiabatic. - The expansion of gas in an aerosol spray can is adiabatic and this leads to the can cooling as it sprays (though often there is also evaporation which adds to the cooling) - Cloud formation on adiabatic expansion induced by shockwaves - A fog or smoke formed when opening a champagne bottle - A bicycle pump gets warm due to rapid pumping (compression) When a gas expands rapidly, there may be no time for heat transfer to take place. The expansion is then adiabatic. Pressure, volume and temperature will all change and work will be done. Using the ideal gas law, for a general expansion when $p$, $V$ and $T$ may change. Initially: $ PV=mR_{gas}T $ and finally: $ (P+\Delta P)(V+\Delta V)=mR_{gas}(T+\Delta T) $ Expanding the brackets, ignoring the second order small quantities and subtracting the first equation gives: $ PV+P\Delta V+V\Delta P+\Delta P\Delta V=mR_{gas}T+mR_{gas}\Delta T $ $ P\Delta V+V\Delta P=mR_{gas}\Delta T $ Using $ R_{gas}=c_{v}(\gamma-1) $ Where: - $c_{v}$ is specific heat at constant volume - $\gamma$ is ?? We have: $ P\Delta V+V\Delta P=mc_{v}(\gamma-1)\Delta T $ In an adiabatic expansion, $\Delta Q=0$ so the First Law $\Delta Q=\Delta W+\Delta U$ becomes: $ \Delta W+\Delta U=0 \qquad \text{or} \qquad \Delta U=-\Delta W $ Using the previous results, we have $\Delta U=mc_{v}\Delta T$ and $\Delta W=p\Delta V$, so $\Delta U=-\Delta W$ becomes: $ mc_{v}\Delta T=-p\Delta V $ ### Example 1 1 litre of air at 20C and 100kPa expands adiabatically to a volume of 2 litres. What is its final temperature and pressure? For air, $\gamma=1.4$. Initially: - $p_{1}=100\times_{1}0^3$ - $V_{1}=0.001 m^3$ - $T_{1}=20+273=293K$ Given: $ p_{1}V_{1}^{\gamma}=p_{2}V_{2}^{\gamma} $ $ p_{2}=\frac{p_{1}V_{1}^{\gamma}}{V_{2}^{\gamma}} $ $ p_{2}=\frac{(100\times 10^3)(0.001)^{1.4}}{(0.002)^{1.4}}=37829\ Pa $ For $T_{2}$, we use: $ T_{1}V_{1}^{\gamma-1}=T_{2}V_{2}^{\gamma-1} $ $ T_{2}=\frac{T_{1}V_{1}^{\gamma-1}}{V_{2}^{\gamma-1}} $ $ T_{2}=\frac{(293)(0.001)^{1.4-1}}{(0.002)^{1.4-1}}=222.1\ K $ Past exam question - $p_{1}=3\times 10^5$ - $V_{1}=0.1m^3$ - $T_1=973K$ $ p_{2}=64.4\times_{1}0^{3} $ $ T_{2}=354C $ $ P_{3}=193.32\times 10^3 $