19.6: The Temperature of a Gas Decreases in a Reversible Adiabatic Expansion

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We can make the same argument for the heat along C. If we do the three processes A and B+C only to a tiny extent we can write:

And now we can integrate from \(V_1\) to \(V_2\) over the reversible adiabatic work along B and from \(T_1\) to \(T_2\) for the reversible isochoric heat along C. To separate the variables we do need to bring the temperature to the right side of the equation.:

The latter expression is valid for a reversible adiabatic expansion of a monatomic ideal gas (say Argon) because we used the \(C_v\) expression for such a system. We can use the gas law \(PV=nRT\) to translate this expression in one that relates pressure and volume see Eq 19.23

We can mathematically show that the temperature of a gas decreases during an adiabatic expansion. Assuming an ideal gas, the internal energy along an adiabatic path is:

\[\begin{split} d\bar{U}&= \delta q+\delta w \\ &= 0-Pd\bar{V}\\ &= -Pd\bar{V} \end{split} \nonumber \]

The constant volume heat capacity is defined as:

\[{\bar{C}}_V=\left(\frac{\partial\bar{U}}{\partial T}\right)_V \nonumber \]

We can rewrite this for internal energy:

\[d\bar{U}={\bar{C}}_VdT \nonumber \]

Combining these two expressions for internal energy, we obtain:

\[{\bar{C}}_VdT=-Pd\bar{V} \nonumber \]

Using the ideal gas law for pressure of an ideal gas:

\[{\bar{C}}_VdT=-\frac{RT}{\bar{V}}d\bar{V} \nonumber \]

Separating variables:

\[\frac{\bar{C}_V}{T}dT=-\frac{R}{\bar{V}}d\bar{V} \nonumber \]

This is an expression for an ideal path along a reversible, adiabatic path that relates temperature to volume. To find our path along a PV surface for an ideal gas, we can start in TV surface and convert to a PV surface. Let's go from (\(T_1,V_1\)) to (\(T_2,V_2\)).

\[\int_{T1}^{T_2}{\frac{\bar{C}_V}{T}dT=-\int_{\bar{V}_1}^{\bar{V}_2}{\frac{R}{\bar{V}}d\bar{V}}} \nonumber \]

\[\bar{C}_V\ln{\left(\frac{T_2}{T_1}\right)}=-R\ln{\left(\frac{{\bar{V}}_2}{{\bar{V}}_1}\right)}=R\ln{\left(\frac{{\bar{V}}_1}{{\bar{V}}_2}\right)} \nonumber \]

\[\ln{\left(\frac{T_2}{T_1}\right)}=\frac{R}{\bar{C}_V}\ln{\left(\frac{\bar{V}_1}{\bar{V}_2}\right)} \nonumber \]

\[\left(\frac{T_2}{T_1}\right)=\left(\frac{\bar{V}_1}{\bar{V}_2}\right)^{\frac{R}{\bar{C}_V}} \nonumber \]

We know that:

\[R={\bar{C}}_P-{\bar{C}}_V \nonumber \]

\[\frac{R}{\bar{C}_V}=\frac{\bar{C}_P-\bar{C}_V}{\bar{C}_V}=\frac{\bar{C}_P}{\bar{C}_V}-1 \nonumber \]

\[\frac{R}{{\bar{C}}_V}=\gamma-1 \nonumber \]

Therefore:

\[\left(\frac{T_2}{T_1}\right)=\left(\frac{\bar{V}_1}{\bar{V}_2}\right)^{\gamma-1} \nonumber \]

This expression shows that volume and temperature are inversely related. That is, as the volume increase from \(V_1\) to \(V_2\), the temperature must decrease from \(T_1\) to \(T_2\).