📂Thermal Physics

Isothermal Expansion of an Ideal Gas

Formulas

The number of moles is $1$, and for an ideal gas undergoing isothermal expansion, the thermal energy is $Q$, the temperature is $T$, the volume before expansion is $V_{1}$, and the volume after expansion is $V_{2}$. Then, the following equation holds:

$$\Delta Q = RT \ln \dfrac{V_{2}}{V_{1}}$$

Explanation

Isothermal expansion refers to expansion under a condition where the temperature does not change. In this case, the change in thermal energy can conveniently be calculated using only the change in volume. Since it is an expansion, $V_{2} > V_{1}$, and from $\Delta Q > 0$, the thermal energy increases, which aligns with intuition.

Proof

The First Law of Thermodynamics

$$d U = \delta Q + \delta W$$

According to the first law of thermodynamics, $dU(T,V)$ is a perfect differential, and the following holds:

$$dU = \dfrac{\partial U}{\partial T} dT + \dfrac{\partial U}{\partial V} dV$$

Average kinetic energy of gas molecules

$$\left< E_{K} \right> = \dfrac{3}{2} k_{B} T$$

Since the average kinetic energy of the gas molecules is as above, the total energy is equal to this times the number of molecules $N$.

$$U = \dfrac{3}{2} N K_{B} T$$

Therefore, we obtain $\dfrac{\partial U}{\partial V} = 0$. And since $\displaystyle C_{V} = \dfrac{\partial U}{\partial T}$(../631), $dU = C_{V} dT$ holds. Also, because the temperature does not change, the following holds:

$$\Delta T = 0 \implies dT = 0 \implies dU = C_{V} dT =0$$

Inserting this into the first law of thermodynamics, we get:

$$0 = \delta Q + \delta W \implies \delta Q = - \delta W$$

However, since $\delta W = - p d V$(../629) holds, we get the following equation:

$$\Delta Q = \int \delta Q = - \int \delta W = \int_{V_{1} }^{V_{2}} p dV$$

If we are dealing with a gas that has moles $n=1$, the ideal gas law is $p = \dfrac{nRT}{V} = \dfrac{RT}{V}$. By substituting this, we get:

$$\Delta Q = \int_{V_{1}}^{V_{2}} \dfrac{RT}{V} dV = RT \ln \dfrac{V_{2}}{V_{1}}$$

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