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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