Question

One mole of Ar initially at \(310 .\) K undergoes an adiabatic expansion against a pressure \(P_{e x t e r n a l}=0\) from a volume of \(8.5 \mathrm{L}\) to a volume of \(82.0 \mathrm{L}\). Calculate the final temperature using the ideal gas and van der Waals equations of state. Assume \(C_{V, m}=3 R / 2\).

Short answer

The final temperature of argon gas using the ideal gas equation is approximately 83.77 K, and using the van der Waals equation is approximately 83.88 K.

Step-by-step solution

Adiabatic process relationship for ideal gas

In an adiabatic process, for an ideal gas, we have the following relation between initial and final states: \(T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}\) where γ is the ratio of heat capacities, Cp,m/Cv,m. For a monatomic gas, like argon, γ = 1 + 2/3 = 5/3

Calculate the final temperature for the ideal gas

Use the adiabatic process relationship from Step 1 and the given data to calculate the final temperature, T2. We have \(T_1 = 310 \, K\), \(V_1 = 8.5 \, L\), \(V_2 = 82.0 \, L\), and \(\gamma = \frac{5}{3}\). \(310 \times (8.5)^{\frac{2}{3}} = T_2 \times (82.0)^{\frac{2}{3}}\) Solve for T2: \(T_2 = 310 \times \frac{(8.5)^{\frac{2}{3}}}{(82.0)^{\frac{2}{3}}} = 83.772 \, K\) The final temperature of argon gas using the ideal gas equation is approximately 83.77 K. ###van der Waals Equation### The van der Waals equation for a real gas is given by: \[\left[P + a \left(\frac{n^2}{V^2}\right)\right] \left(V - nb\right) = nRT\] Where P is pressure, V is volume, T is temperature, n is the number of moles, a and b are van der Waals constants. Given the process is an adiabatic expansion against 0 pressure, we can rewrite the equation as: \[a \left(\frac{n^2}{V^2}\right) (V - nb) = nRT\] For argon, the van der Waals constants are given by: \(a = 1.355\, L^2 \,\mathrm{atm} \,\mathrm{mol}^{-2}\), \(b = 0.03201\, L \,\mathrm{mol}^{-1}\)

Calculate the work done during the adiabatic process

Since the expansion is against 0 external pressure, the work done is simply the change in internal energy. Using the first law of thermodynamics, we have: \(W = nC_{V,m}(T_2 - T_1)\) where \(C_{V, m} = \frac{3}{2}R = 12.471 \, \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}} \)

Calculate the final temperature for the van der Waals gas

Use the work done from Step 3 and substitute it into the van der Waals equation. Solve for the final temperature, T2. \(12.471 \, \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}} (T_2 - 310 \, K) = 1.355\, L^2 \,\mathrm{atm} \,\mathrm{mol}^{-2} \left(\frac{1}{82.0 \, L^2} - \frac{1}{8.5 \, L^2}\right) (1)(82.0 \, L - 0.03201 \, L)\) \(12.471(T_2 - 310) = 0.018197\, \mathrm{atm} \,\mathrm{L} \,\mathrm{mol}^{-1} \cdot 81.968\, L \) Now, we need to convert the units from L atm to J: 1 L atm = 101.325 J/mol \(12.471(T_2 - 310) = 0.018197 \cdot 81.968 \cdot 101.325\) Solve for T2: \(T_2 = 310 + \frac{0.018197 \cdot 81.968 \cdot 101.325}{12.471} = 83.875 \, K\) The final temperature of argon gas using the van der Waals equation is approximately 83.88 K.

Key concepts

Ideal Gas Equation

Understanding the ideal gas law is critical for comprehending various states of a gas under different conditions. The ideal gas equation is represented as \( PV = nRT \) where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. This equation is based on the assumption that the gas particles are point particles with no volume and no intermolecular forces.

Using the ideal gas equation, we can relate changes in pressure, volume, and temperature when the number of particles remains constant. For instance, in an adiabatic process, as seen in the exercise, where no heat is exchanged with the environment, the ideal gas law combines with the adiabatic process relationship to determine changes in temperature as the gas expands or contracts.

In the given exercise, the volume increases, so according to the ideal gas law, if the number of moles and external pressure remain constant, the temperature should decrease, reflecting an inverse relationship between volume and temperature in an adiabatic expansion.

van der Waals Equation

The van der Waals equation is an adjusted version of the ideal gas law that accounts for the volume of gas particles and the intermolecular forces between them. This equation is more accurate for real gases, especially under high pressure and low temperature. It is expressed as \[\left(P + a \left(\frac{n^2}{V^2}\right)\right) \left(V - nb\right) = nRT\] where \(a\) and \(b\) are van der Waals constants that differ for each gas.

In the context of the exercise, using this equation allows for a more precise calculation of the final temperature of argon after adiabatic expansion, considering the real behavior of gases. However, the effect of \(a\) and \(b\) in such a process is relatively small due to the zero external pressure, simplifying the equation and bringing the van der Waals result closer to that obtained using the ideal gas law.

Heat Capacities

The heat capacity of a substance is a measure of how much heat energy is required to raise its temperature by one degree. There are two types of heat capacities: specific heat capacity at constant volume (\(C_{V}\)) and constant pressure (\(C_{P}\)). The ratio of these two values (\(\gamma = \frac{C_{P}}{C_{V}}\)) is crucial in understanding adiabatic processes.

Monatomic gases, such as argon in our exercise, have fixed heat capacities because they have only translational kinetic energy. The given heat capacity, \(C_{V,m} = 3R/2\), reflects this simple energy structure. When considering the adiabatic process, the heat capacity at constant volume helps us determine the internal energy change, which in turn is used to find the final temperature after expansion.

Monatomic Gas Behavior

Monatomic gases, like argon, neon, and helium, consist of single atoms. These gases follow the simplest thermodynamic behavior since they only possess translational kinetic energy, and conform closely to the ideal gas model under many conditions.

For monatomic gases, due to their atomic structure and lack of molecular bonds, the heat capacity values are nearly constant, allowing for straightforward calculations of changes in thermodynamic state properties. The exercise uses these standard heat capacities and the monatomic nature of argon to apply adiabatic equations correctly and predict the final temperature after expansion. The predictable behavior of monatomic gases makes them a perfect starting point for students to learn gas laws and thermodynamics.