Question

A quantity of ideal monatomic gas consists of \(n\) moles initially at temperature \(T_{1}\). The pressure and volume are then slowly doubled in such a manner as to trace out a straight line on the \(p V\) diagram. In terms of \(n, R\), and \(T_{1}\), find \((\) a) \(W,(b)\) \(\Delta E_{\text {int }}\), and \((c) Q .(d)\) If one were to define an equivalent specific heat for this process, what would be its value?

Short answer

The work done by the gas is \(W = p_{1}V_{1}\), the change in internal energy is \(Δ E_{int} = n * 3/2 * R * T_{1}\), the heat transferred is \(Q = p_{1}V_{1} + n * 3/2 * R * T_{1}\) and the equivalent specific heat for this process is \(C_{eq} = 5/2 R\).

Step-by-step solution

Find the Work Done (W)

Let's consider that the initial volume is \(V_{1}\) and final volume is \(V_{2} = 2V_{1}\). Similarly, initial pressure is \(p_{1}\) and final pressure is \(p_{2} = 2p_{1}\). The work \(W\) done by the gas during expansion is given by the integral of \(p dV\) from \(V_{1}\) to \(V_{2}\) under the process path, which forms a straight line on the \(pV\) diagram. The work can therefore be calculated as \(W = \int_{V_{1}}^{V_{2}} pdV\). The line equation on the \(pV\) graph is \(p = p_{1} + (p_{1}/V_{1})V\). After integral calculation, one finds \(W = p_{1}V_{1}\).

Calculate Change in Internal Energy (ΔEint)

The change in internal energy for an ideal gas is given by \(Δ E_{int} = nC_{v}ΔT\), where \(n\) is the number of moles, \(C_{v}\) is the heat capacity at constant volume for monatomic gas, and ΔT is the change in temperature. Since we are given initial temperature, \(T_{1}\), and know that \(C_{v} = 3/2 R\) for an ideal monatomic gas, we need to determine the final temperature \(T_{2}\), after the process. By using the formula for ideal gas \(p_{2}V_{2} = nRT_{2}\), we find that \(T_{2} = 2T_{1}\). Substituting these values we get \(Δ E_{int} = nC_{v}(T_{2} - T_{1}) = n * 3/2 * R * T_{1}\).

Find Heat Transferred (Q)

The first law of thermodynamics states that the heat supplied \(Q\) to the gas is equal to the sum of the work done by the gas \(W\) and the change in its internal energy \(ΔE_{int}\). Hence, one can write \(Q = W + ΔE_{int}\). Substituting the earlier obtained values, we find \(Q = p_{1}V_{1} + n * 3/2 * R * T_{1}\).

Define the Equivalent Specific Heat

The equivalent specific heat is defined as \(C_{eq} = Q / (nΔT)\). Substituting the earlier obtained equations for \(Q\) and \(ΔT = T_{2} - T_{1} = T_{1}\), we find \(C_{eq} = 2R + 3/2 R = 5/2 R\).

Key concepts

First Law of Thermodynamics

The first law of thermodynamics is a statement of conservation of energy, specifically for thermodynamic systems. It establishes the relationship between heat transferred to a system, the work done by the system, and the change in internal energy of the system. Mathematically, it can be expressed as \[\Delta U = Q - W\], where \(\Delta U\) is the change in internal energy, \(Q\) is the heat supplied to the system and \(W\) is the work done by the system. In our example with the monatomic gas, the law is applied to relate the heat added to the gas \(Q\), the work done \(W\) during the expansion, and the resulting change in internal energy \(\Delta U\).

Understanding this law is crucial as it forms the basis for many calculations in thermodynamics, such as those in the provided exercise. It helps us predict how energy will flow in a given physical process and lets us account for the energy transformations that occur. Clear comprehension of this principle allows students to solve problems involving heat and work transactions within thermodynamic systems effectively.

Work Done by Gas

When a gas expands or contracts, it can do work on its surroundings. The work done by a gas during a thermodynamic process depends on the path taken in the process, which can be visualized on a pressure-volume \(pV\) diagram. In the case of the monatomic gas from the exercise, the work done by the gas, \(W\), is calculated by integrating the pressure with respect to the volume as the gas expands.

As seen in the solution, the work was found to be \(W = p_1V_1\) after integrating along a straight line path on the \(pV\) diagram. This is a crucial concept for students to understand, as it shows how the external parameters of the gas system (pressure and volume) are linked directly to the mechanical energy (work) exchanged with the surroundings. Recognizing that work is an energy transfer allows students to appreciate that when a gas expands against an external pressure, it must use energy to do so.

Change in Internal Energy

The internal energy of an ideal gas is related to its temperature and the number of moles of gas present. It is a measure of the total kinetic energy of all the molecules in the system. When dealing with monatomic ideal gases, the change in internal energy \(\Delta E_{\text{int}}\) can be calculated using the expression \(\Delta E_{\text{int}} = nC_v\Delta T\), where \(n\) is the number of moles, \(C_v\) is the heat capacity at constant volume, and \(\Delta T\) is the temperature change.

In the example provided, after doubling the temperature during the process, we compute the change in internal energy as \(\Delta E_{\text{int}} = n \frac{3}{2} R T_1\), where \(R\) is the gas constant. Students need to grasp that the internal energy change is independent of the path taken and depends solely on the initial and final states of the system. This concept emphasizes the link between the microscopic motion of particles and the macroscopic properties of the gas.

Heat Capacity

Heat capacity is a physical quantity that represents the amount of heat required to change the temperature of a substance by a given amount. For a monatomic ideal gas, the heat capacity at constant volume \(C_v\) can be expressed as \(C_v = \frac{3}{2} R\), where \(R\) is the ideal gas constant. The heat capacity tells us about the gas's ability to store energy.

To further explore this concept, the solution shows how to find the equivalent specific heat \(C_{eq}\) for the non-standard process described in the exercise, yielding \(C_{eq} = \frac{5}{2} R\). Students often find the concept of heat capacity to be abstract because it involves understanding how molecular structure affects a substance's thermal properties. However, it is a fundamental aspect of material characteristics, which has real-world implications in understanding how different substances heat up and cool down.