Question

A cylinder with a frictionless, movable piston like that shown in Fig. 19.5 contains a quantity of helium gas. Initially the gas is at 1.00 \(\times\) 10\(^5\) Pa and 300 K and occupies a volume of 1.50 L. The gas then undergoes two processes. In the first, the gas is heated and the piston is allowed to move to keep the temperature at 300 K. This continues until the pressure reaches 2.50 \(\times\) 10\(^4\) Pa. In the second process, the gas is compressed at constant pressure until it returns to its original volume of 1.50 L. Assume that the gas may be treated as ideal. (a) In a \(pV\)-diagram, show both processes. (b) Find the volume of the gas at the end of the first process, and the pressure and temperature at the end of the second process. (c) Find the total work done by the gas during both processes. (d) What would you have to do to the gas to return it to its original pressure and temperature?

Short answer

Volume at end of first process is 6.00 L; final temperature is 75 K. Total work done is -8,500 J.

Step-by-step solution

Understand Given Data

We begin by noting the initial conditions: The initial pressure \(P_i\) is \(1.00 \times 10^5 \text{ Pa}\), the initial temperature \(T_i\) is 300 K, and the initial volume \(V_i\) is 1.50 L. The gas is treated as ideal.

Process 1 - Isothermal Expansion

The first process is isothermal (constant temperature at 300 K). According to the ideal gas law, \(P_i V_i = P_f V_f\) for an isothermal process. We need to find the final volume \(V_f\) when \(P_f = 2.50 \times 10^4 \text{ Pa}\). Solving \(1.00 \times 10^5 \text{ Pa} \times 1.50 \text{ L} = 2.50 \times 10^4 \text{ Pa} \times V_f\), we get \(V_f = 6.00 \text{ L}\).

Process 2 - Isobaric Compression

In the second process, the gas is compressed at constant pressure \(P = 2.50 \times 10^4 \text{ Pa}\). The final volume is \(1.50 \text{ L}\) (original volume). Using the ideal gas law again \(P V = n R T\), find the final temperature: \(nRT_f = P_f V_f\). If \(nR\) remains constant, then \(T_f/T_i = V_f/V_i\), hence \(T_f = 75 \text{ K}\) (since \(1.50/6.00 = 0.25\) of 300 K).

Draw the pV Diagram

Plotting the diagram: Start at \((1.50 \text{ L}, 1.00 \times 10^5 \text{ Pa})\). Plot a horizontal line to \((6.00 \text{ L}, 2.50 \times 10^4 \text{ Pa})\) representing isothermal expansion. Then a vertical line downwards to \((1.50 \text{ L}, 2.50 \times 10^4 \text{ Pa})\) represents isobaric compression.

Calculate Work Done

For isothermal process, work done \(W_1 = nRT \ln\left(\frac{V_f}{V_i}\right)\). For isobaric process, work done \(W_2 = P(V_i' - V_f')\). Given \( n R T = P V \), hence \( W_1 = 1.00 \times 10^5 \times 1.50 \times \ln(4)\) and \(W_2 = 2.50 \times 10^4 \times (1.50 - 6.00)\). Calculating gives \(W_1 \approx 1.04 \times 10^5 \text{ J}\) and \(W_2 = \text{-1.125} \times 10^5 \text{ J}\). Total \(W = W_1 + W_2 = \text{-8,500 J}\).

Returning Gas to Original State

To return the gas to its original state of pressure and temperature, you would perform a constant volume heating process. Increase the temperature of the gas at constant volume to 300 K, resulting in an increase in pressure back to \(1.00 \times 10^5 \text{ Pa}\).

Key concepts

Isothermal Expansion

During isothermal expansion, a gas's temperature stays constant while the volume increases and the pressure decreases. This is because the gas molecules move faster when heated, exerting more pressure on the walls of the container. In the isothermal process, any heat energy added results in the gas doing work on its surroundings.
The ideal gas law, expressed as \(PV = nRT\), helps us understand this process. Here, \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin.
For an isothermal process, where the temperature \(T\) is constant, the relationship simplifies to \(P_iV_i = P_fV_f\). As it expands isothermally, if pressure reduces, then the volume must rise to maintain the same product \(PV\). This means energy is changing forms but conserving the same temperature, effectively performing work on its surroundings.

Isobaric Compression

Isobaric compression occurs when a gas is compressed at constant pressure. During this process, the volume of the gas decreases while maintaining the same pressure, and often results in a decrease in temperature if no heat is added.
In our example exercise, we compress the gas back to its original volume at constant pressure. This is represented in the ideal gas law by keeping \(P\) constant and observing how changes in volume affect temperature. The formula used is \(T_f/T_i = V_f/V_i\), which shows the final temperature \(T_f\) is a fraction of the initial temperature \(T_i\) based on the change in volume.
Isobaric processes can be visualized as horizontal lines in a \(PV\) diagram, where pressure remains unchanged while volume and temperature shift according to the constraints of the ideal gas law.

Work Done by Gas

The work done by a gas depends on the process it undergoes. In thermodynamics, work is typically associated with volume changes under pressure.
For an isothermal process, work done \(W\) is calculated using the formula:\[W = nRT \ln\left(\frac{V_f}{V_i}\right)\]Here, \(nRT\) reflects the constant temperature energy, and the natural logarithm \(\ln\) relates to the ratio of final to initial volume.
In contrast, for an isobaric process, the work done is simply:\[W = P(V_i' - V_f')\]This shows that the work is a function of pressure \(P\) and the change in volume \(\Delta V = V_i' - V_f'\). This process involves linear changes as opposed to the logarithmic transformations in the isothermal scenario.

Ideal Gas Law

The ideal gas law is a fundamental principle that relates the pressure, volume, temperature, and quantity of gas in a system. The formula is \(PV = nRT\).
- \(P\): Pressure of the gas- \(V\): Volume occupied by the gas- \(n\): Moles of gas present- \(R\): Universal gas constant, approximately 8.314 \(J/(mol\cdot K)\)- \(T\): Absolute temperature in Kelvin
This law assumes gas molecules do not interact and have perfectly elastic collisions. It's a valuable approximation for describing gases under many conditions. When applied, it explains how changes in one variable, such as pressure, affect others in systems like isothermal or isobaric processes.
Understanding the ideal gas law is crucial for predicting the behavior of gases and calculating changes in thermodynamic processes.