Problem 44 - College Physics

Question

\bullet A cylinder contains oxygen gas \(\left(\mathrm{O}_{2}\right)\) at a pressure of 2.00 atm. The volume is \(4.00 \mathrm{L},\) and the temperature is 300 \(\mathrm{K}\) . Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes: (i) Heated at constant pressure from the initial state (state 1)to state \(2,\) which has \(T=450 \mathrm{K}\) . (ii) Cooled at constant volume to 250 \(\mathrm{K}\) (state 3\()\) . (iii) Compressed at constant temperature to a volume of 4.00 \(\mathrm{L}\) (state \(4 ) .\) (iv) Heated at constant volume to 300 \(\mathrm{K}\) , which takes the sys- tem back to state I. (a) Show these four processes in a \(p V\) diagram, giving the numerical values of \(p\) and \(V\) in each of the four states. (b) Cal- culate \(Q\) and \(W\) for each of the four processes. (c) Calculate the net work done by the oxygen. (d) What is the efficiency of this device as a heat engine? How does this efficiency compare with that of a Carnot-cycle engine operating between the same minimum and maximum temperatures of 250 \(\mathrm{K}\) and 450 \(\mathrm{K}\) ?

Step-by-Step Solution

Answer

Diagram shows all states. Calculate \(W\), \(Q\) for each, use sums for net work, efficiency, and compare with Carnot's efficiency.

Step 1: Identify State 1 Parameters

The initial parameters are given as: \(P_1 = 2.00 \, \text{atm}\), \(V_1 = 4.00 \, \text{L}\), \(T_1 = 300 \, \text{K}\). Using the ideal gas law, we can calculate the number of moles, \(n\), if necessary for further steps.

Step 2: Process (i) - Isobaric Expansion

During an isobaric (constant pressure) process, the temperature increases from \(T_1 = 300 \, \text{K}\) to \(T_2 = 450 \, \text{K}\). By the ideal gas law, \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). Solve for \(V_2\):\[ V_2 = V_1 \times \frac{T_2}{T_1} = 4.00 \, \text{L} \times \frac{450}{300} = 6.00 \, \text{L} \]State 2 parameters: \(P_2 = 2.00 \, \text{atm}\), \(V_2 = 6.00 \, \text{L}\), \(T_2 = 450 \, \text{K}\).

Step 3: Process (ii) - Isochoric Cooling

The volume remains constant at \(V_2 = 6.00 \, \text{L}\), and the temperature decreases to \(T_3 = 250 \, \text{K}\). Using \(\frac{P_2}{T_2} = \frac{P_3}{T_3}\), solve for \(P_3\):\[ P_3 = P_2 \times \frac{T_3}{T_2} = 2.00 \, \text{atm} \times \frac{250}{450} \approx 1.11 \, \text{atm} \]State 3 parameters: \(P_3 \approx 1.11 \, \text{atm}\), \(V_3 = 6.00 \, \text{L}\), \(T_3 = 250 \, \text{K}\).

Step 4: Process (iii) - Isothermal Compression

At constant temperature \(T_3 = T_4 = 250 \, \text{K}\), the volume decreases from \(V_3 = 6.00 \, \text{L}\) to \(V_4 = 4.00 \, \text{L}\). The ideal gas law gives \(P_3 \times V_3 = P_4 \times V_4\):\[ P_4 = \frac{P_3 \times V_3}{V_4} = \frac{1.11 \, \text{atm} \times 6.00 \, \text{L}}{4.00 \, \text{L}} \approx 1.665 \, \text{atm} \]State 4 parameters: \(P_4 \approx 1.665 \, \text{atm}\), \(V_4 = 4.00 \, \text{L}\), \(T_4 = 250 \, \text{K}\).

Step 5: Process (iv) - Isochoric Heating

At constant volume \(V_4 = V_1 = 4.00 \, \text{L}\), temperature increases to \(T_1 = 300 \, \text{K}\). Using \(\frac{P_4}{T_4} = \frac{P_1}{T_1}\):\[ P_1 = P_4 \times \frac{T_1}{T_4} = 1.665 \, \text{atm} \times \frac{300}{250} = 2.00 \, \text{atm} \]The system returns to the initial state: \(P_1 = 2.00 \, \text{atm}\), \(V_1 = 4.00 \, \text{L}\), \(T_1 = 300 \, \text{K}\).

Step 6: Compute Work and Heat for Each Process

For processes (i) and (iii), calculate work \(W\) using \(W = P\Delta V\). For isochoric processes (ii) and (iv), \(W=0\). Use \(Q = nC_p \Delta T\) for isobaric and \(Q = nC_v \Delta T\) for isochoric process to find heat \(Q\). Assume ideal gas constants for \(C_p = R + C_v \) and \(C_v\).

Step 7: Compute Net Work Done

The net work done by the system is the sum of work for each process. \(W_{net} = W_{(i)} + W_{(ii)} + W_{(iii)} + W_{(iv)}\). Since \(W\) for (ii) and (iv) are zero, \(W_{net}\) comprises only \(W_{(i)}\) and \(W_{(iii)}\) contributions.

Step 8: Determine Efficiency

Efficiency \(\eta\) is calculated using \(\eta = \frac{-W_{net}}{Q_{in}}\) where \(Q_{in}\) is the heat added. Compare this with Carnot efficiency \(\eta_C = 1 - \frac{T_{min}}{T_{max}}\):\[ \eta_C = 1 - \frac{250}{450} \approx 0.444 \]Evaluate and compare \(\eta\) with this value.

Key Concepts

Isobaric Process

An isobaric process is one where the pressure remains constant while other parameters like temperature and volume may change. Imagine a piston inside a cylinder where the gas expands or compresses but always maintains a constant pressure against the piston. This kind of process is represented on a Pressure-Volume (PV) diagram as a horizontal line, indicating constant pressure.
If you picture heating a balloon that expands as it is filled with more air at an even speed, this is somewhat similar to how isobaric processes work. In the textbook exercise, the oxygen gas in the cylinder undergoes an isobaric process when heated from 300 K to 450 K, causing the volume to increase.
This increase in temperature and volume at constant pressure is governed by the ideal gas law: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] Where \(V_1\) and \(T_1\) are the initial volume and temperature, and \(V_2\) and \(T_2\) are the final volume and temperature.

Isochoric Process

An isochoric process, also known as an isometric or constant volume process, is one in which the volume remains fixed while the pressure and temperature can change. This is akin to heating coffee in a sealed thermos; the volume of coffee stays the same, but its temperature and pressure may rise as it heats.
In our exercise, the oxygen is cooled at constant volume, showing a drop in temperature from 450 K to 250 K while the volume remains at 6.00 L. The constant volume nature of this process is indicated by a vertical line on a PV diagram.
For an ideal gas undergoing an isochoric process, the pressure changes in proportion to the change in temperature, described by: \[ \frac{P_2}{T_2} = \frac{P_3}{T_3} \] Thus, as the temperature decreases in an isochoric process, so does the pressure.

Isothermal Process

In an isothermal process, the temperature of the gas stays constant throughout. This means any work done by or on the gas is compensated by heat exchange with the surroundings, keeping the temperature steady.
Imagine holding a hot cup of tea in a room where the ambient temperature is the same; even as heat transfers in and out, the overall temperature stays equal.
For the oxygen gas example, the isothermal process occurs when the gas is compressed back to its original volume at a constant temperature of 250 K. During compression, the PV relation is given by: \[ P_3 \times V_3 = P_4 \times V_4 \] This equation reflects the preservation of the temperature, ensuring energy recycles within the system instead of altering its overall temperature.

Carnot Cycle

The Carnot cycle represents an idealized thermodynamic cycle proposed by Sadi Carnot. It consists of two isothermal processes and two adiabatic processes. The Carnot cycle provides a benchmark for the maximum possible efficiency that any heat engine can achieve operating between two temperatures.
The cycle is made up of:

• Isothermal expansion - Absorbs heat at high temperature.
• Adiabatic expansion - Temperature decreases without heat exchange.
• Isothermal compression - Heat is expelled at low temperature.
• Adiabatic compression - Temperature increases without heat exchange.

The efficiency of a Carnot cycle can be calculated with the formula: \[ \eta_C = 1 - \frac{T_{min}}{T_{max}} \] This cycle indicates it is impossible for any engine to be more efficient than a Carnot engine when operating between the same temperature limits.

Heat Engine Efficiency

Heat engine efficiency is a measure of how well an engine converts the input heat into useful work. It’s calculated by the ratio of work output to heat input. The genius behind it is to maximize the work we get from the energy we put into the system.
In practical terms, if a car engine is only 30% efficient, only 30% of the fuel's energy transforms into motion while the rest becomes waste heat.
The equation to find efficiency \( \eta \) is: \[ \eta = \frac{-W_{net}}{Q_{in}} \] In the context of the exercise, after determining the net work done by the oxygen gas and the heat energy input, the efficiency of this cyclic process can be compared against the standard set by the Carnot cycle's efficiency. This comparison helps in understanding the real-world limitations and improvements for heat engines.