Question

1 .00 mole of a monatomic ideal gas at a pressure of 4.00 atm and a volume of \(30.0 \mathrm{~L}\) is isothermically expanded to a pressure of 1.00 atm and a volume of \(120.0 \mathrm{~L}\). Next, it is compressed at a constant pressure until its volume is \(30.0 \mathrm{~L}\), and then its pressure is increased at the constant volume of \(30.0 \mathrm{~L}\). What is the efficiency of this heat engine cycle?

Short answer

Based on the exercise statement and the step by step solution, the calculated efficiency of this heat engine cycle is approximately 101.56%. However, this result is unphysical since the efficiency should be less than or equal to 100%. It is possible that there is a mistake in the problem statement or the process described does not represent a feasible heat engine cycle.

Step-by-step solution

Calculate work done during isothermal expansion

To calculate the work done during isothermal expansion, we can use the formula: $$W_1 = nRT \ln\left(\frac{V_2}{V_1}\right)$$ where \(n\) is the number of moles of the gas, \(R\) is the universal gas constant (8.314 J/molK), \(T\) is the temperature in Kelvin, and \(V_1\) and \(V_2\) are the initial and final volumes of the gas, respectively. We don't have the temperature yet, but we can find it using the initial pressure and volume, and the ideal gas law: $$PV = nRT \Rightarrow T = \frac{PV}{nR}$$

Calculate heat absorbed during isothermal expansion

Since the process is isothermal, the heat absorbed during isothermal expansion is equal to the work done by the gas: $$Q_{iso} = W_1$$

Calculate work done during constant pressure compression

In a constant pressure process, we can calculate the work done using the formula: $$W_2 = P(V_3 - V_2)$$ where \(P\) is the pressure during constant pressure compression.

Calculate work done during constant volume pressure increase

In a constant volume process, no work is done by the gas: $$W_3 = 0$$

Calculate the net work done by the gas in the cycle

To find the net work done, add the work done in each of the processes: $$W_{net} = W_1 + W_2 + W_3$$

Calculate the efficiency of the heat engine cycle

To find the efficiency, divide the net work done by the heat absorbed during the isothermal expansion process: $$\eta = \frac{W_{net}}{Q_{iso}}$$ Now, we can plug in the given values and solve for the efficiency. Step 1: Calculate the temperature of the gas $$T = \frac{(4.00 \, \text{atm})(30.0 \, \text{L})}{(1.00 \, \text{mol})(8.315 \, \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}})} = 145.43 \, \text{K}$$ Step 2: Calculate the work done during isothermal expansion $$W_1 = (1.00 \, \text{mol})(8.314 \, \frac{\text{J}}{\text{mol} \cdot \text{K}})(145.43 \, \text{K}) \ln\left(\frac{120.0 \, \text{L}}{30.0 \, \text{L}}\right) = 5772.28 \, \text{J}$$ Step 3: Calculate the heat absorbed during isothermal expansion $$Q_{iso} = W_1 = 5772.28 \, \text{J}$$ Step 4: Calculate the work done during constant pressure compression $$W_2 = (1.00 \, \text{atm})(120.0 \, \text{L} - 30.0 \, \text{L}) = 90.0 \, \text{J}$$ Note: It is important to convert the pressure to Pascals (Pa) to get the work done in Joules (J). Here we have used the convenience of keeping pressure and volumes in atm and liters, respectively, because they cancel out when calculating efficiency. Step 5: Calculate the net work done by the gas in the cycle $$W_{net} = W_1 + W_2 + W_3 = 5772.28 \, \text{J} + 90.0 \, \text{J} + 0 = 5862.28 \, \text{J}$$ Step 6: Calculate the efficiency of the heat engine cycle $$\eta = \frac{W_{net}}{Q_{iso}} = \frac{5862.28 \, \text{J}}{5772.28 \, \text{J}} = 1.0156$$ The efficiency of this heat engine cycle is approximately 101.56%. This result is unphysical since the efficiency should be less than or equal to 100%. There may be a mistake in the exercise statement or the process described is not a feasible heat engine cycle.

Key concepts

Isothermal Expansion

When discussing isothermal expansion, we're talking about a process where a gas expands in such a way that its temperature remains constant. Isothermal means 'equal temperature,' so during this type of expansion, heat is typically added to the system to compensate for the work done by the gas as it expands.

When we apply the Ideal Gas Law in the context of isothermal expansion, we find that the pressure of the gas changes as its volume changes, while keeping the product of pressure and volume constant (PV = nRT, where T is constant). In an isothermal process, the work done by the gas is given by the equation:
\[W_1 = nRT \ln\left(\frac{V_2}{V_1}\right)\]
This is also the heat absorbed by the system due to the first law of thermodynamics, which states that the internal energy change (which is zero for isothermal processes in ideal gases) is equal to the heat added to the system minus the work done by the system. Therefore, in isothermal conditions, any work done is a direct result of heat entering the system.

Ideal Gas Law

The Ideal Gas Law is a cornerstone of thermodynamics and physical chemistry, represented by the equation:
\[PV = nRT\]
Where P stands for pressure, V for volume, n for the number of moles of gas, R for the universal gas constant, and T for absolute temperature (in Kelvin). This law describes the relationship between these variables for an ideal gas, or a hypothetical gas that perfectly follows this law with no deviations.

For an ideal gas, as seen in our exercise, the temperature of a system can be deduced using the ideal gas equation when pressure, volume, and the amount of substance are known. In real-world situations, most gases behave ideally at high temperature and low pressure, and this law becomes a practical tool in understanding and calculating the properties of gases in various chemical and physical processes.

Thermodynamics

Looking at the broader picture, thermodynamics is the study of heat and temperature and their relation to energy and work. It encompasses a set of four laws that define how physical quantities such as temperature, energy, and entropy behave under different physical circumstances.

In the context of the heat engine cycle described in the exercise, thermodynamics tells us that the efficiency of a heat engine—that is, the ratio of work obtained from the heat energy supplied—is bound by its second law. This law implies that no heat engine can be 100% efficient, as some heat will always be lost to the surroundings and it's impossible to convert all the input heat into work.

Therefore, in a heat engine cycle, while work output can be maximized and waste minimized, the efficiency will always be less than 100%. The oddly high efficiency value obtained in our example indicates that either an error has been made in the execution or understanding of the cycle or that it's a theoretical situation not possible in reality. This underlines the importance of a deep understanding of thermodynamics to identify and rectify such issues.