In an ideal heat-engine cycle a fixed mass of air is taken through the following four processes. A → B Isothermal compression from an initial pressure of 1.0 × 10⁵ Pa and a volume of 9.0 × 10⁻² m³ to a pressure of 2.2 × 10⁵ Pa. The work done by the air is 7100 J. B → C Increase in volume at constant pressure to a volume of 5.9 × 10³ m³; C → D Isothermal expansion to a pressure of 1.0 × 10⁵ Pa and a volume of 13 × 10⁻² m³. The work done by the air is 10 300 J. D → A A reduction in volume at constant pressure to the original volume. Figure 3 shows the cycle. Figure 3 Show, by calculation, that the volume at B is 4.1 × 10² m³. The temperature of the air between A and B is 295 K. Show that the temperature of the air at C is about 420 K. An ideal engine based on this cycle uses a device called an economiser. The economiser stores all the energy transferred in the cooling process D → A and gives up all this energy to the air in process B → C. This means that an external source supplies energy to the air by heating only in process C → D. Complete Table 1 to show the values of work done W and energy transfer Q in each of the four processes. Use the space below Table 1 for any calculations. Use a consistent sign convention.

Photo AI

In an ideal heat-engine cycle a fixed mass of air is taken through the following four processes - AQA - A-Level Physics - Question 4 - 2019 - Paper 6

Question 4

In an ideal heat-engine cycle a fixed mass of air is taken through the following four processes. A → B Isothermal compression from an initial pressure of 1.0 × 10⁵ ... show full transcript

Worked Solution & Example Answer:In an ideal heat-engine cycle a fixed mass of air is taken through the following four processes - AQA - A-Level Physics - Question 4 - 2019 - Paper 6

Step 1

Show, by calculation, that the volume at B is 4.1 × 10² m³.

96%

114 rated

Answer

Using the proportionality, apply the gas law: $P_A V_A = n R T_A$ Substituting values: $(1.0 × 10^5 Pa) (9.0 × 10^{-2} m^3) = n (8.31 J/mol·K)(295 K)$ Solving for n gives: $n = \frac{1.0 × 10^5 Pa × 9.0 × 10^{-2} m^3}{8.31 J/mol·K × 295 K}≈ 3.69 ext{ moles}$

Now substituting for volume at point B: $2.2 × 10^5 Pa × V_B = 3.69 ext{ moles} × 8.31 J/mol·K × 295 K$ Solving yields: $V_B = \frac{3.69 × 8.31 × 295}{2.2 × 10^5} ≈ 4.1 × 10^{-2} m^3$

Step 2

Show that the temperature of the air at C is about 420 K.

99%

104 rated

Answer

To find the temperature at point C, we apply the ideal gas law again considering that the pressure at C is 1.0 × 10⁵ Pa and the volume is 13 × 10⁻² m³.

Using the previously calculated moles from point B:

We know that: $P_C V_C = n R T_C$ Substituting values: $(1.0 × 10^5 Pa)(13 × 10^{-2} m^3) = 3.69 ext{ moles} × 8.31 J/mol·K × T_C$

Rearranging gives us: $T_C = \frac{(1.0 × 10^5)(13 × 10^{-2})}{3.69 × 8.31}$ Calculating this gives: $T_C ≈ 420 K$

Step 3

Complete Table 1 to show the values of work done W and energy transfer Q in each of the four processes.

96%

101 rated

Answer

Based on the processes detailed:

Process	Work done W (J)	Energy Transfer Q (J)
A → B	-7100	-7100
B → C	4000	4000
C → D	10300	10300
D → A	-14000	-14000

Step 4

Explain why W is equal to Q in process A → B and in process C → D.

98%

120 rated

Answer

In processes A → B and C → D, the system is undergoing isothermal processes. During isothermal expansion or compression:

The internal energy of an ideal gas remains constant.
Any heat (Q) absorbed or released by the system is entirely converted into work (W) done by or on the gas.
Therefore, for these processes, W = Q holds true due to the first law of thermodynamics, which states that the change in internal energy equals the heat added to the system minus the work done by the system. In this situation, with internal energy unchanged, W equals Q.

Step 5

Deduce whether this claim is true.

97%

117 rated

Answer

The claim states that the efficiency of this engine cycle is the same as the maximum theoretical efficiency of a heat engine operating between the same temperatures.
To analyze this, we consider that the ideal efficiency $ext{efficiency} = 1 - \frac{T_{C}}{T_{H}}$ where T_C is the cold reservoir and T_H is the hot reservoir.
Given that the actual heat engine is not 100% efficient, we can deduce that, while it might approach the theoretical maximum, the real engine is likely to have additional losses due to factors such as friction, heat loss, and other inefficiencies, hence this claim may not hold true.

Step 6

Discuss one problem that would be faced by an engineer designing a real engine based on this cycle.

97%

121 rated

Answer

One significant problem that an engineer would face is managing the heat exchange and thermal efficiency of the cycle. Real engines cannot achieve the idealized conditions of constant temperature and pressure. The heat losses during the processes can lead to suboptimal performance, requiring more energy input to achieve the desired output. Additionally, materials and mechanical limitations could affect the capability to operate at the ideal or desired temperatures, potentially leading to material degradation or failure.

In an ideal heat-engine cycle a fixed mass of air is taken through the following four processes - AQA - A-Level Physics - Question 4 - 2019 - Paper 6

Worked Solution & Example Answer:In an ideal heat-engine cycle a fixed mass of air is taken through the following four processes - AQA - A-Level Physics - Question 4 - 2019 - Paper 6

Show, by calculation, that the volume at B is 4.1 × 10² m³.

Show that the temperature of the air at C is about 420 K.

Complete Table 1 to show the values of work done W and energy transfer Q in each of the four processes.

Explain why W is equal to Q in process A → B and in process C → D.

Deduce whether this claim is true.

Discuss one problem that would be faced by an engineer designing a real engine based on this cycle.

Join the A-Level students using SimpleStudy...