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In an ideal heat-engine cycle a fixed mass of air is taken through the following four processes - AQA - A-Level Physics - Question 3 - 2019 - Paper 6

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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 3 - 2019 - Paper 6

Step 1

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

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Answer

To find the volume at point B, we can use Boyle's Law, which states that for a fixed mass of gas at constant temperature, the product of pressure and volume is a constant. Using the data for points A and B:

o A: P_A = 1.0 × 10⁵ Pa, V_A = 9.0 × 10⁻³ m³

o B: P_B = 2.2 × 10⁵ Pa

Using Boyle's Law:

demonstrate: P_A V_A = P_B V_B

Substituting known values and solving for V_B:

o 1.0 × 10⁵ Pa × 9.0 × 10⁻³ m³ = 2.2 × 10⁵ Pa × V_B
2.2 × 10⁵ Pa × V_B = 9.0 × 10⁻³ m³ × 1.0 × 10⁵ Pa
V_B = \frac{(9.0 × 10⁻³ m³ × 1.0 × 10⁵ Pa)}{2.2 × 10⁵ Pa}

Calculating gives:

V_B = 4.1 × 10⁻² m³.

Thus, the calculation shows that the volume at B is indeed 4.1 × 10² m³.

Step 2

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

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Answer

To find the temperature at point C, we can use the ideal gas law:

PV=nRTPV = nRT

where:

  • P is the pressure,
  • V is the volume,
  • n is the number of moles,
  • R is the ideal gas constant (approximately 8.31 J/(mol·K)),
  • T is the temperature in Kelvin.

For process C:

  • P_C = 1.0 × 10⁶ Pa,
  • V_C = 13 × 10⁻³ m³.

Since the number of moles and R are constants, we can express the relationship across A, B, and C:

Using the data for point B where T_B = 295 K:

PAVA=nRTAextandPCVC=nRTCP_A V_A = nRT_A ext{ and } P_C V_C = nRT_C

Setting the ratios: PCVCPBVB=TCTB\frac{P_C V_C}{P_B V_B} = \frac{T_C}{T_B}

Substituting values: (1.0×106Pa)×(13×103m3)(2.2×105Pa)×(4.1×102m3)=TC295K\frac{(1.0 × 10⁶ Pa) × (13 × 10⁻³ m³)}{(2.2 × 10⁵ Pa) × (4.1 × 10⁻² m³)} = \frac{T_C}{295 K}

Now, calculate for T_C:

o T_C = 295 K × (\frac{(1.0 × 10⁶) × (13 × 10⁻³)}{(2.2 × 10⁵) × (4.1 × 10⁻²)})

After calculating, T_C is found to be approximately 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.

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Answer

Based on the conditions of each process, the values can be summarized as:

ProcessWork done W (J)Energy transfer Q (J)
A → B-7100-7100
B → C4000+4000
C → D10300+10300
D → A0-1400

Note:

  • Negative values indicate work done by the system, while positive values indicate work done on the system.

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