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Figure 4 shows a gas strut supporting the lid of a trailer - AQA - A-Level Physics - Question 3 - 2017 - Paper 6

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Figure 4 shows a gas strut supporting the lid of a trailer. A fixed mass of nitrogen gas is sealed into the cylinder of the strut. The gas is initially at a pressu... show full transcript

Worked Solution & Example Answer:Figure 4 shows a gas strut supporting the lid of a trailer - AQA - A-Level Physics - Question 3 - 2017 - Paper 6

Step 1

Calculate the pressure and temperature of the gas at the end of the compression assuming the compression to be an adiabatic process.

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Answer

To find the final pressure and temperature of the gas, we can use the adiabatic relations for an ideal gas:

  1. The pressure-volume relation: P_1 V_1^{ rac{eta}{eta-1}} = P_2 V_2^{ rac{eta}{eta-1}} where ( \beta = \gamma = 1.4 ) for nitrogen.

  2. The temperature-volume relation:
    T1V1β1=T2V2β1T_1 V_1^{\beta-1} = T_2 V_2^{\beta-1}

Let's denote:

  • Initial pressure, ( P_1 = 1.2 \times 10^6 \text{ Pa} )
  • Initial volume, ( V_1 = 9.0 \times 10^{-5} \text{ m}^3 )
  • Final volume, ( V_2 = 6.8 \times 10^{-5} \text{ m}^3 )
  • Initial temperature, ( T_1 = 290 ext{ K} )

Now we can calculate the final pressure, ( P_2 ): P2=P1(V1V2)β=1.2×106(9.0×1056.8×105)1.4P_2 = P_1 \left( \frac{V_1}{V_2} \right)^{\beta} = 1.2 \times 10^6 \left( \frac{9.0 \times 10^{-5}}{6.8 \times 10^{-5}} \right)^{1.4} After computing the above expression, we find:

  • Final pressure: approximately 1.68×1061.68 \times 10^6 Pa.

Next, calculate the final temperature, ( T_2 ): T2=T1(V1V2)β1=290(9.0×1056.8×105)0.4T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\beta-1} = 290 \left( \frac{9.0 \times 10^{-5}}{6.8 \times 10^{-5}} \right)^{0.4} This yields a final temperature of approximately 328328 K.

Step 2

Explain why the rapid compression of the gas can be assumed to be an adiabatic process.

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Answer

The rapid compression of the gas can be assumed to be an adiabatic process because during the quick closing of the lid, there is not enough time for heat exchange between the gas and its surroundings. In an adiabatic process, by definition, no heat transfer occurs (( Q = 0 )). As the compression happens swiftly, the temperature of the gas increases due to work done on it without any heat lost or gained.

Step 3

When the lid is closed slowly, the compression can be assumed to be isothermal.

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Answer

When the lid is closed slowly, the gas has enough time to exchange heat with its surroundings. This means that during compression, the temperature of the gas remains constant, which characterizes an isothermal process. In an isothermal process, the internal energy of the gas remains unchanged, and any work done on the gas is offset by heat transfer into the gas. Therefore, the compression can be treated as isothermal when the lid is closed slowly.

Step 4

Compare without calculation the work done in each process.

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Answer

In an adiabatic process, the work done on the gas results in an increase in internal energy and temperature, as there is no heat transfer. Consequently, the work done during adiabatic compression is greater than that done during isothermal compression, where the temperature remains constant and some work is used to transfer heat out of the gas. Thus, while the gas behaves differently in each case, the work done in the adiabatic case is greater due to the inability to dissipate heat.

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