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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 P2P_2 and temperature T2T_2, we can use the following equations for an adiabatic process:

  1. For pressure:

    P_1 V_1^{eta} = P_2 V_2^{eta}

    where eta = rac{C_p}{C_v} = γ = 1.4.

    Substituting the values:

    1.2×106(9.0×105)1.4=P2(6.8×105)1.41.2 \times 10^6 (9.0 \times 10^{-5})^{1.4} = P_2 (6.8 \times 10^{-5})^{1.4}

    Solving this yields:

    P22.56×106 PaP_2 \approx 2.56 \times 10^6 \text{ Pa}

  2. For temperature, using the relation:

    T2T1=(V1V2)(γ1)\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{(γ-1)}

    Substituting the values:

    T2=290(9.0×1056.8×105)0.4T_2 = 290 \left(\frac{9.0 \times 10^{-5}}{6.8 \times 10^{-5}}\right)^{0.4}

    This simplifies to:

    T2328 KT_2 \approx 328 \text{ 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 considered adiabatic because there is not enough time for heat transfer to occur between the gas and its surroundings. In such a short timeframe, any heat that might be lost or gained is negligible compared to the work done on the gas during compression.

Step 3

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

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Answer

In isothermal compression, the gas is allowed to exchange heat with its surroundings, keeping the temperature constant throughout the process. Since the initial and final states are the same, the work done can also be analyzed using the relationship:

W=nRTln(VfVi)W = nRT \ln\left(\frac{V_f}{V_i}\right)

where nn is the number of moles, RR is the gas constant, and VfV_f and ViV_i are the final and initial volumes, respectively. In comparison, since no heat transfer occurs in the adiabatic process, the work done will be larger due to the increase in internal energy.

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