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

To find the final pressure and temperature after adiabatic compression, we can use the formulas for adiabatic processes:

1. The pressure-volume relationship: P_1 V_1^{rac{eta}{eta-1}} = P_2 V_2^{rac{eta}{eta-1}}
2. The temperature-volume relationship: T_1 V_1^{eta - 1} = T_2 V_2^{eta - 1}

Where:

• $P_1 = 1.2 \times 10^6 \, \text{Pa}$ (initial pressure)
• $V_1 = 9.0 \times 10^{-5} \, \text{m}^3$ (initial volume)
• $V_2 = 6.8 \times 10^{-3} \, \text{m}^3$ (final volume)
• $T_1 = 290 \, \text{K}$ (initial temperature)
• eta = 1.4 (adiabatic index for nitrogen)

Step 1: Calculate final pressure $P_2$:

$P_2 = P_1 \left( \frac{V_1}{V_2} \right)^{\beta}$ $P_2 = 1.2 \times 10^6 \left( \frac{9.0 \times 10^{-5}}{6.8 \times 10^{-3}} \right)^{1.4}$

Calculating this gives: $P_2 \approx 2.05 \times 10^6 \, \text{Pa}$

Step 2: Calculate final temperature $T_2$:

$T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\beta - 1}$ $T_2 = 290 \left( \frac{9.0 \times 10^{-5}}{6.8 \times 10^{-3}} \right)^{0.4}$

Calculating this gives: $T_2 \approx 336 \, \text{K}$

Thus, the final pressure is approximately $2.05 \times 10^6 \, \text{Pa}$ and the final temperature is approximately $336 \, \text{K}$.