A number of assumptions are made when explaining the behaviour of a gas using the molecular kinetic theory model - AQA - A-Level Physics - Question 1 - 2017 - Paper 2

Using the ideal gas equation, we have:

$N = \frac{P V}{k T}$

Where:

• P = 1.0 × 10^5 Pa
• V = 0.50 × 10^{-3} m³
• k = 1.38 × 10^{-23} J/K (Boltzmann constant)
• T = 300 K (27 °C in Kelvin)

Calculating:

$N = \frac{(1.0 × 10^5) (0.50 × 10^{-3})}{1.38 × 10^{-23} (300)} ≈ 1.17 × 10^{20} \text{ molecules}$

The pressures and volumes for states A and B can be obtained from Figure 1. Assuming:

• At A: $P_A = 1 \times 10^5 \text{ Pa}, V_A = 0.50 \times 10^{-3} \text{ m}^3$
• At B: $P_B = 2 \times 10^5 \text{ Pa}, V_B = 0.30 \times 10^{-3} \text{ m}^3$

Using the relationship defined by the ideal gas law and manipulating it we find:

$T_B = \frac{P_B V_B}{k N}$

Thus, calculating the temperatures gives us:

$T_A = 300 K, T_B ≈ 215 K$

Therefore, the change in temperature is:

$ΔT = T_B - T_A = 215 - 300 = -85 K$

The temperature of the gas does change from B to C. Given that the pressure at B is higher than at A, and if we deduce from gas laws that volume is inversely proportional to pressure under constant mass, it can be inferred that the temperature will not remain constant and will likely decrease further.

The work done on the gas can be determined using the area under the curves in Figure 1. For the transformation from A to B:

$W_{AB} = \text{Area under AB curve}$

And for B to C:

$W_{BC} = \text{Area under BC curve}$

Since the area under the curve represents the work done, quantitatively comparing these two areas will show that the work done on the gas during the compression from A to B is greater than the work done during B to C due to how pressure changes while volume changes.