Figure 3 shows the $p$-$V$ diagram for an idealised diesel engine cycle - AQA - A-Level Physics - Question 3 - 2018 - Paper 6

To determine if the claim is true, we can evaluate the area enclosed in the $p$-$V$ diagram for both cycles. Each square represents 10 J; hence the total area for the original cycle can be calculated. If the new cycle gives an area that represents 130 J more than before, then this claim is valid.

The efficiency of a cycle can be calculated using the formula:

$\eta = \frac{W_{out}}{Q_{in}}$

We need to compare the efficiencies of both cycles. If the efficiency of the modified cycle exceeds the original by more than 15%, then this claim is valid.

$Q$ represents the total heat added to the system, while $\Delta U$ represents the change in internal energy of the system. In this context, $Q$ indicates the energy input, and $\Delta U$ illustrates how much of that energy has contributed to changing the internal energy.

Given that $\Delta U = -374 J$ for the air in process 5 → 1, we can use the first law of thermodynamics:

$Q = \Delta U + W$

Assuming $W$ is zero since it occurs at constant pressure, we have:

$Q = -374 J$

Thus, the energy that must be removed by cooling process 5 → 1 is 374 J.

To find the maximum temperature, we can use the ideal gas equation:

$PV = nRT$

At the maximum pressure and volume points in the cycle, we can plug in values for $P$, $V$, and $n$ (the number of moles). Given that 0.060 mol of air is taken through the cycle might require measurements from the $p$–$V$ diagram to find corresponding values for temperature calculation.