Answer
Absolute zero $=-233.33 \mathrm{N}$
Work Step by Step
since the Neptunians have a celsius type scale, the absolute temperature
can be written as,
$$
\mathrm{T}_{\mathrm{abs}}=\mathrm{T}_{\mathrm{N}}+\mathrm{K} \quad\quad\quad\quad(1)
$$
where,
$\mathrm{T}_{\mathrm{abs}}$ is the absolute temperature.
$\mathrm{T}_{\mathrm{N}}$ is the temperature in the Neptunian scale
$\mathrm{K}$ is the constant that needs to be added to convert the temperature from
absolute scale to Neptunian scale
Also given the perfect gas law is followed in the Neptunian environment i.e.
$p \cdot V=n \cdot R \cdot T \quad\quad\quad\quad\quad(2)$
From the question, we have two scenarios where the values for $\mathrm{PV}$ and $\mathrm{T}$
have been provided. Since the gas is the same and no new mass is added we can
equate the number of moles of the gas in both the scenarios. i.e.,
$\begin{aligned} \frac{p_{1} \cdot V_{1}}{T_{1}} &=\frac{p_{2} \cdot V_{2}}{T_{2}} \\ \frac{28}{K+0} &=\frac{40}{K+100} \\ 28 \times(K+100) &=40 \times(K+0) \\ K &=233.33 \end{aligned}$
Hence the value at absolute zero
$$
\begin{aligned} \mathrm{T}_{\mathrm{abs}} &=\mathrm{T}_{\mathrm{N}}+\mathrm{K} \\ 0 &=\mathrm{T}_{\mathrm{N}}+233.33 \\ \mathrm{T}_{\mathrm{N}} &=-233.33 \mathrm{N} \end{aligned}
$$
