One relation, and the constraint that earns each shortcut

Write the law in either of its two forms:

$$PV = nRT \qquad \text{or} \qquad PV = Nk_BT,$$

with $R = 8.31$ J/(mol$\cdot$K) counting moles and $k_B = 1.38\times10^{-23}$ J/K counting particles. In both, $T$ is in kelvins, because the relation predicts $P \to 0$ and $V \to 0$ as $T \to 0$, and there is nothing special about the temperature at which water happens to freeze.

Ratios are where Celsius does real damage. A sealed rigid can taken from $20^\circ$C to $40^\circ$C looks like a doubling and is not:

$$\frac{T_2}{T_1} = \frac{313\ \text{K}}{293\ \text{K}} = 1.07,$$

so the pressure rises by about $7\%$ and the can survives. The same slip shows up on a pressure-versus-temperature graph, where the extrapolated zero-pressure intercept sits at about $-273^\circ$C. That intercept is what defines the zero of the absolute scale, and it is why only that scale makes a ratio mean anything.

For a fixed amount of gas moving between two states:

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}.$$

Now read the apparatus and see what it pins.

1. Rigid sealed container. $V$ cannot change, so $V$ cancels and $P/T$ is constant.
2. Piston free to slide against the atmosphere. The gas pressure is set by the load, so $P$ cancels and $V/T$ is constant.
3. Held at one temperature. $T$ cancels and $PV$ is constant.
4. Gas added or removed. Nothing above applies: $n$ changed, so go back to $PV = nRT$ for each state separately.

The tell for this error is easy to spot in a problem statement. Count the quantities the problem says changed. If it names three and your solution tracks two, you assumed a constraint the apparatus never provided.

The ideal gas model makes three assumptions and it is worth keeping them apart:

1. Particle volume is negligible next to the volume of the container.
2. No appreciable force acts between particles between collisions.
3. Collisions, with each other and with the walls, are elastic and frequent.

Assumption 2 is the one that gets misread. "No forces between the particles" is a statement about the gaps, not about the collisions. Delete the collisions and you have deleted the mechanism for pressure, since pressure is momentum delivered to the wall, and you have deleted the mechanism that shares energy through the gas and brings it to one temperature.

The model degrades exactly where those assumptions do: compress a gas hard enough that the particles' own volume matters, or cool it far enough that the weak attractions between them stop being negligible, and real gases start to deviate. That is the same regime in which they condense.

The two forms of the law differ only in how the amount is bookkept:

$$N = n N_A, \qquad N_A = 6.02\times10^{23} \ \text{mol}^{-1}, \qquad R = N_A k_B.$$

Use $nRT$ when the problem gives you moles or grams and a molar mass, and $Nk_BT$ when it gives you a particle count or when you are heading toward $K_{\text{avg}} = \tfrac{3}{2}k_BT$ and want the same constant on both sides.

One consequence deserves stating on its own, because problems lean on it: at a given $P$ and $T$, a fixed volume of any ideal gas holds the same number of particles. Helium and argon in identical flasks at identical conditions differ in mass by a factor of ten and in particle count not at all.