Combination of several unit symbols when finding new pressure from ideal gas law

As a car is driven, the tires heat up and the volume and the pressure change. Calculate the new pressure of a 35.0 L tire initially at 235 kPa that expands to 36.5 L when it heats up from 15.0 °C to 43.0 °C. (4 marks)

$$P_1 = \pu{235 kPa}$$
$$P_2 = ?$$
$$V_1 = \pu{35.0 L}$$
$$V_2 = \pu{36.5 L}$$
$$T_1 = \pu{15.0 °C} = 15.0 + 273 = \pu{288 K}$$
$$T_2 = \pu{43.0 °C} = 43.0 + 273 = \pu{316 K}$$

The number of moles of gas in the balloon is not changing, so $$n_1 = n_2$$.

$$P_1V_1/n_1T_1 = P_2V_2/n_2T_2$$

(since $$n_1 = n_2$$ we can cancel them from both sides)

$$P_1V_1/T_1 = P_2V_2/T_2$$

(cross multuple to allow for isolation of $$P_2$$)

$$P_1V_1T_2 = P_2V_2T_1$$

(divide both sides by $$V_2T_1$$ to isolate $$P_2$$)

$$P_2 = P_1V_1T_2/V_2T_1$$

(now plug in values)

\begin{align} P_2 &= (\pu{235 kPa})(\pu{35.0 L})(\pu{316 K})/(\pu{36.5 L})(\pu{288 K})\\ &= 2 599 100 / 10 512 \\ &= \pu{247 kPa} \end{align}

The new pressure of the tire would be 247 kPa.

I like keeping track of the units throughout solving, but not sure how to do that with three units...

Can I write $$P_2 = 2599100\cdot\pu{kPa/L/K}/10512\cdot\pu{L/K}$$ (L and K cancel out and leave me with kPa), or is it recommended not to record the units in that step and only write the final unit with the final answer?

Section 5.1 Unit symbols in SI Brochure [1, p. 130] has an answer:

In forming products and quotients of unit symbols the normal rules of algebraic multiplication or division apply. Multiplication must be indicated by a space or a half-high (centred) dot (⋅), since otherwise some prefixes could be misinterpreted as a unit symbol. Division is indicated by a horizontal line, by a solidus (oblique stroke, /) or by negative exponents. When several unit symbols are combined, care should be taken to avoid ambiguities, for example by using brackets or negative exponents. A solidus must not be used more than once in a given expression without brackets to remove ambiguities.

$$\pu{m kg/(s3 A)}$$,
or $$\pu{m kg s–3 A–1}$$,
but not $$\pu{m kg/s3/A}$$,
nor $$\pu{m kg/s3 A}$$.

I personally prefer to use powers of the units — subjectively, this makes the expressions clearer and also reserves a slash "/" for separating the symbol of the quantity from the units on plots and in the table headers. However, in your case you don't need any of that. According to the ideal gas law

$$pV = nRT$$

and

$$n_iR = \text{const} \quad \implies \quad \frac{p_iV_i}{T_i} = \text{const},$$

so you have a simple proportion

$$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \quad \implies \quad p_2 = \frac{V_1T_2}{V_2T_1}p_1.$$

Obviously, there is no need to juggle around with the units as all of them except for the pressure units are cancelled out (note how I also grouped homogeneous quantities within the fraction one under another, and wrote pressure separately for clarity):

$$p_2 = \frac{V_1T_2}{V_2T_1}p_1 = \frac{\pu{35.0 L}\cdot\pu{316 K}}{\pu{36.5 L}\cdot\pu{288 K}}\cdot\pu{235 kPa} = \pu{247 kPa}.$$

And yes, generally it is considered a good practice not to do any intermediate calculations unless you are explicitly asked so, and solve the problem algebraically in the first place.

References

1. BIPM. Le Système International d’unitès / The International System of Units (“The SI Brochure”), 8th ed.; Bureau international des poids et mesures: Sèvres, 2006.