Talked-through calculations
Every numerical example uses the same pattern: list the quantities (convert first), write the equation, substitute, unit, sense-check. $g=10\,\text{N/kg}$.
Before you read my working
Cover the solution. Spend two minutes on the question. Then uncover one step at a time. If your method is different but your physics is the same, that is fine — I care about the conversion, the equation, the substitution, and the unit.
Example 1 · $\rho=m/V$ with $\text{cm}^3$ · 5.3
A pebble has mass $180\,\text{g}$. Its volume, found with a eureka can, is $75\,\text{cm}^3$. Calculate the density in $\text{kg/m}^3$.
What I write first
I refuse to divide $180$ by $75$ and call it $\text{kg/m}^3$. Convert both.
$$m = 180 \div 1000 = 0.180\,\text{kg}$$ $$V = 75 \div 10^6 = 7.5\times 10^{-5}\,\text{m}^3$$ $$\rho = \frac{m}{V} = \frac{0.180}{7.5\times 10^{-5}} = 2400\,\text{kg/m}^3$$Sense-check: denser than water, so it sinks. Typical rock. If you had $2.4$, you answered in $\text{g/cm}^3$ — convert $\times 1000$ or redo in SI.
Example 2 · $p=F/A$ · 5.5
A punch exerts a force of $400\,\text{N}$ on an area of $0.0020\,\text{m}^2$. Calculate the pressure.
$$F=400\,\text{N},\quad A=0.0020\,\text{m}^2$$ $$p = \frac{F}{A} = \frac{400}{0.0020} = 2.0\times 10^5\,\text{Pa}$$Sense-check: about twice atmospheric pressure. A small area and a large force should give a large $p$. Unit is Pa, not N.
Example 3 · $p=h\rho g$ without atmosphere · 5.7
Calculate the pressure due to the water at a depth of $4.0\,\text{m}$. $\rho_{\text{water}}=1000\,\text{kg/m}^3$.
$$h=4.0\,\text{m},\quad \rho=1000\,\text{kg/m}^3,\quad g=10\,\text{N/kg}$$ $$p = h\rho g = 4.0 \times 1000 \times 10 = 4.0\times 10^4\,\text{Pa}$$The question said “due to the water”. I do not add $1.0\times 10^5\,\text{Pa}$. Sense-check: $10\,\text{m}$ of water is roughly one atmosphere, so $4\,\text{m}$ should be about $0.4$ of an atmosphere — $4\times 10^4\,\text{Pa}$ is right.
Example 4 · $p=h\rho g$ plus atmosphere · 5.7
A diver is $2.5\,\text{m}$ below the surface of the sea. $\rho=1020\,\text{kg/m}^3$. Atmospheric pressure is $1.00\times 10^5\,\text{Pa}$. Calculate (a) the pressure difference due to the seawater and (b) the total pressure on the diver.
(a)
$$p = h\rho g = 2.5 \times 1020 \times 10 = 2.55\times 10^4\,\text{Pa}$$(b)
$$p_{\text{total}} = 2.55\times 10^4 + 1.00\times 10^5 = 1.255\times 10^5\,\text{Pa}$$Part (a) is the extra pressure. Part (b) is what a pressure gauge at that depth would read if it were an absolute gauge. Missing the add-on costs the last mark, not the first two.
Example 5 · °C $\leftrightarrow$ K · 5.17
A gas is heated from $20^\circ\text{C}$ to $100^\circ\text{C}$. Write both temperatures in kelvin. State absolute zero in °C and in K.
$$T_1 = 20 + 273 = 293\,\text{K}$$ $$T_2 = 100 + 273 = 373\,\text{K}$$Absolute zero is $-273^\circ\text{C}$ or $0\,\text{K}$. Notice that $100^\circ\text{C}$ is not five times $20^\circ\text{C}$ on the kelvin scale: $373/293\approx 1.27$, not $5$. That is why $p/T$ with Celsius numbers is nonsense.
Example 6 · Boyle · 5.22
Air is trapped in a syringe. $p_1=100\,\text{kPa}$, $V_1=40\,\text{cm}^3$. The piston is pushed in slowly until $V_2=20\,\text{cm}^3$. The temperature is constant. Calculate $p_2$.
List: constant $T$, fixed mass $\rightarrow$ $p_1V_1=p_2V_2$. I keep kPa and $\text{cm}^3$ because they are consistent on both sides.
$$p_2 = \frac{p_1 V_1}{V_2} = \frac{100 \times 40}{20} = 200\,\text{kPa}$$Sense-check: volume halved, pressure doubled. If the temperature had risen as well (a fast compression), Boyle would not apply on its own.
Example 7 · $p/T$ · 5.21
A rigid can contains gas at $120\,\text{kPa}$ and $20^\circ\text{C}$. The can is heated to $100^\circ\text{C}$. The volume does not change. Calculate the new pressure.
Convert first. Then $p_1/T_1=p_2/T_2$.
$$T_1=293\,\text{K},\quad T_2=373\,\text{K},\quad p_1=120\,\text{kPa}$$ $$p_2 = 120 \times \frac{373}{293} = 152.8\ldots = 153\,\text{kPa}\ (3\,\text{s.f.})$$Sense-check: $T$ rose by about $27\%$, so $p$ should rise by about $27\%$ of $120$, which is roughly $32$, giving $152$. Using $20$ and $100$ gives $600\,\text{kPa}$ — throw that away.
Example 8 · 4-mark collision explain · 5.15, 5.20
The volume of air in a car tyre stays almost constant. On a hot day the temperature of the air increases. Explain, in terms of particles, why the pressure in the tyre increases. [4]
The air particles move randomly and collide with the inside walls of the tyre [1]. Kelvin temperature is proportional to the average kinetic energy, so the particles move faster [1]. They collide with the walls more frequently [1]. They also collide with a greater force, so the force on each unit of area increases and the pressure rises [1].
Phrases that score nothing on their own: “the particles expand”, “the air needs more space”, “heat makes pressure”. I need collisions, frequency and/or force, and a link to $p=F/A$.