IBDP Physics B.3 Gas laws IB Style Question Bank : HL Paper 2


Pressure \(p\), volume \(V\) and temperature \(T\) are measured for a fixed mass of gas. \(T\) is measured in degrees Celsius. The graph shows the variation of \(p V\) with \(T\). The mass of a molecule of the gas is \(4.7 \times 10^{-26} \mathrm{~kg}\).

(a) State the unit for \(p V\) in fundamental SI units.[1]

(b) Deduce, using the graph, whether the gas acts as an ideal gas.[3]

(c) Calculate, in g, the mass of the gas.[3]



a\(\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-2} \checkmark\)

Graph shown is a straight line/linear
expected graph should be a straight line/linear
If ideal then \(T\) intercept must be at \(T=-273^{\circ} \mathrm{C}\)
Use of \(y=m x+c\) to show that \(x=-273^{\circ} \mathrm{C}\) when \(y=0\) v (hence ideal)
Calculates \(\frac{p V}{T}\) for two different points \(\checkmark\)
Obtains \(1.50 \ll \mathrm{J} \mathrm{K}^{-1}\) w for both \(\checkmark\)
States that for ideal gas \(\frac{p V}{T}=n R\) which is constant and concludes that gas is ideal \(\checkmark\)

c.Use of \(n=\frac{p V}{R T}\) OR \(N=\frac{p V}{k T} \checkmark\)
Mass of gas \(=n \times N_{\mathrm{A}} \times\) mass of molecule
Mass of gas \(=N \times\) mass of molecule
\(5.1 \ll g n \checkmark\)


The air in a kitchen has pressure 1.0 × 105 Pa and temperature 22°C. A refrigerator of internal volume 0.36 m3 is installed in the kitchen.

    1. With the door open the air in the refrigerator is initially at the same temperature and pressure as the air in the kitchen. Calculate the number of molecules of air in the refrigerator. [2]

    2. The refrigerator door is closed. The air in the refrigerator is cooled to 5.0°C and the number of air molecules in the refrigerator stays the same.

(i) Determine the pressure of the air inside the refrigerator. [2]

(ii) The door of the refrigerator has an area of 0.72 m2. Show that the minimum force needed to open the refrigerator door is about 4 kN. [2]

(iii) Comment on the magnitude of the force in (b)(ii). [2]



a N = \(\frac{pv}{kT}\) OR N = \(\frac{1.0\times 10\times ^5\times 0.36}{1.38\times 10^{-23\times 295}}\) N =  8.8 ×10 24

b i use of \(\frac{P}{T}\) = constant OR P = \(\frac{nRT}{V}\) OR \(\frac{NkT}{V}\) p = 9.4× 104 «Pa »

b ii F = A p = ×∆ F =  0.72 × (1.0 – 0.94) ×105 OR 4.3 ×103 «N »

b iii force is «very» large  there must be a mechanism that makes this force smaller OR assumption used to calculate the force/pressure is unrealistic

Scroll to Top